Chapter 1 Systems of Linear Equations and Matrices

Chapter 1Systems of Linear Equations andMatricesSystem of linear algebraic equations and their solution constitute one of the major topicsstudied in the course known as “linear algebra.” In the first section we shall introducesome basic terminology and discuss a method for solving such system.1.1 Systems of Linear EquationsLinear EquationsAny straight line in the xy-plane can be represented algebraically by an equation of theforma 1 x+a 2 y = bwhere a 1 , a 2 , and b are real constants and a 1 and a 2 are not both zero. An equation ofthis form is called a linear equation in the variables x and y. More generally, we definea linear equation in the n veriables x 1 ,x 2 ,...,x n to be one that can be expressed inthe forma 1 x 1 +a 2 x 2 +···+a n x n = bwhere a 1 ,a 2 ,...,a n , and b are real constants. The variables in a linear equation aresometimes called unknowns. For example, the equationsare linear. The equations3x+y = 7, y = 1 5 x+2z +4, and x 1 +3x 2 −2x 3 +5x 4 = 7√ x+3y = 2, 3x−2y −5z +yz = 4, and y = cosxare not linear.Observe that a linear equation does not involve any products or roots of variables. Allvariables occur only to the first power and do not appear as arguments for trigonometric,logarithmic, or exponential functions.1

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 2A solution of a linear equation a 1 x 1 + a 2 x 2 + ··· + a n x n = b is a sequence of nnumbers s 1 , s 2 , ..., s n such that the equation satisfied when we substitute x 1 = s 1 ,x 2 = s 2 , ..., x n = s n . The set of all solutions of the equation is called its solution setor sometimes the general solution of the equation.Example 1.1 Find the solution set of(a) 5x−2y = 3 and(b) x 1 −3x 2 +4x 3 = 2.Solution .........Linear SystemsA finite set of linear equations in the variables x 1 , x 2 , ..., x n is called system of linearequations or a linear system. A sequence of numbers s 1 , s 2 , ..., s n is called asolution of the system if x 1 = s 1 , x 2 = s 2 , ..., x n = s n is a solution of every equationin the system. For example, the system4x 1 −x 2 +3x 3 = −13x 1 +x 2 +9x 3 = −4the the solution x 1 = 1, x 2 = 2, and x 3 = −1 since these values satisfy both equations.However, x 1 = 1, x 2 = 8, x 3 = 1 is not a solution since these values satisfy only the firstequation in the system. Thus, not all system of linear equations have solutions.A system of equations that has no solutions is said to be inconsistent; if there isat least one solution of the system, it is called consistent. To illustrate the possibilitiesthat can occur in solving systems of linear equations, consider a general system of twolinear equations in the unknowns x and y:a 1 x+b 1 y = c 1a 2 x+b 2 y = c 2(a 1 ,b 1 not both zero)(a 2 ,b 2 not both zero)The graphs of these equations are lines; call them l 1 and l 2 . Since a point (x,y) lies ona line if and only if the number x and y satisfy the equation of the line, the solutions ofthe system of equations correspond to points of the intersection of l 1 and l 2 . There arethree possibilities, illustrated in Figure 1.1• The lines l 1 and l 2 may be parallel, in which case there is no intersection andconsequently no solution to the system.• The lines l 1 and l 2 may intersect at only one point, in which case the system hasexactly one solution.• The lines l 1 and l 2 may coincide, in which case there are infinitely many points ofintersection and consequently infinitely many solutions to the system.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 3yyyl 1 l 2l 1 l 2l 1 and l 2xxx(a) No solution(b) One solutionFigure 1.1:(c) Infinitely many solutionsAlthough we have considered only two equations with two unknowns here, we willshow later that the same three possibilities hold for arbitrary linear systems:Every system of linear equations has no solutions, or has exactly one solution,or has infinitely many solutions.An arbitrary system of m linear equations in n unknowns can be written asa 11 x 1 +a 12 x 2 +···+a 1n x n = b 1a 21 x 1 +a 22 x 2 +···+a 2n x n = b 2 (1.1)....a m1 x 1 +a m2 x 2 +···+a mn x n = b mwhere x 1 , x 2 , ..., x n are the unknowns and the subscripted a’s and b’s denote constants.For example, a general system of three linear equations in four unknowns can be writtenasa 11 x 1 +a 12 x 2 +a 13 x 3 +a 14 x 4 = b 1a 21 x 1 +a 22 x 2 +a 23 x 3 +a 24 x 4 = b 2a 31 x 1 +a 32 x 2 +a 33 x 3 +a 34 x 4 = b 3The double subscripting on the coefficients of the unknowns is a useful device thatis used to specify the location of the coefficient in the system. The first subscript onthe coefficient a ij indicates the equation in which the coefficient occurs, and the secondsubscript indicates which unknown it multiplies. Thus, a 12 is in the first equation andmultiplies unknown x 2 .

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 4Augmented MatricesA system of m linear equations in n unknowns can be abbreviated by writing only therectangular array of numbers⎡⎤a 11 a 12 ··· a 1n b 1a 21 a 22 ··· a 2n b 2⎢⎥⎣ . . . . ⎦ .a m1 a m2 ··· a mn b mThis is called the augmented matrix for the system. For example, the augmentedmatrix for the system of equationsisx 1 +x 2 +2x 3 = 92x 1 +4x 2 −3x 3 = 13x 1 +6x 2 −5x 3 = 0⎡⎣1 1 2 92 4 −3 13 6 −5 0Remark When constructing an augmented matrix, we must write the unknowns in thesame order in each equation, and the constants must be on the right.Exercise 1.11. Which of the following are linear equations in x 1 , x 2 and x 3 ?(a) x 1 +5x 2 − √ 2x 3 = 1 (b) x 1 +3x 2 +x 1 x 3 = 2(c) x 1 = −7x 2 +3x 3⎤⎦.(d) x −21 +x 2 +8x 3 = 5(e) x 3/51 −2x 2 +x 3 = 4 (f) πx 1 − √ 2x 2 + 1 3 x 3 = 7 1/32. Find the solution set of each of the following linear systems.(a) 7x−5y = 3 (b) 3x 1 −5x 2 +4x 3 = 7(c) −8x 1 +2x 2 −5x 3 +6x 4 = 1 (d) 3v −8w +2x−y +4z = 03. Which of the following are linear systems?(a) x 1 −3x 2 = x 3 −4x 4 = 1−x 1x 1 +x 4 +x 3 −2 = 0(c) 3x−xy = 1x+2xy −y = 0(e) 2x 1 −sinx 2 = 3x 2 = x 1 +x 3−x 1 +x 2 −3x 3 = 0(b) 2x− √ y +3z = −1x + 2y − z = 24x − y = −1(d) y = 2x−1y = −x(f) x 1 + x 2 +x 3 = 1−2x 2 1 − 2x 3 = −13x 2 −x 3 = 2

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 54. Find the augmented matrix for each of the following systems of linear equations.(a) 3x 1 −2x 2 = −14x 1 +5x 2 = 37x 1 +3x 2 = 2(c) x 1 = 1x 2 = 2x 3 = 3(b) 2x 1 +2x 3 = 13x 1 −x 2 +4x 3 = 76x 1 +x 2 − x 3 = 0(d) x 1 + 2x 2 − x 4 + x 5 = 13x 2 + x 3 − x 5 = 2x 3 + 7x 4 = 15. Find a system of linear equations corresponding to the augmented matrix.⎡(a) ⎣(c)2 0 03 −4 00 1 1⎤⎦[ 7 2 1 −3 51 2 4 0 1]⎡(b) ⎣(d)⎡⎢⎣3 0 −2 57 1 4 −30 −2 1 71 0 0 0 70 1 0 0 −20 0 1 0 30 0 0 1 4⎤⎦⎤⎥⎦6. (a) Find a linear equation in the variables x and y that has the general solution(b) Show thatx = 5+2t, y = t.x = t, y = 1 2 t− 5 2is also the general solution of the equation in part (a).7. Consider the system of equationsax+by = kcx+dy = lex+fy = mIndicate what we can say about the relative positions of the line ax + by = k,cx+dy = l, and ex+fy = m when(a) the system has no solutions.(b) the system has exactly one solution.(c) the system has infinitely many solutions.8. Show that the linear systemax+by = ecx+dy = fwhere a,b,c,d,e,f are constants, has a unique solution if ad−bc ≠ 0. Express thissolution in terms of a,b,c,d,e, and f.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 61. (a), (c), (f)Answer to Exercise 1.12. (a) x = 3 7 + 5 7 t, y = t (b) x 1 = 5 3 s− 4 3 t− 1 8 , x 2 = s, x 3 = t(c) x 1 = 1 4 r − 5 8 s+ 3 4 t− 1 8 , x 2 = r, x 3 = s, x 4 = t(d) v = 8 q − 2 r + 1 s− 4 t, w = q, x = r, y = s, z = t3 3 3 33. (a), (d)⎡ ⎤3 −2 −1⎡2 0⎤2 1⎡4. (a) ⎣ 4 5 3 ⎦ (b) ⎣ 3 −1 4 7 ⎦ (c) ⎣7 3 2 6 1 −1 0⎡ ⎤1 2 0 −1 1 1(d) ⎣ 0 3 1 0 −1 2 ⎦0 0 1 7 0 15. (a) 2x 1 = 03x 1 −4x 2 = 0x 2 = 1(b) 3x 1 −2x 3 = 57x 1 +x 2 +4x 3 = −3−2x 2 +x 3 = 7(c) 7x 1 +2x 2 +x 3 −3x 4 = 5x 1 +2x 2 +4x 3 = 1(d) x 1 = 7x 2 = −2x 3 = 3x 4 = 41 0 0 10 1 0 20 0 1 36. (a) x−2y = 5 (b) Let x = t; then t−2y = 5. Solving for y yields y = 1 2 t− 5 2 .7. (a) The line have no common point of intersection.(b) The line intersect in exactly one point.(c) The three lines coincide.1.2 Elementary OperationsElementary Row OperationsThe basic method for solving a system of linear equations is to replace the given systemby a new system that has the same solution set but is easier to solve. This new system isgenerally obtained in a series of steps by applying the following three types of operationsto eliminate unknowns systematically:⎤⎦

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 71. Multiply any equation, say the ith, by a nonzero constant α. This is denoted byαr i .2. Interchange of two equations, say the ith and jth. This is denoted by r i ↔ r j .3. Add α times the ith equation to the jth equation. This is denoted by r j + αr i .(This keeps the ith equation unchanged.)Since the rows of an augmented matrix correspond to the equations in the associatedsystem, these three operations correspond to the following operations on the rows of theaugmented matrix:1. Multiply any row by a nonzero constant: αr i .2. Interchange of two rows: r i ↔ r j .3. Add a multiple of one row to another: r j +αr i .These are called elementary row operations. The following illustrates how theseoperations can be used to solve system of linear equations.Example 1.2 Solve the following system by using elementary row operations.x+ y +2z = 92x+4y −3z = 13x+6y −5z = 0Solution In the left column we solve a system of linear equations by operating on theequations in the system, and in the right column we solve the same system by operatingon the rows of the augmented matrix.x+ y +2z = 92x+4y −3z = 13x+6y −5z = 0⎡⎣1 1 2 92 4 −3 13 6 −5 0⎤⎦Add −2 times the first equationto the second to obtainx+ y +2z = 92y −7z = −173x+6y −5z = 0Add −3 times the first equationto the third to obtainx+ y + 2z = 92y − 7z = −173y −11z = −27Multiply the second equation Multiplyby 1 to obtain2Add −2 times the first rowto the second to obtain⎡ ⎤1 1 2 9⎣ 0 2 −7 −17 ⎦3 6 −5 0Add −3 times the first rowto the third to obtain⎡ ⎤1 1 2 9⎣ 0 2 −7 −17 ⎦0 3 −11 −27the second rowby 1 to obtain2

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 8x+ y + 2z = 9y − 7 2 z = −17 23y −11z = −27⎡⎣1 1 2 90 1 − 7 2− 17 20 3 −11 −27⎤⎦Add −3 times the second equationto the third to obtainx+ y +2z = 9y − 7 2 z = −17 2− 1 2 z = −3 2Multiply the third equationby −2 to obtainx+ y +2z = 9y − 7 2 z = −17 2z = 3Add −1 times the second equationto the first to obtainx + 11 2 z = 352y − 7 2 z = −17 2z = 3Add − 11 times the third equation2to the first and 7 times the third2equation to the second to obtainx = 1y = 2z = 3Add −3 times the second rowto the third to obtain⎡1 1 2 9⎤⎢⎣ 0 1 − 7 2− 17 ⎥2 ⎦0 0 − 1 2− 3 2Multiply the third rowby −2 to obtain⎡ ⎤1 1 2 9⎣ 0 1 − 7 − 17 ⎦2 20 0 1 3Add −1 times the second rowto the first to obtain⎡ ⎤11 351 02 2⎣ 0 1 − 7 − 17 ⎦2 20 0 1 3Add − 11 times the third row2to the first and 7 times the third2row to the second to obtain⎡ ⎤1 0 0 1⎣ 0 1 0 2 ⎦0 0 1 3The solution x = 1, y = 2, z = 3 is now evident. ♣Echelon FormsIn previous example, we solved a linear system in the unknowns x, y, and z by reducingthe augmented matrix to the form⎡ ⎤1 0 0 1⎣ 0 1 0 2 ⎦.0 0 1 3This is an example of a matrix that is in reduced row-echelon form. To be of thisform, a matrix must have the following properties:1. If the row does not consist entirely of zeros, then the first nonzero number in therow is a 1. We call this a leading 1.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 92. If there are any rows that consist entirely of zeros, then they are grouped togetherat the bottom of the matrix.3. In any two successive rows that do not consist entirely of zeros, the leading 1 inthe lower row occurs farther to the right than the leading 1 in the higher row.4. Each column that contains a leading 1 has zeros everywhere else in that column.A matrix that has the first three properties is said to be in row-echelon matrix.(Thus, a matrix in reduced row-echelon form is of necessity in row-echelon form, but notconversely.)The following matrices are in reduced row-echelon form.⎡ ⎤⎡ ⎤ ⎡ ⎤ 0 1 −2 0 01 0 0 4 1 0 0[ ]⎣0 1 0 7 ⎦, ⎣0 1 0⎦,⎢0 0 0 1 3⎥ 0 0⎣0 0 0 0 0⎦ , 0 00 0 1 −1 0 0 10 0 0 0 0The following matrices are in row-echelon form.⎡ ⎤ ⎡ ⎤1 4 −3 7 1 0 0⎣0 1 6 2 ⎦, ⎣0 1 0⎦,0 0 1 −1 0 0 0⎡ ⎤0 1 2 6 0⎣0 0 1 −1 0⎦0 0 0 0 1Example 1.3 Suppose that the augmented matrix for a linear equations has been reducedby row operations to the given reduced row-echelon form. Solve the system.⎡ ⎤ ⎡ ⎤1 0 0 21 0 0 3 −5(a) ⎣ 0 1 0 4 ⎦ (b) ⎣ 0 1 0 4 −1 ⎦0 0 1 −30 0 1 2 6⎡(c) ⎣1 0 2 −30 1 1 40 0 0 0Solution .........Elimination Methods⎤⎦⎡(d) ⎣1 0 4 00 1 −5 00 0 0 1We have just seen how easy it is to solve a system of linear equations once its augmentedmatrix is in reduced row-echelon form.• The process of using row operations to transform a linear system into one whoseaugmented matrix is in row-echelon form is called Gaussian elimination.• The process of using row operations to transform a linear system into one whoseaugmented matrix is in reduced row-echelon form is called Gauss-Jordan elimination.⎤⎦

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 10Remark It can be show that every matrix has a unique reduced row-echelon form.Example 1.4 Solve by Gauss-Jordan elimination.x 1 +3x 2 −2x 3 +2x 5 = 02x 1 +6x 2 −5x 3 −2x 4 +4x 5 −3x 6 = −15x 3 +10x 4 +15x 6 = 52x 1 +6x 2 +8x 4 +4x 5 +18x 6 = 6Solution .........Back-SubstitutionIt is sometimes preferable to solve a system of linear equations by using Gaussian eliminationto bring the augmented matrix into row-echelon form without continuing all theway to the reduced row-echelon form. When this is done, the corresponding system ofequations can be solved by a technique called back-substitution. The next exampleillustrates the idea.Example 1.5 From Example 1.4, solve by Gaussian elimination and back-substitution.Solution .........Example 1.6 Solvex+y +2z = 92x+4y −3z = 13x+6y −5z = 0by Gaussian elimination and back-substitution.Solution .........Homogeneous Linear SystemsA system of linear equations is said to be homogeneous if the constant terms are allzero; that is the system has the forma 11 x 1 +a 12 x 2 +···+a 1n x n = 0a 21 x 1 +a 22 x 2 +···+a 2n x n = 0..a m1 x 1 +a m2 x 2 +···+a mn x n = 0Every homogeneous system of linear equations is consistent, since all such systemshave x 1 = 0, x 2 = 0, ..., x n = 0 as a solution. This solution is called the trivialsolution; if there are other solutions, they are called nontrivial solution.Because a homogeneous linear system always has the trivial solution, there are onlytwo possibilities for its solutions:..

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 11• The system has only the trivial solution.• The system has infinitely many solutions in addition to the trivial solution.In the special case of a homogeneous linear system of two equations in two unknowns,saya 1 x+b 1 y = 0 (a 1 ,b 1 not both zero)a 2 x+b 2 y = 0 (a 2 ,b 2 not both zero)the graphs of the equations are lines through the origin, and the trivial solution correspondsto the point of intersection at the origin.yya 1 x+b 1 y = 0xa 2 x+b 2 y = 0(a) Only the trivial solutionxa 1 x+b 1 y = 0anda 2 x+b 2 y = 0(b) Infinitely many solutionsThere is one case in which a homogeneous system is assured of having nontrivialsolution — namely, whenever the system involves more unknowns than equations. Tosee why, consider the following example of four equations in five unknowns.Example 1.7 Solve the following homogeneous system of linear equations by usingGauss-Jordan elimination.Solution .........2x 1 +2x 2 −x 3 +x 5 = 0−x 1 −x 2 +2x 3 −3x 4 +x 5 = 0x 1 +x 2 −2x 3 −x 5 = 0x 3 +x 4 +x 5 = 0Theorem 1.1 A homogeneous system of linear equations with more unknowns thanequations has infinitely many solutions.Remark Note that Theorem 1.1 applied only to homogeneous systems. A nonhomogeneoussystem with more unknowns than equations need not be consistent; however, ifthe system is consistent, it will have infinitely many solutions.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 12Exercise 1.21. Which of the following 3×3 matrices are in reduced row-echelon form?⎡ ⎤1 0 0(a) ⎣0 1 0⎦0 0 1⎡ ⎤1 0 0(e) ⎣0 0 0⎦0 0 1⎡ ⎤0 1 0(b) ⎣1 0 0⎦0 0 0⎡ ⎤1 1 0(f) ⎣0 1 0⎦0 0 0⎡ ⎤0 1 0(c) ⎣0 0 1⎦0 0 0⎡ ⎤1 0 2(g) ⎣0 1 3⎦0 0 0⎡ ⎤1 0 0(d) ⎣0 0 1⎦0 0 0⎡ ⎤0 0 0(h) ⎣0 0 0⎦0 0 02. Which of the following 3×3 matrices are in row-echelon form?⎡ ⎤1 0 0(a) ⎣0 1 0⎦0 0 1⎡ ⎤1 4 6(e) ⎣0 0 1⎦0 1 3⎡ ⎤1 3 0(b) ⎣0 0 1⎦0 0 0⎡ ⎤1 1 0(f) ⎣0 1 0⎦0 0 0⎡ ⎤1 1 1(c) ⎣0 1 2⎦0 0 3⎡ ⎤1 3 4(g) ⎣0 0 1⎦0 0 0⎡ ⎤1 0 0(d) ⎣0 1 0⎦0 2 0⎡ ⎤1 2 3(h) ⎣0 0 0⎦0 0 13. In each part determine whether the matrix is in row-echelon form, reduced rowechelonform, both, or neither.⎡ ⎤ ⎡ ⎤0 1 3 5 7 1 0 0 5 [ ](a) ⎣0 0 1 2 3⎦(b) ⎣0 0 1 3⎦1 0 3 1(c)0 1 2 40 0 0 0 0 0 1 0 4(d)[ ]1 −7 5 50 1 3 2⎡ ⎤1 0 0 1 2(e) ⎣0 1 0 2 4⎦0 0 1 3 6⎡ ⎤0 1(f) ⎣0 0⎦0 04. Find the reduced row-echelon form of A when⎡⎤2 2 −1 0 1 0A = ⎢−1 −1 2 −3 1 0⎥⎣ 1 1 −2 0 −1 0⎦ .0 0 1 1 1 05. Find the reduced row-echelon form of B when⎡ ⎤0 0 −2 0 7 12B = ⎣2 4 −10 6 12 28⎦.2 4 −5 6 −5 −16. In each part suppose that the augmented matrix for a system of linear equationshas been reduced by row operations to the given reduced row-echelon form. Solvethe system.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 13⎡(a) ⎣(c) ⎣(e)⎡⎡⎢⎣1 0 0 −30 1 0 00 0 1 71 −3 0 20 0 1 −20 0 0 0⎤⎦⎤⎦1 −6 0 0 3 −20 0 1 0 4 70 0 0 1 5 80 0 0 0 0 0⎤⎥⎦⎡(b) ⎣(d)1 0 0 −7 80 1 0 3 20 0 1 1 −5[ 1 2 0 1 50 0 1 3 4⎡(f) ⎣1 −3 0 00 0 1 00 0 0 1]⎤⎦⎤⎦7. Solve each of the following systems by Gauss-Jordan elimination.(a) x 1 + x 2 +2x 3 = 8−x 1 −2x 2 +3x 3 = 13x 1 −7x 2 +4x 3 = 10(c) 2x 1 −3x 2 = −22x 1 + x 2 = 13x 1 +2x 2 = 1(e) 4x 1 −8x 2 = 123x 1 −6x 2 = 9−2x 1 +4x 2 = −6(b) 2x 1 +2x 2 +2x 3 = 0−2x 1 +5x 2 +2x 3 = 18x 1 + x 2 +4x 3 = −1(d) −2b+3c = 13a+6b−3c = −26a+6b+3c = 5(f) x 1 +3x 2 + x 3 + x 4 = 32x 1 −2x 2 + x 3 +2x 4 = 83x 1 + x 2 +2x 3 − x 4 = −1(g)x− y +2z − w = −12x+ y −2z −2w = −2−x+2y −4z + w = 13x −3w = −3(h)3x 1 +2x 2 − x 3 = −155x 1 +3x 2 +2x 3 = 03x 1 + x 2 +3x 3 = 11−6x 1 −4x 2 +2x 3 = 30(i) x 1 + x 2 +x 3 + x 4 = 02x 1 + x 2 −x 3 +3x 4 = 0x 1 −2x 2 +x 3 + x 4 = 0(j) 10y −4z + w = 1x+ 4y − z + w = 23x+2y + z +2w = 5−2x−8y +2z −2w = −4x−6y +3z = 18. Solve each of the systems in Exercise 2 by Gaussian elimination.9. Determine which of the following homogeneous systems have nontrivial solutions.(a) 2x 1 −3x 2 +4x 3 − x 4 = 07x 1 + x 2 −8x 3 +9x 4 = 02x 1 +8x 2 + x 3 − x 4 = 0(c) a 11 x 1 +a 12 x 2 +a 13 x 3 = 0a 21 x 1 +a 22 x 2 +a 23 x 3 = 0(b) x 1 +3x 2 − x 3 = 0x 2 −8x 3 = 04x 3 = 0(d) 3x 1 −2x 2 = 06x 1 −4x 2 = 0

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 1410. Solve the following homogeneous systems of linear equations by any method.(a) 2x 1 +x 2 +3x 3 = 0x 1 +2x 2 = 0x 2 + x 3 = 0(c) −x 1 +x 2 − x 3 +3x 4 = 03x 1 +x 2 − x 3 − x 4 = 02x 1 −x 2 −2x 3 − x 4 = 0(e) 2x+2y +4z = 0w − y −3z = 02w+3x+ y + z = 0−2w + x+3y −2z = 0(b) 3x 1 +x 2 +x 3 +x 4 = 05x 1 −x 2 +x 3 −x 4 = 0(d) − 1 x 2 1 + 1 x 3 2 + 1 x 2 3 = 01x 4 1 − 2x 3 2 + 1x 4 3 = 01x 4 1 + 1x 3 2 − 3x 4 3 = 0(f) 2x− y −3z = 0−x+2y −3z = 0x+ y +4z = 011. Consider a linear system whose augmented matrix is of the form⎡ ⎤1 2 1 1⎣ −1 4 3 2 ⎦2 −2 α 3For what values of α will the system have a unique solution?12. Consider a linear system whose augmented matrix is of the form⎡ ⎤1 2 1 0⎣ 2 5 3 0 ⎦−1 1 β 0(a) Is it possible for the system to be consistent? Explain.(b) For what values of β will the system have infinitely many solutions?13. Consider a linear system whose augmented matrix is of the form⎡ ⎤1 1 3 2⎣ 1 2 4 3 ⎦1 3 α β(a) For what values of α and β will the system have infinitely many solutions?(b) For what values of α and β will the system be inconsistent?14. Solve the system2x 1 −x 2 = λx 12x 1 −x 2 + x 3 = λx 2−2x 1 +2x 2 +x 3 = λx 3for x 1 , x 2 , and x 3 in the two cases λ = 1, λ = 2.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 15Answer to Exercise 1.21. (a), (b), (d), (g), (h) 2. (a), (b), (c), (f), (g)3. (a) Row-echelon (b) Neither (c) Both (d) Row-echelon (e) Both (f) Both⎡ ⎤1 1 0 0 1 0 ⎡ ⎤4. ⎢0 0 1 0 1 01 2 0 3 0 7⎥⎣0 0 0 1 0 0⎦ 5. ⎣0 0 1 0 0 1⎦0 0 0 0 1 20 0 0 0 0 06. (a) x 1 = 3, x 2 = 0, x 3 = 7 (b) x 1 = 7t+8, x 2 = −3t+2, x 3 = −t−5, x 4 = t(c) x 1 = 2+3t, x 2 = t, x 3 = −2 (d) x 1 = 5−2s−t, x 2 = s, x 3 = 4−3t, x 4 = t(e) x 1 = 6s−3t−2, x 2 = s, x 3 = −4t+7, x 4 = −5t+8, x 5 = t(f) Inconsistent7. (a) x 1 = 3, x 2 = 1, x 3 = 2 (b) x 1 = − 1 7 − 3 7 t, x 2 = 1 7 − 4 7 t, x 3 = t(c) Inconsistent (d) Inconsistent (e) x 1 = 3+2t, x 2 = t(f) x 1 = 3 4 − 5 8 t, x 2 = − 1 4 − 1 8 t, x 3 = t, x 4 = 3(g) x = t−1, y = 2s, z = s, w = t (h) x 1 = −4, x 2 = 2, x 3 = 7(i) x 1 = − 4 3 t, x 2 = 0, x 3 = 1 3 t, x 4 = t(j) x = 14 − 6 t, y = − 7 + 3 t− 1 s, z = −2+t, w = t5 5 10 10 109. (a), (c), (d)10. (a) x 1 = 0, x 2 = 0, x 3 = 0(b) x 1 = −s, x 2 = −t−s, x 3 = 4s, x 4 = t(c) x 1 = t, x 2 = −t, x 3 = t (d) x 1 = 5 3 t, x 2 = t, x 3 = t(e) w = t, x = −t, y = t, z = 0 (f) x = 0, y = 0, z = 011. α ≠ −2 12. β = 2 13. (a) α = 5, β = 4, (b) α = 5, β ≠ 414. If λ = 1, then x 1 = x 2 = s, x 3 = 0. If λ = 2, then x 1 = − 1 2 s, x 2 = 0, x 3 = s1.3 Matrices and Matrix OperationsIn this section we begin our study of matrix theory by giving some of the fundamentaldefinitions of the subject. We shall see how matrices can be combined through thearithmetic operations of addition, subtraction, and multiplication.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 16Matrix Notation and TerminologyDefinition 1.1 A matrix is a rectangular array of numbers, each of which is called anentry of the matrix.For example,⎡ ⎤2 1⎣−1 2⎦ (1.2)0 5( )1 −1 0(1.3)0 5 0and [1√2 π](1.4)Each horizontal line of number is called a row of the matrix; each vertical one iscalled a column. For example, in matrix (1.2) the third row consist of the entries 0 and5, whereas the entries in the first column are 2, −1, and 0.If the given matrix has m rows and n columns, it is called m×n matrix (read, mby n matrix), and the numbers m and n are the sizes of the matrix. The matrices (1.2),(1.3), and (1.4), are respectively, 3×2, 2×3, and 1×3 matrices.A matrix with only one column or m ×1 matrix (with m > 1) is called a columnmatrix (or a column vector), and a matrix with only one row or 1×n matrix (withn > 1) is called a row matrix (or a row vector). For example, the matrix (1.4) iscalled a row matrix or a row vector.Capital (uppercase) letters, such as A, B, and C, will be used to denote matrices,whereas the entries of a matrix will be denoted by the corresponding lowercase letterwith “double subscripts.” The entry that occurs in row i and column j of a matrix Awill be denoted by a ij . Thus, a general 2×4 matrix might be written as[ ]a11 aA = 12 a 13 a 14a 21 a 22 a 23 a 24and a general m×n matrix as⎡ ⎤a 11 a 12 ··· a 1na 21 a 22 ··· a 2nA = ⎢ ⎥⎣ . . . ⎦ .a m1 a m2 ··· a mnWhen compactness of notation is desired, the preceding matrix can be written as[a ij ] m×n and [a ij ]the first notation being used when it is important in the discussion to know the size, andthe second being used when the size need not be emphasized.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 17The entry in row i and column j of a matrix A is also commonly denoted by thesymbol (A) ij . Thus, for the matrix⎡ ⎤3 6 −1A = ⎣4 2 0 ⎦1 −3 −5we have (A) 12 = a 12 = 6, (A) 21 = a 21 = 4, and (A) 33 = a 33 = 5.Row and column matrices are of special importance, and it is common practice todenote them by boldface lowercase letters rather than capital letters. For such matrices,double subscripting of the entries is unnecessary. Thus a general 1×n row matrix a anda general m×1 column matrix b would be written as⎡ ⎤b 1a = [ ]b 2a 1 a 2 ··· a n and b = ⎢ ⎥⎣ . ⎦b nA matrix A with n rows and n columns is called a square matrix of order n, andthe entries a 11 , a 22 , ..., a nn in (1.5) are said to be on the main diagonal of A.⎡ ⎤a 11 a 12 ··· a 1na 21 a 22 ··· a 2n⎢ ⎥⎣ . . . ⎦ . (1.5)a n1 a n2 ··· a nnDefinition 1.2 A square matrix in which all the entries above the main diagonal arezero is called lower triangular, and a square matrix in which all the entries belowthe main diagonal are zero is called upper triangular. A matrix that is either uppertriangular or lower triangular is called triangular.Operations on MatricesDefinition 1.3 Given matrices A and B, the entries a ij and b kp are correspondingentries if and only if i = k and j = p. In other words, an entry of A corresponds toan entry of B if and only if they occupy the same position in their respective matrices.For example, ifA =[ ] 1 2 34 5 6and B =[ ] 7 8 9−1 −2 −3then 1 and 7, 2 and 8, and 4 and −1 are all pairs of corresponding entries. We shallspeak of corresponding entries only when the two matrices in question have the samesize.Definition 1.4 Equality of matrices. Two matrices A and B are equal, writtenA = B, if they have the same size and their corresponding entries are equal.In matrix notation, if A = [a ij ] and B = [b ij ] have the same size, then A = B if andonly if a ij = b ij for all i and j.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 18Example 1.8 Which of the following matrices are equal:[ ] [ ]1 2 −1 1 2A = , B = ,4 0 1+2 4 0Solution .........[ ] 14−1C = 2 , D =4 0 3[ ] 1 2 −1?4 0 1Definition 1.5 Sum of matrices. Let A and B be matrices with the same size. Thesum of A and B, written A+B, is the matrix obtained by adding corresponding entriesof A and BIn matrix notation, if A = [a ij ] and B = [b ij ] have the same size, then C = A+B ifand only if c ij = a ij +b ij for all i and j.[ ] 2 −1 5Example 1.9 Let A = , B =3 0 −4A+B and A+C, if definedSolution .........[ ] −3 2 7, and C =5 −6 2[ ] 3 5. Find−2 −1Definition 1.6 Multiplication by a scalar. Let A be a matrix and c be a scalar.Then the product cA is the matrix obtained by multiplying each entry of A by scalar c.The matrix cA is said to be a scalar multiple of A.In matrix notation, if A = [a ij ], then B = cA if and only if b ij = ca ij for all i and j.Definition 1.7 The negative of a matrix B, written −B, is the matrix (−1)B. It isobtained from B by simply reversing the sign of every entry.Example 1.10 LetFind 3A and (−1)A.A =[ ]1 −1 0 3.2 6 −4 8Solution .........Definition 1.8 Difference of matrices. Let A and B be matrices with the same size.The difference of A and B, written A−B, is the matrix C given by C = A+(−B).Note The matrix A−B can be obtained by simply subtracting each entry of B fromthe corresponding entry of A.⎡ ⎤ ⎡ ⎤5 4 −3 0Example 1.11 Find C = 2A−B where A = ⎣−1 7 ⎦ and B = ⎣ 4 2 ⎦.9 −3 5 −7

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 19Solution .........If A 1 ,A 2 ,...,A n are matrices of the same size and c 1 ,c 2 ,...,c n are scalars, then anexpression of the formc 1 A 1 +c 2 A 2 +···+c n A nis called a linear combination of A 1 ,A 2 ,...,A n with coefficients c 1 , c 2 , ..., c n . Forexample, If[ ] [ ] [ ]2 3 4 0 2 7 9 −6 3A = , B = , and C = ,1 3 1 −1 3 −5 3 0 12then2A−B + 1 [ ] 2 3 43 C = 2 1 3 1[ ] 4 6 8= +2 6 2[ ] 7 2 2=4 3 11[ 0 2 7+(−1)−1 3 −5][ 0 −2 −71 −3 5+]+ 1 [ ] 9 −6 33 3 0 12[ ] 3 −2 11 0 4is the linear combination of A, B, and C with scalar coefficients 2, −1, and 1 3 .Definition 1.9 Matrix multiplication. If A is an m×r matrix and B is an r ×nmatrix, then the product AB is the m × n matrix whose entries are determined asfollows. To find the entry in row i and column j of AB, single out row i from matrixA and column j from matrix B. Multiply the corresponding entries from the row andcolumn together, and then add up the resulting products.In other words, if C = AB, thenor, using summation notation,c ij = a i1 b 1j +a i2 b 2j +···+a in b njc ij =n∑a ik b kj .k=1Example 1.12 Given the pairs of matrices A and B, find the products AB and BA, ifdefined.⎡ ⎤[ ] 4 1 4 31 2 4(a) A = , B = ⎣0 −1 3 1⎦2 6 02 7 5 2⎡ ⎤ ⎡ ⎤1 2 3 x(b) A = ⎣4 5 6⎦, B = ⎣y⎦7 8 9 zSolution .........Note It should be apparent from Example 1.12 that multiplication of matrices “actdifferently” frommultiplicationofnumbers. Thelatterhasthepropertyofcommutativity(for all numbers a and b, ab = ba), but matrix multiplication does not.One situation in which the products AB and BA are both defined (although theystill need not be equal) is when A and B have the same size and are square.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 20Partitioned MatricesA matrix can be subdivided or partitioned into small matrices by inserting horizontaland vertical rules between selected rows and columns. For example, the following arethree possible partitions of a general 3×4 matrix A:• partition of A into four submatrices A 11 , A 12 , A 21 , and A 22⎡ ⎤a 11 a 12 a 13 a 14[ ]A = ⎣ a 21 a 22 a 23 a 24⎦ A11 A= 12Aa 31 a 32 a 33 a 21 A 2234• partition of A into its row matrices r 1 , r 2 , and r 3⎡ ⎤ ⎡ ⎤a 11 a 12 a 13 a 14 r 1A = ⎣ a 21 a 22 a 23 a 24⎦ = ⎣r 2⎦a 31 a 32 a 33 a 34 r 3• partition of A into its column matrices c 1 , c 2 , c 3 , and c 4⎡ ⎤a 11 a 12 a 13 a 14A = ⎣ a 21 a 22 a 23 a 24⎦ = [ ]c 1 c 2 c 3 c 4a 31 a 32 a 33 a 34Matrix Multiplication by Columns and by RowsSometimes it may bedesirable tofind aparticular rowor columnof a matrixproduct ABwithout computing the entire product. The following results are useful for that purpose:ith row matrix of AB = [ ith row matrix of A ] B (1.6)jth column of AB = A [ jth column of B ] (1.7)Example 1.13 If A and B are the matrices in Example 1.12(a), then find the first rowand second column of ABSolution .........If a 1 , a 2 , ..., a m denote the row matrices of A and b 1 , b 2 , ..., b n denote the columnmatrices of B, then it follows from Formula (1.6) and (1.7) that⎡ ⎤ ⎡ ⎤a 1 a 1 Ba 2AB = ⎢ ⎥⎣ . ⎦ B = a 2 B⎢ ⎥(1.8)⎣ . ⎦a m a m B(AB computed row by row)AB = A [ b 1 b 2 ··· b n]=[Ab1 Ab 2 ··· Ab n](1.9)(AB computed column by column)

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 21Matrix Products as Linear CombinationsSuppose thatThen⎡ ⎤ ⎡ ⎤a 11 a 12 ··· a 1n x 1a 21 a 22 ··· a 2nA = ⎢ ⎥⎣ . . . ⎦ and x = x 2⎢ ⎥⎣ . ⎦a m1 a m2 ··· a mn x n⎡⎤a 11 x 1 +a 12 x 2 +···+a 1n x na 21 x 1 +a 22 x 2 +···+a 2n x nAx = ⎢⎥⎣ . . . ⎦a m1 x 1 +a m2 x 2 +···+a mn x n⎡ ⎤ ⎡ ⎤ ⎡ ⎤a 11 a 12 a 1na 21= x 1 ⎢ ⎥⎣ . ⎦ +x a 222⎢⎥⎣ . ⎦ +···+x a 2nn⎢⎥⎣ . ⎦a m1 a m2 a mn(1.10)In words, (1.10) tell us that the product Ax of the matrix A with a column matrix xis a linear combination of the column matrices of A with the coefficients coming fromthe matrix x. Moreover, the product yA of a 1×m matrix y with an m×n matrix A isa linear combination of the row matrices of A with scalar coefficients coming from y.Example 1.14 The matrix product⎡ ⎤⎡⎤ ⎡ ⎤2 −1 3 −1 5⎣ 1 3 −2⎦⎣2 ⎦ = ⎣−1⎦−3 2 1 3 10can be written as the linear combination of the column matrices⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤2 −1 3 5−1⎣1 ⎦+2⎣3 ⎦+3⎣−2⎦ = ⎣−1⎦−3 2 1 10The matrix product⎡ ⎤[ ]2 −1 35 −1 10 ⎣ 1 3 −2⎦ = [ −21 12 27 ]−3 2 1can be written as the linear combination of the row matrices5 [ 2 −1 3 ] −1 [ 1 3 −2 ] +10 [ −3 2 1 ] = [ −21 12 27 ] ♣It follows from (1.8) and (1.10) that the jth column matrix of a product AB is alinear combination of the column matrices of A with the coefficients coming from the jthcolumn of B.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 22Example 1.15 We showed in Example 1.12(a) that⎡ ⎤ ⎡ ⎤3 −2 [ ] −14 1 −3AB = ⎣2 4 ⎦ −2 1 3= ⎣ 12 6 30 ⎦4 1 61 −3 −14 −2 −15The column matrices of AB can be expressed as linear combinations of the column matricesof A as follows:⎡ ⎤ ⎡ ⎤ ⎡ ⎤−14 3 −2⎣ 12 ⎦ = −2⎣2⎦+4⎣4 ⎦−14 1 −3⎡ ⎤ ⎡ ⎤ ⎡ ⎤1 3 −2⎣ 6 ⎦ = 1⎣2⎦+1⎣4 ⎦−2 1 −3⎡ ⎤ ⎡ ⎤ ⎡ ⎤−3 3 −2⎣ 30 ⎦ = 3⎣2⎦+6⎣4 ⎦ ♣−15 1 −3Matrix Form of a Linear SystemMatrix multiplication has an important application to system of linear equations Considerany system of m linear equations in n unknowns.a 11 x 1 +a 12 x 2 +···+a 1n x n = b 1a 21 x 1 +a 22 x 2 +···+a 2n x n = b 2 (1.11)....a m1 x 1 +a m2 x 2 +···+a mn x n = b mSince two matrices are equal if and only if their corresponding entries are equal, wecan replace the m equations in this system by the single matrix equation⎡⎤ ⎡ ⎤a 11 x 1 +a 12 x 2 +···+a 1n x n b 1a 21 x 1 +a 22 x 2 +···+a 2n x n⎢⎥⎣ . . . ⎦ = b 2⎢ ⎥⎣ . ⎦a m1 x 1 +a m2 x 2 +···+a mn x n b mThe m×1 matrix on the left side of this equation can be written as a product to give⎡ ⎤⎡⎤ ⎡ ⎤a 11 a 12 ··· a 1n x 1 b 1a 21 a 22 ··· a 2nx 2⎢ ⎥⎢⎥⎣ . . . ⎦⎣. ⎦ = b 2⎢ ⎥⎣ . ⎦ .a m1 a m2 ··· a mn x m b mIf we designate these matrices A, x, and b, respectively, then the original system of mequations in n unknowns has been replaced by the single matrix equationAx = b

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 23The matrix A in this equation is called the coefficient matrix of the system. Theaugmented matrix for the system is obtained by adjoining b to A as the last column;thus the augmented matrix is⎡⎤a 11 a 12 ··· a 1n b 1[ ] a 21 a 22 ··· a 2n b 2A b = ⎢⎥⎣ . . . . ⎦ .a m1 a m2 ··· a mn b mTranspose of a MatrixDefinition 1.10 Transpose of a matrix. The transpose of the m×n matrix A, writtenA T , is the n×m matrix B whose jth column is the jth row of A. Equivalently, B = A Tif and only if b ij = a ji for each i and j.Example 1.16 Find the transpose of(a) A =[ ]2 7−8 0(b) B =[ ]1 2 −74 5 6(c) C = [ −1 3 4 ]Solution .........Definition 1.11 If A is a square matrix, then trace of A, denoted by tr(A), is definedto be the sum of the entries on the main diagonal of A. The trace of A is undefined if Ais not a square matrix.Example 1.17 Find the trace of matrices A and B.⎡ ⎤⎡ ⎤ −1 2 7 0a 11 a 12 a 13A = ⎣a 21 a 22 a 23⎦, B = ⎢ 3 5 −8 4⎥⎣ 1 2 7 −3⎦a 31 a 32 a 334 −2 1 0Solution .........Exercise 1.31. Suppose that A,B,C,D, and E are matrices with the following dimensions:A B C D E(4×5) (4×5) (5×2) (4×2) (5×4)Determine which of the following matrix expression are defined. For those that aredefined, give the dimension of the resulting matrix.(a) BA (b) AC +D (c) AE +B (d) AB +B(e) E(A+B) (f) E(AC) (g) E T A (h) (A T +E)D2. If a matrix A is 3×5 and the product AB is 3×7, what is the size of B?

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 243. Solve the following matrix equation for a,b,c, and d[ ] [ ]a−b b+c 8 1=3d+c 2a−4d 7 64. Consider the matrices⎡ ⎤3 0A = ⎣−1 2⎦, B =1 1Compute the following (where possible).[ ] 4 −1, C =0 2⎡ ⎤ ⎡ ⎤1 5 2 6 1 3D = ⎣−1 0 1⎦, E = ⎣−1 1 2⎦3 2 4 4 1 3[ ] 1 4 2,3 1 5(a) 2A T +C (b) D T −E T (c) (D −E) T (d) B T +5C T(e) 1 2 CT − 1 4 A (f) B −BT (g) 2E T −3D T (h) (2E T −3D T ) T5. Using the matrices in Exercise 4, compute the following (where possible).(a) AB (b) BA (c) (3E)D (d) (AB)C(e) A(BC) (f) CC T (g) (DA) T (h) (C T B)A T(i) tr(DD T ) (j) tr(4E T −D) (k) tr(C T A T +2E T )[ ] [ ]1 −2 −1 2 −16. Let A = and AB = . Find the first and second columns−2 5 6 −9 3of B.7. LetUse the method of Example 1.13 to find⎡ ⎤ ⎡ ⎤1 2 1 1 0 2A = ⎣3 3 5⎦ and B = ⎣2 1 1⎦2 4 1 5 4 1(a) the first row of AB(c) the second column of AB(e) the third row of AA(b) the third row of AB(d) the first column of BA(f) the third column of AA8. Let A and B be the matrices Exercise7. Use the method of Example 1.15 to(a) express each column matrix of AB as a linear combination of the columnmatrices of A(b) express each column matrix of BA as a linear combination of the columnmatrices of B

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 259. Ineachpart,expressthegivensystemoflinearequationsasasinglematrixequationAx = b.(a) 3x 1 +2x 2 = 12x 1 −3x 2 = 5(b) x 1 + x 2 = 52x 1 + x 2 − x 3 = 63x 1 −2x 2 +2x 3 = 7(c) 2x 1 + x 2 + x 3 = 4x 1 − x 2 +2x 3 = 23x 1 −2x 2 − x 3 = 0(d) 4x 1 + −3x 3 + x 4 = 15x 1 + x 2 −8x 4 = 32x 1 −5x 2 +9x 3 − x 4 = 03x 2 − x 3 +7x 4 = 210. In each part, express the matrix equation as a system of linear equations.⎡ ⎤⎡⎤ ⎡ ⎤⎡ ⎤⎡⎤ ⎡ ⎤ 3 −2 0 1 w 03 −1 2 x 1 2(a) ⎣ 4 3 7⎦⎣x 2⎦ = ⎣−1⎦(b) ⎢ 5 0 2 −2⎥⎢x⎥⎣ 3 1 4 7 ⎦⎣y⎦ = ⎢0⎥⎣0⎦−2 1 5 x 3 4−2 5 1 6 z 011. In each part, find a 6×6 matrix [a ij ] that satisfies the state condition. Make youranswers as general as possible by using letters rather than specific numbers for thenonzero entries.(a) a ij = 0 if i ≠ j(b) a ij = 0 if i > j(c) a ij = 0 if i < j (d) a ij = 0 if |i−j| > 112. Find the matrix A such that⎡ ⎤ ⎡ ⎤x x+yA⎣y⎦ = ⎣x−y⎦z 0for all choices of x, y, and z?13. Indicate whether the statement is always true or sometimes false. Justify youranswer with a counterexample.(a) The expressions tr(AA T ) and tr(A T A) are always defined, regardless of thesize of A.(b) tr(AA T ) = tr(A T A) for every matrix A.(c) If the first column of A has all zeros, then so does the first column of everyproduct AB.(d) If the first row of A has all zeros, then so does the first row of every productAB.(e) If A is a square matrix with two identical rows, then AA has two identicalrows.(f) If A is a square matrix and AA has a column of zeros, then A must have acolumn of zeros.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 26(g) If B is an n×n matrix whose entries are positive even integers, and if A is ann×n matrix whose entries are positive integers, then the entries of AB andBA are positive even integers.(h) If the matrix sum AB +BA is defined, then A and B must be square.Answer to Exercise 1.31. (a) Undefined (b) 4×2 (c) Undefined (d) Undefined (e) 5×5 (f) 5×2(g) Undefined (h) 5×22. 5×7 3. a = 5,b = −3,c = 4,d = 1⎡ ⎤ ⎡ ⎤[ ] −5 0 −1 −5 0 −17 2 44. (a) (b) ⎣ 4 −1 1 ⎦ (c) ⎣ 4 −1 1 ⎦ (d) Undefined3 5 7−1 −1 1 −1 −1 1⎡ ⎤ ⎡ ⎤ ⎡ ⎤− 1 3 [ ] 9 1 −1 9 −13 04 2(e) ⎣ 90⎦ 0 −14(f) (g) ⎣−13 2 −4⎦ (h) ⎣ 1 2 1 ⎦3 9 1 00 1 −6 −1 −4 −64 4⎡ ⎤ ⎡ ⎤ ⎡ ⎤12 −342 108 75 3 45 95. (a) ⎣−4 5 ⎦ (b) Undefined (c) ⎣12 −3 21⎦ (d) ⎣11 −11 17⎦4 136 78 63 7 17 13⎡ ⎤⎡ ⎤3 45 9 [ ] [ ] 12 6 9(e) ⎣11 −11 17⎦ 21 17 0 −2 11(f) (g) (h) ⎣48 −20 14⎦17 35 12 1 87 17 1324 8 16[ [ ]7 −8(i) 61 (j) 35 (k) 28 6. b 1 = , b4]2 =−5⎡ ⎤ ⎡ ⎤⎡ ⎤7. (a) [ 10 6 5 ] (b) [ 15 8 9 ] 6 5(c) ⎣23⎦ (d) ⎣7⎦ (e) [ 16 20 23 ] 12(f) ⎣23⎦8 1923⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤10 1 2 18. (a) ⎣34⎦ = 1⎣3⎦+2⎣3⎦+5⎣5⎦15 2 4 1⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤6 1 2 1⎣23⎦ = 0⎣3⎦+1⎣3⎦+4⎣5⎦8 2 4 1⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤5 1 2 1⎣14⎦ = 2⎣3⎦+1⎣3⎦+1⎣5⎦9 2 4 1⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤5 1 0 2(b) ⎣7⎦ = 1⎣2⎦+3⎣1⎦+2⎣1⎦19 5 4 1

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 279. (a)⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤10 1 0 2⎣11⎦ = 2⎣2⎦+3⎣1⎦+4⎣1⎦26 5 4 1⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤3 1 0 2⎣8⎦ = 1⎣2⎦+5⎣1⎦+1⎣1⎦26 5 4 1[ ][ ] [ 3 2 x1 1=2 −3 x 2 5]⎡2 1 1(c) ⎣1 −1 23 −2 −110. (a) 3x 1 − x 2 +2x 3 = 24x 1 +3x 2 +7x 3 = −1−2x 1 + x 2 +5x 3 = 4⎡ ⎤⎡⎤ ⎡ ⎤1 1 0 x 1 5(b) ⎣2 1 −1⎦⎣x 2⎦ = ⎣6⎦3 −2 2 x 3 7⎡ ⎤⎡⎤ ⎡ ⎤⎤⎡⎤ ⎡ ⎤ 4 0 −3 1 xx 1 41 1⎦⎣x 2⎦ = ⎣2⎦ (d) ⎢5 1 0 −8⎥⎢x 2⎥⎣2 −5 9 −1⎦⎣x x 3 03⎦ = ⎢3⎥⎣0⎦0 3 −1 7 x 4 2(b) 3w−2x + z = 05w +2y −2z = 03w+ x+4y +7z = 0−2w +5x+ y +6z = 0⎡⎤ ⎡⎤a 11 0 0 0 0 0 a 11 a 12 a 13 a 14 a 15 a 160 a 22 0 0 0 00 a 22 a 23 a 24 a 25 a 2611. (a)0 0 a 33 0 0 0⎢ 0 0 0 a 44 0 0(b)0 0 a 33 a 34 a 35 a 36⎥ ⎢ 0 0 0 a 44 a 45 a 46⎥⎣ 0 0 0 0 a 55 0 ⎦ ⎣ 0 0 0 0 a 55 a 56⎦0 0 0 0 0 a 66 0 0 0 0 0 a 66⎡⎤ ⎡⎤a 11 0 0 0 0 0 a 11 a 12 0 0 0 0a 21 a 22 0 0 0 0a 21 a 22 a 23 0 0 0(c)a 31 a 32 a 33 0 0 0⎢a 41 a 42 a 43 a 44 0 0(d)0 a 32 a 33 a 34 0 0⎥ ⎢ 0 0 a 43 a 44 a 45 0⎥⎣a 51 a 52 a 53 a 54 a 55 0 ⎦ ⎣ 0 0 0 a 54 a 55 a 56⎦a 61 a 62 a 63 a 64 a 65 a 66 0 0 0 0 a 65 a 66⎡ ⎤1 1 012. ⎣1 −1 0⎦0 0 0[ ] [ ] 0 2 1 213. (a) True (b) True (c) False; for example, A = and B =0 4 3 4[ ]1 −1(d) True (e) True (f) False; for example, A = (g) True (h) True1 −1

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 281.4 Inverses; Rules of Matrix ArithmeticIn this section we shall discuss some properties of the arithmetic operations on matrices.We shall see that many of the basic rules of arithmetic for real numbers also hold formatrices, but a few do not.Properties of Matrix OperationsFor real numbers a and b, we always have ab = ba, which is called the commutative lawfor multiplication. For matrices, however, AB and BA need not be equal.Example 1.18 Consider the matrices[ ]1 0A = , B =2 3[ ]1 23 4Multiplying givesThus, AB ≠ BA ♣AB =[ ] [ ]1 2 5 6, BA =11 16 11 16Theorem 1.2 (Properties of Matrices Arithmetic) Let A, B, and C be matricesand a and b be scalars. Assume that the sizes of the matrices are such that each operationis defined. Then(a) A+B = B +A(Commutative law for addition)(b) A+(B +C) = (A+B)+C (Associative law for addition)(c) A(BC) = (AB)C(d) A(B +C) = AB +AC(e) (B +C)A = BA+CA(f) A(B −C) = AB −AC(h) a(B +C) = aB +aC(j) (a+b)C = aC +bC(l) (ab)C = a(bC)(Associative law for multiplication)(Left distributive laws)(Right distributive laws)(g) (B −C)A = BA−CA(i) a(B −C) = aB −aC(k) (a−b)C = aC −bC(m) a(BC) = (aB)C = B(aC)Example 1.19 Let⎡ ⎤1 0 −1⎡ ⎤−1 2 3⎡ ⎤−2 1 1A = ⎣2 1 −1⎦, B = ⎣ 0 1 0⎦, C = ⎣ 1 1 0 ⎦3 −2 0 −2 −1 0 3 0 −3Find A(BC), (AB)C, A(B +C), and AB +AC.Solution .........

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 29Zero MatricesDefinition 1.12 The matrix analog of the number 0 is called a zero matrix. A zeromatrix, denoted by 0, is a matrix of any sizes consisting of only zero entries.For example, the matrices⎡ ⎤0 0 0⎣0 0 0⎦,0 0 0[ ] 0 0 0,0 0 0⎡ ⎤0⎣0⎦, and0[ 0 0 0 0]are all zero matrices.Since we already know that some of the rules of arithmetic for real number do notcarryover to matrix arithmetic. Forexample, consider thefollowing two standardresultsin the arithmetic of real numbers.• If ab = ac and a ≠ 0, then b = c. (This is called the cancellation law.)• If ad = 0, then at least one of the factors on the left is 0As the next example shows, the corresponding results are not generally true in matrixarithmetic.Example 1.20 Consider the matrices[ ] [ ]0 1 1 1A = , B = , C =0 2 3 4ThenAB = AC =[ ]3 46 8[ ]2 5, D =3 4AD =[ ]0 00 0[ ]3 70 0Thus, although A ≠ 0, it is incorrect to cancel the A from both sides of the equationAB = AC and write B = C. Also, AD = 0, yet A ≠ 0 and D ≠ 0. Thus, thecancellation law is not valid for matrix multiplication, and it is possible for a product ofmatrices to be zero without either factor being zero. ♣Theorem 1.3 (Properties of Zero Matrices) Assuming that the sizes of the matricesare such that the indicated operations can be performed, the following rules of matrixarithmetic are valid.(a) A+0 = 0+A = A(b) A−A = 0(c) 0−A = −A(d) A0 = 0; 0A = 0

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 30Identity MatricesOf special interest are square matrices with 1’s on the main diagonal and 0’s off the maindiagonal, such as⎡ ⎤⎡ ⎤[ ]1 0 0 01 0 0 1 0, ⎣0 1 0⎦,⎢0 1 0 0⎥0 1 ⎣0 0 1 0⎦ .0 0 10 0 0 1Amatrixofthisformiscalledanidentity matrix andisdenotedbyI. Ifitisimportantto emphasize the size, we shall write I n for the n×n identity matrix.The identity matrix is so named because it has the property that AI n = A andI m A = A for any m × n matrix A. Thus it is the matrix analog of the number 1, themultiplicative identity for numbers.Example 1.21 Consider the matrices⎡ ⎤ ⎡ ⎤3 4 1 b 11 b 12A = ⎣2 6 3⎦ B = ⎣b 21 b 22⎦0 1 8 b 31 b 32Thenand⎡ ⎤⎡⎤ ⎡ ⎤1 0 0 3 4 1 3 4 1I 3 A = ⎣0 1 0⎦⎣2 6 3⎦ = ⎣2 6 3⎦ = A0 0 1 0 1 8 0 1 8⎡ ⎤ ⎡ ⎤b 11 b 12[ ] bBI 2 = ⎣b 21 b 22⎦ 1 0 11 b 12= ⎣b 0 1 21 b 22⎦ = B ♣b 31 b 32 b 31 b 32Theorem 1.4 If R is the reduced row-echelon form of an n×n matrix A, then eitherR has a row of zeros or R is the identity matrix I n .Definition 1.13 Let A be a square matrix. If there exist a matrix B such that AB =BA = I, we say that A is invertible (or nonsingular) and that B is the inverse ofA. If A has no inverse, it is said to be noninvertible (or singular).Example 1.22 Show that the matrix B =Solution .........Properties of Inverses[3 51 2]is an inverse of A =Theorem 1.5 If B and C are both inverses of the matrix A, then B = C.[ ]2 −5.−1 3Asaconsequence ofTheorem 1.5, wecansaynowspeakof the inverse ofaninvertiblematrix. If A is invertible, then its inverse will be denoted by the symbol A −1 . Thus,AA −1 = I and A −1 A = I

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 31Theorem 1.6 The matrixA =[ ]a bc dis invertible if ad−bc ≠ 0, in which case the inverse is given by the formulaA −1 =1ad−bc[ ] d −b=−c a⎡⎢⎣dad−bc− cad−bc− b ⎤ad−bc⎥a ⎦.ad−bcTheorem 1.7 If A and B are invertible matrices of the same dimension, then AB isinvertible and(AB) −1 = B −1 A −1Theorem 1.8 If A 1 ,A 2 ,...,A n are invertible matrices of the same dimension, thenA 1 A 2···A n is invertible andExample 1.23 Let A =Solution .........Powers of a Matrix(A 1 A 2···A n ) −1 = A −1n A−1 n−1···A−1 2 A−1 1][ ] 7 −3and B =5 −2[ 4 13 1. Show that (AB) −1 = B −1 A −1 .Definition 1.14 If A is a square matrix, then we define the nonnegative integer powersof A to beA 0 = I A n = AA···A } {{ } (n > 0)nfactorsMoreover, if A is invertible, then we define the negative integer powers to beA −n = (A −1 ) n = A −1 −1} A−1···A {{ }nfactorsNote In general, if A is a square matrix, and r and s are positive integers, we haveA r A s = A r+s . It is also true that under the same condition, (A r ) s = A rs .[ ] [ ]3 −2 1 2Example 1.24 Let A = and A−1 1−1 = . Find the matrices A1 33 and A −3 .Solution .........Theorem 1.9 (Laws of Exponents) If A is invertible matrix, then:(a) A −1 is invertible and (A −1 ) −1 = A .(b) A n is invertible and (A n ) −1 = (A −1 ) n for n = 0,1,2,....

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 32(c) For any nonzero scalar k, the matrix kA is invertible and (kA) −1 = 1 k A−1 .Example 1.25 If( 13 A ) −1=Solution .........Properties of the Transpose[ ] 2 −1, then find A.−5 3Theorem 1.10 If the sizes of the matrices are such that the state operations can beperformed, then(a) (A T ) T = A(b) (A+B) T = A T +B T and (A−B) T = A T −B T(c) (kA) T = kA T , where k is any scalar(d) (AB) T = B T A TInvertibility of a TransposeTheorem 1.11 If A is an invertible matrix, then A T is also invertible and(A T ) −1 = (A −1 ) T .[ ]−3 1Example 1.26 Let A = . Show that (A5 −2T ) −1 = (A −1 ) T .Solution .........1. LetShow that(a) A+(B +C) = (A+B)+C(c) (a+b)C = aC +bC(e) a(BC) = (aB)C = B(aC)(g) (B +C)A = BA+CA(i) (A T ) T = A(k) (aC) T = aC TExercise 1.4⎡ ⎤ ⎡ ⎤1 −2 0 6 0 2A = ⎣3 5 −1⎦, B = ⎣ 4 1 −1⎦2 3 4 −3 8 5⎡ ⎤4 −5 3C = ⎣ 5 7 −2⎦, a = 3, b = −5−3 2 −1(b) (AB)C = A(BC)(d) a(B −C) = aB −aC(f) A(B −C) = AB −AC(h) a(bC) = (ab)C(j) (A+B) T = A T +B T(l) (AB) T = B T A T

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 332. Let 0 denote a 2×2 matrix, each of whose entries is zero. Find(a) 2×2 matrix A such that A ≠ 0 and AA = 0 and(b) 2×2 matrix A such that A ≠ 0 and AA = A.3. Use Theorem 1.6 to compute the inverses of the following matrices.[ ] 1 3(a) A =2 −4[ ]−4 −5(c) C =5 6[ ] 2 3(b) B =1 1[ ]3 −7(d) D =−6 134. Use the matrices A, B, and C in Exercise3 to verify that(a) (A −1 ) −1 = A(b) (B T ) −1 = (B −1 ) T(c) (AB) −1 = B −1 A −1 (d) (ABC) −1 = C −1 B −1 A −1[ ] 1 −15. Let A = . Find the power A−1 22 and A 3 .⎡ ⎤1 0 06. Let A = ⎣2 1 0⎦. Find the power A 2 and A 3 .3 2 1[ ]1 17. Let A = . Find a general formula for A0 1n .8. In each part, use the given information to find A.[ ][ ]2 −1−3 7(a) A −1 =(b) (7A)3 5−1 =1 −2[ ][ ]−3 −1−1 2(c) (5A T ) −1 =(d) (I +2A)5 2−1 =4 5[ ] cosθ sinθ9. Find the inverse of .−sinθ cosθ[ 1]10. Find the inverse of2 (ex +e −x 1)2 (ex −e −x )12 (ex −e −x 1)2 (ex +e −x .)11. Consider the matrix⎡ ⎤a 11 0 ··· 00 a 22 ··· 0A = ⎢ ⎥⎣ . . . ⎦0 0 ··· a nnwhere a 11 a 22···a nn ≠ 0. Show that A is invertible and find its inverse.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 3412. Let A, B, and 0 be 2×2 matrices. Assuming that A is invertible, find a matrix Csuch that [ A−10C A −1 ]is the inverse of the partitioned matrix[ A 0B A13. Use the result in Exercise 12 to find the inverses of the following matrices.]⎡ ⎤1 1 0 0(a) ⎢−1 1 0 0⎥⎣ 1 1 1 1⎦1 1 −1 1⎡ ⎤1 1 0 0(b) ⎢0 1 0 0⎥⎣0 0 1 1⎦0 0 0 114. (a) Find a nonzero 3×3 matrix A such that A is symmetric.(b) Find a nonzero 3×3 matrix A such that A is skew-symmetric.15. A square matrix A is called symmetric if A T = A and skew-symmetric ifA T = −A. Show that if B is a square matrix, then(a) BB T and B +B T are symmetric,(b) B −B T is skew-symmetric.16. Indicate whether the statement is always true or sometimes false. Justify youranswer with a counterexample.(a) (AB) 2 = A 2 B 2 , where A are B are n×n matrices.(b) (A−B) 2 = (B −A) 2 , where A are B are n×n matrices.(c) (A+B)(A−B) = A 2 −B 2 , where A are B are n×n matrices.(d) (AB −1 )(BA −1 ) = I n , where A are B are n×n matrices.(e) Let A and B be square matrices such that AB = 0. Show that if A isinvertible, then B = 0.(f) If a square matrix A satisfies A 2 −3A+I = 0, then A −1 = 3I −A.(g) If A and B are m×n matrices, then AB T and A T B are defined.(h) If AB = BA and if A is invertible, then A −1 B = BA −1 .(i) If A T is singular, then A is singular.(j) A matrix with a row of zeros cannot have an inverse.(k) A matrix with a column of zeros cannot have an inverse.(l) If A is invertible and AB = AC, then B = C.(m) Let A be an n×n matrix and let x and y be n×1 matrices. If Ax = Ay andx ≠ y, then A is singular.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 352. (a)(n) IfAisinvertible andk isanynonzero scalar, then(kA) n = k n A n forall integervalues of n.(o) Sum of two invertible matrices is invertible.(p) If AB is invertible, then B is invertible.[ ] [ ] a b x y(q) + is symmetric.b a y x[ ]0 10 03. (a) A −1 =(d) D −1 =[ ]1 0(b)0 0][2 35 101− 15 10[−133− 7 3−2 −1]Answer to Exercise 1.4(b) B −1 =[ ]−1 31 −2(c) C −1 =[ ] [ ]2 −3 5 −85. A 2 = , A−3 53 =−8 13⎡ ⎤ ⎡ ⎤1 0 0 1 0 06. A 2 = ⎣4 1 0⎦, A 3 = ⎣6 1 0⎦ 7. A n =10 4 1 21 6 1[ ] [ ] [ 5 1 2113 13 7 −28. (a) A = (b) A = (c) A =(d) A =− 313[−9213113 13213− 613]1737[ ]1 n0 1[ ] [ cosθ −sinθ 1]9. 102 (ex +e −x ) − 1 2 (ex −e −x )sinθ cosθ − 1 2 (ex −e −x 1)2 (ex +e −x )⎡ ⎤1a 110 ··· 010a11. A = ⎢22··· 0⎥⎣ . . . ⎦ 12. C = −A−1 BA −110 0 ···a nn⎡ ⎤1− 1 0 0 ⎡ ⎤2 2 1 11 −1 0 00 013. (a)2 2 ⎢ 1⎣ 0 0 − 1 ⎥2 2⎦ (b) ⎢0 1 0 0⎥⎣0 0 1 −1⎦1 1 0 0 0 1−1 02 2− 1 51535][ ]6 5−5 −4

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 36⎡ ⎤ ⎡ ⎤1 2 3 0 −1 −114. (a) ⎣2 1 4⎦ (b) ⎣1 0 −1⎦3 4 5 1 1 0[ ] [ ] 0 1 1 216. (a) False; for example, A = and B =0 2 3 4[ ] [ ]0 1 1 2(c) False; for example, A = and B =0 2 3 4(b) True(d) True (e) True(f) True (g) True (h) True (i) True (j) True (k) True (l) True (m) True[ ] [ ]1 0 −1 0(n) True (o) False; for example, A = and B = (p) True0 1 0 −1(q) True1.5 Elementary Matrices and a Method for FindingA −1In this section we shall develop an algorithm for finding the inverse of an invertiblematrix. We shall also discuss some of the basic properties of invertible matrices.Definition 1.15 An n×n matrix is called an elementary matrix if it can be obtainedfrom the n×n identity matrix I n by performing a single elementary row operation.List below are three elementary matrices and the operations that produce them.[ ] 1 0Multiply the second row of I0 −52 by −5.⎡ ⎤1 0 0 0⎢0 0 0 1⎥⎣0 0 1 0⎦ Interchange the second and fourth rows of I 4.0 1 0 0⎡ ⎤1 0 −2⎣0 1 0 ⎦ Add −2 times the third row of I 3 to the first row.0 0 1Theorem 1.12 If the elementary matrix E results from performing a certain row operationon I m and A is an m×n matrix, then the product EA is the matrix that resultswhen this same row operation is performed on A.Example 1.27 Let⎡ ⎤ ⎡ ⎤ ⎡ ⎤1 0 0 0 1 0 1 0 0E 1 = ⎣ 0 1 0⎦, E 2 = ⎣1 0 0⎦, E 3 = ⎣0 1 0⎦−4 0 1 0 0 1 0 0 5

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 37andFind the product E 1 A, E 2 A, and E 3 A.Solution .........⎡ ⎤a b cA = ⎣d e f⎦g h iIf an elementary row operation is applied to an identity matrix I to produce anelementary matrix E, then there is a second row operation that, when applied to E,produces I back again. For example, if E is obtained by multiplying the ith row of Iby a nonzero constant k, then I can be recovered if the ith row of E is multiplied by1. The various possibilities are listed in the following table. The operations on the rightkside of this table are called the inverse operations of the corresponding operations onthe left.Row Operation on IThat Produces EMultiply row i by k ≠ 0Row Operation on EThat Reproduces IMultiply row i by 1 kInterchange rows i and j Interchange rows i and jAdd k times row i to row j Add −k times row i to row jExample 1.28 In each of the following, an elementary row operation is applied to the2×2 identity matrix to obtain an elementary matrix E, then E is restored to the identitymatrix by applying the inverse row operation.[ 1 00 1[1 00 1]−→↑Multiply thesecond row by 3]−→↑Interchange the firstand second rows[ 1 00 1]−→↑Add −7 times thesecond row tothe first[ 1 00 3[0 11 0]−→[ ] 1 00 1↑Multiply thesecond row by 1 3][ 1 −70 1−→[1 00 1]↑Interchange the firstand second rows]−→[ 1 00 1↑Add −7 times thesecond row tothe firstThe next theorem gives an important property of elementary matrices.Theorem 1.13 Every elementary matrix is invertible, and the inverse is also an elementarymatrix.]

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 38Theorem 1.14 If A is an n×n matrix, then the following statements are equivalent,that is, all true or all false.(a) A is invertible.(b) Ax = 0 has only the trivial solution.(c) The reduced row-echelon form of A is I n .(d) A is expressible as a product of elementary matrices.By Theorem 1.14(c), we can find elementary matrices E 1 , E 2 , ..., E k−1 , E k such thatE k E k−1···E 2 E 1 A = I n (1.12)By Theorem 1.13, E 1 , E 2 , ..., E k−1 , E k are invertible. Multiplying both sides ofEquation (1.12) on the left successively by E −1k , E−1 k−1, ..., E−1 2 , E1 −1 we obtainA = E1 −1 E2 −1 ···E −1k−1 E−1 k I n = E1 −1 E2 −1 ···E −1k−1 E−1 kBy Theorem 1.13, this equation expresses A as a product of elementary matrices.Row EquivalenceIf a matrix B can be obtained from a matrix A by performing a finite sequence ofelementary row operations, then obviously we can get from B back to A by performingthe inverses of these elementary row operations in reverse order. Matrices that can beobtained from one another by a finite sequence of elementary row operations are said tobe row equivalent.In other words, a matrix B is row equivalent to A if there exists a finite sequence E 1 ,E 2 , ..., E k−1 , E k of elementary matrices such thatExample 1.29 LetB = E k E k−1···E 2 E 1 A.⎡ ⎤1 2 −3A = ⎣1 −2 1 ⎦.5 −2 −3Find the elementary matrices E 1 , E 2 , ..., E k−1 , E k such thatE k E k−1···E 2 E 1 A = A rwhere A r is the reduced row-echelon form of A.Solution .........

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 39A Method for Inverting MatricesAs our first application of Theorem 1.14, we shall establish a method for determiningthe inverse of an invertible matrix. Multiplying (1.12) on the right by A −1 yields(E k E k−1···E 2 E 1 A)A −1 = I n A −1E k E k−1···E 2 E 1 I n = A −1which tells us that A −1 can be obtained by multiplying I n successively on the left by theelementary matrices E 1 , E 2 , ..., E k−1 , E k . Since each multiplication on the left by oneof these elementary matrices performs a row operation, it follow that the sequence of rowoperations that reduces A to I n will reduce I n to A −1 . Thus we have the following result:To find the inverse of an invertible matrix A we must find a sequence ofelementary row operations that reduces A to identity and then performthis sequence of operation on I n to obtain A −1To accomplish this we shall adjoint the identity matrix to the right side of A, therebyproducing a matrix of the form[A|I].Then we shall apply row operations to this matrix until the left side is reduced to I;these operations will convert the right side to A −1 , so the final matrix will have the form[I|A −1 ]Example 1.30 Find the inverse of⎡ ⎤1 −2 1A = ⎣2 −5 2 ⎦3 2 −1Solution .........Often it will not be known in advance whether a given matrix is invertible. If an n×n(matrix A is not invertible,)then it cannot be reduced to I n by elementary row operationpart (c) of Theorem 1.14 . Stated another way, the reduced row-echelon form of A hasat least one row of zeros. Thus, if the procedure in the last example is attempted on amatrix that is not invertible, then at some point in the computations a row of zeros willoccur on the left side. It can then be concluded that the given matrix is not invertible,and the computations can be stopped.⎡ ⎤1 3 4Example 1.31 Let A be the matrix ⎣−2 −5 −3⎦. Determine whether A is invertible.1 4 9Solution .........

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 40Example 1.32 In Example 1.30 we show that⎡ ⎤1 −2 1A = ⎣2 −5 2 ⎦3 2 −1is an invertible matrix. From Theorem 1.14, it follows that the homogeneous systemhas only the trivial solution. ♣x 1 −2x 2 + x 3 = 02x 1 −5x 2 +2x 3 = 03x 1 +2x 2 − x 3 = 0Exercise 1.51. Which of the following are elementary matrices?[ ] [ ] [ ]0 1 −3 1 1 0(a) (b) (c)1 0 1 0 0 √ 2⎡ ⎤⎡ ⎤ ⎡ ⎤ 2 0 0 21 1 0 0 1 0(e) ⎣0 0 1⎦ (f) ⎣1 0 0⎦ (g) ⎢0 1 0 0⎥⎣0 0 1 0⎦0 0 0 0 0 10 0 0 1⎡ ⎤1 0 0(d) ⎣0 1 0⎦5 0 12. Find a row operation that will restore the given elementary matrix to an identitymatrix.⎡ ⎤[ ]1 0 01 0(a)(b) ⎣0 1 0⎦−3 10 0 5⎡ ⎤1 0 0 0(c) ⎢0 0 0 1⎥⎣0 0 1 0⎦0 1 0 0⎡ ⎤1 0 − 1 04 (d) ⎢0 1 0 0⎥⎣0 0 1 0⎦0 0 0 13. For each of the following pairs of matrices, find an elementary matrix E such thatEA = B.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 41[ ] [ ]2 −1 −4 2(a) A =B =5 35 3⎡ ⎤2 1 3(b) A = ⎣−2 4 5⎦ ⎡ ⎤2 1 3B = ⎣ 3 1 4⎦3 1 4 −2 4 5⎡ ⎤4 −2 3⎡ ⎤4 −2 3(c) A = ⎣ 1 0 2⎦ B = ⎣1 0 2⎦−2 3 1 0 3 5⎡3 4⎤1⎡3 4⎤1(d) A = ⎣2 −7 −1⎦ B = ⎣2 −7 −1⎦2 −7 3 8 1 54. Given⎡ ⎤ ⎡ ⎤ ⎡ ⎤1 2 4 1 2 4 1 2 4A = ⎣2 1 3⎦, B = ⎣2 1 3⎦, C = ⎣0 −1 −3⎦1 0 2 2 2 6 2 2 6(a) Find an elementary matrix E such that EA = B.(b) Find an elementary matrixF such that FB = C.(c) Is C row equivalent to A? Explain.5. Given⎡ ⎤2 1 1A = ⎣6 4 5⎦4 1 3Find elementary matrices E 1 , E 2 , E 3 such thatwhere U is an upper triangular matrix.E 3 E 2 E 1 A = U6. Use the method shown in Example 1.30 and Example 1.31 to find the inverse ofthe given matrix if the matrix is invertible.[ ] [ ] [ ]1 4 −3 6 6 −4(a) (b)(c)2 7 4 5 −3 2⎡ ⎤1 0 1(d) ⎣3 3 4⎦⎡ ⎤1 1 1(e) ⎣0 1 1⎦⎡ ⎤2 0 5(f) ⎣0 3 0⎦2 2 3 0 0 1 1 0 3⎡ ⎤ ⎡ ⎤ ⎡ √ √ ⎤1 12 1 4− 2 5 5 5 2 3 2 0(g) ⎣3 2 5⎦(h) ⎣11 1 ⎦5 5 10(i) ⎣−4 √2 2 0 ⎦0 −1 10 0 115− 4 5110

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 42⎡ ⎤1 0 0 0(j) ⎢1 3 0 0⎥⎣1 3 5 0⎦1 3 5 7⎡ ⎤1−8 17 232(k) ⎢ 4 0 −95 ⎥⎣ 0 0 0 0 ⎦−1 13 4 2⎡ ⎤0 0 2 0(l) ⎢1 0 0 1⎥⎣0 −1 3 0 ⎦2 1 5 −37. Find all values of a such that the following matrices are invertible:⎡ ⎤[ ]1 a 02 a(a) A =(b) A = ⎣−1 0 1⎦3 40 1 18. Givencompute A −1 and use it to:A =[ ] 3 15 2B =[ ] 1 23 4(a) Find a 2×2 matrix X such that AX = B.(b) Find a 2×2 matrix Y such that YA = B.9. Consider the matrixA =[ ] 1 0−5 2(a) Find elementary matrices E 1 and E 2 such that E 2 E 1 A = I.(b) Write A −1 as a product of two elementary matrices.(c) Write A as a product of two elementary matrices.10. Express the matrix⎡0⎤1 7 8A = ⎣ 1 3 3 8 ⎦−2 −5 1 −8in the form A = EFGR, where E,F, and G are elementary matrices and R is inrow-echelon form.11. Consider the following given matrix A. Findelementary matrices E 1 , E 2 , ..., E k−1 ,E k such that E k E k−1···E 2 E 1 A = A r where A r is the reduced row-echelon form ofA.⎡ ⎤⎛ ⎞1 0 2(a) ⎣0 1 0⎦0 0 3⎛ ⎞0 0 1 1(c) ⎝1 2 0 0⎠3 0 0 12 0 2 4(b) ⎝1 1 2 −10 0 2 4⎠⎡ ⎤1 1 −1 −1(d) ⎢2 0 0 2⎥⎣0 0 0 3 ⎦2 2 −2 −212. Indicate whether the statement is always true or sometimes false.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 43(a) Every square matrix can be expressed as a product of elementary matrices.(b) The product of two elementary matrices is an elementary matrix.(c) If A is invertible and a multiple of the first row of A is added to the secondrow, then the resulting matrix is invertible.(d) If A is invertible and AB = 0, then it must be true that B = 0.(e) If A is a singular n×n matrix, then Ax = 0 has infinitely many solutions.(f) If A is a singular n×n matrix, then the reduced row-echelon from of A hasat least one row of zeros.(g) If A −1 is expressible as a product of elementary matrices, then the homogeneouslinear system Ax = 0 has only the trivial solution.Answer to Exercise 1.51. (a), (c), (d), (f) 2. (a) r 2 +3r 1 (b) 5r 3 (c) r 2 ↔ r 4 (d) r 1 + 1 r 4 3⎡ ⎤ ⎡ ⎤ ⎡ ⎤[ ] 1 0 0 1 0 0 1 0 0−2 03. (a) (b) ⎣0 0 1⎦ (c) ⎣0 1 0⎦ (d) ⎣0 1 0⎦0 10 1 0 0 2 1 2 0 1⎡ ⎤1 0 0⎡ ⎤1 0 04. (a) E = ⎣0 1 0⎦ (b) F = ⎣0 1 −1⎦1 0 1 0 0 1⎡ ⎤1 0 0⎡ ⎤1 0 0⎡ ⎤1 0 05. E 1 = ⎣−3 1 0⎦, E 2 = ⎣ 0 1 0⎦, E 3 = ⎣0 1 0⎦0 0 1 −2 0 1 0 1 1[ ] [ ]⎡ ⎤−7 4 −5 21 2 −339 136. (a) (b) (c) Not invertible (d) ⎣−1 1 −1⎦2 −1 4 139 130 −2 3⎡ ⎤ ⎡ ⎤ ⎡ ⎤1 −1 0 3 0 −5 −7 5 3(e) ⎣0 1 −1⎦ (f) ⎣ 10 0 ⎦3(g) ⎣ 3 −2 −2⎦0 0 1 −1 0 2 3 −2 −1⎡ ⎤ ⎡ √1 3 1 2 −3 √ ⎤ ⎡ ⎤2 1 0 0 0026 26(h) ⎣ 0 1 −1⎦ ⎢(i) ⎣4 √2 2⎥0⎦ (j) ⎢− 10 03 3 ⎥26 26 ⎣ 0 −−2 2 010⎦5 50 0 1 0 0 − 1 17 7⎡ ⎤(k) Not invertible (l)⎢⎣− 4 537. (a) a ≠ 8 3(b) a ≠ 1 8. (a)3515150 −1 0210 0 024 2− 1 − 1 5 5 5 5][−1 04 2⎥⎦(b)[ ]−8 5−14 9

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 44[ ] [ ]1 0 1 09. (a) E 1 = , E5 1 2 =0 1 (b) A −1 = E 2 E 1 (c) A = E1 −1 E2−12⎡ ⎤⎡⎤⎡⎤⎡⎤0 1 0 1 0 0 1 0 0 1 3 3 810. ⎣1 0 0⎦⎣0 1 0⎦⎣0 1 0⎦⎣0 1 7 8⎦0 0 1 −2 0 1 0 1 1 0 0 0 0⎡ ⎤ ⎡ ⎤1 0 0 1 0 −211. (a) E 1 = ⎣0 1 0⎦, E 2 = ⎣0 1 0 ⎦0 0 1 0 0 13⎡ ⎤ ⎡ ⎤ ⎡ ⎤10 0 1 0 0 1 0 02(b) E 1 = ⎣0 1 0⎦, E 2 = ⎣−1 1 0⎦, E 3 = ⎣0 1 0⎦,0 1 0 0 0 1 0 0 1 2⎡ ⎤ ⎡ ⎤1 0 −1 1 0 0E 4 = ⎣0 1 0 ⎦, E 5 = ⎣0 1 −1⎦0 0 1 0 0 1⎡ ⎤ ⎡ ⎤ ⎡ ⎤0 1 0 1 0 0 1 0 0(c) E 1 = ⎣1 0 0⎦, E 2 = ⎣ 0 1 0⎦, E 3 = ⎣0 0 1⎦,0 0 1 −3 0 1 0 1 0⎡ ⎤ ⎡ ⎤1 0 0 1 −2 0E 4 = ⎣0 − 1 0⎦, E6 5 = ⎣0 1 0⎦0 0 1 0 0 1⎡ ⎤ ⎡ ⎤ ⎡ ⎤1 0 0 0 1 0 0 0 1 0 0 0(d) E 1 = ⎢−2 1 0 0⎥⎣ 0 0 1 0⎦ , E 2 = ⎢ 0 1 0 0⎥⎣ 0 0 1 0⎦ , E 3 = ⎢0 − 1 0 02 ⎥⎣0 0 1 0⎦0 0 0 1 −2 0 0 1 0 0 0 1⎡ ⎤ ⎡ ⎤ ⎡ ⎤1 −1 0 0 1 0 0 0 1 0 −1 0E 4 = ⎢0 1 0 0⎥⎣0 0 1 0⎦ , E 5 = ⎢0 1 0 0⎥⎣0 0 1 0⎦ , E 6 = ⎢0 1 0 0⎥⎣0 0 1 0⎦ ,30 0 0 1 0 0 0 1 0 0 0 1⎡ ⎤1 0 0 0E 7 = ⎢0 1 2 0⎥⎣0 0 1 0⎦0 0 0 112. (a) False (b) True (c) True (d) True (e) True (f) True (g) True1.6 Further Results on Systems of Equations andInvertibilityA Basic TheoremTheorem 1.15 Every system of linear equations has no solutions, or has exactly onesolutions, or has infinitely many solutions.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 45Solving Linear Systems by Matrix InversionSo far, we have studied two methods for solving linear system: Gaussian elimination andGauss-Jordan elimination. The following theorem provides a new method for solvingcertain linear systems.Theorem 1.16 If A is an invertible n × n matrix, then for each n × 1 matrix b, thesystem of equations Ax = b has exactly one solution, namely, x = A −1 b.Example 1.33 Solve the systemSolution .........x 1 −2x 2 + x 3 = 72x 1 −5x 2 +2x 3 = 63x 1 +2x 2 − x 3 = 1Remark Note that the method of Example 1.33 applies only when the system has asmany equations as unknowns, andthe coefficient matrixis invertible. This methodis lessefficient, computationally, than Gaussian elimination, but it is important in the analysisof equations involving matrices.Linear Systems with a Common Coefficient MatrixFrequently, one is concerned with solving a sequence of systemsAx = b 1 , Ax = b 2 , Ax = b 3 ,..., Ax = b keach of which has the same square coefficient matrix A. If A is invertible, then thesolutionsx 1 = A −1 b 1 , x 2 = A −1 b 2 , x 3 = A −1 b 3 ,..., x k = A −1 b kcan be obtained with one matrix inversion and k matrix multiplications. Once again,however, a more efficient method is to form the matrix[A|b 1 |b 2 | ··· |b k ] (1.13)inwhich the coefficient matrix A is “augmented” by all k of the matrices and then reduce(1.13) to reduced row-echelon form by Gauss-Jordan elimination. In this way we cansolve all k systems at once. This method has the added advantage that it applies evenwhen A is not invertible.Example 1.34 Solve the systems(a) 2x 1 −4x 2 = −10x 1 −3x 2 + x 4 = −4x 1 − x 3 +2x 4 = 43x 1 −4x 2 +3x 3 − x 4 = −11

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 46Solution .........Properties of Invertible Matrices(b) 2x 1 −4x 2 = −8x 1 −3x 2 + x 4 = −2x 1 − x 3 +2x 4 = 93x 1 −4x 2 +3x 3 − x 4 = −15Up to now, to show that an n×n matrix A is invertible, it has been necessary to findan n×n matrix B such thatAB = I and BA = IThenext theorem shows thatif we produce ann×nmatrixB satisfying either condition,then the other condition holds automatically.Theorem 1.17 Let A be a square matrix.(a) If B is a square matrix satisfying BA = I, then B = A −1 .(b) If B is a square matrix satisfying AB = I, then B = A −1 .Theorem 1.18 (Equivalent Statements) If A is an n×n matrix, then the followingare equivalent.(a) A is invertible.(b) Ax = 0 has only the trivial solution.(c) The reduced row-echelon form of A is I n .(d) A is expressible as a product of elementary matrices.(e) Ax = b is consistent for every n×1 matrix b.(f) Ax = b has exactly one solution for every n×1 matrix b.Theorem 1.19 Let A and B be square matrices of the same size. If AB is invertible,then A and B must also be invertible.The following fundamental problem will occur frequently in various contexts.A Fundamental Problem: Let A be a fixed m×n matrix. Find allm×1 matrices b such that the system of equations Ax = b is consistent.If Ais aninvertible matrix, Theorem 1.16 completely solves this problem by assertingthat for every m × 1 matrix b, the linear system Ax = b has the unique solutionx = A −1 b. If A is not square, or if A is square but not invertible, then Theorem 1.16does not apply. In this case the matrix b must usually satisfy certain conditions in orderfor Ax = b to be consistent.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 47Example 1.35 What conditions must b 1 , b 2 , and b 3 satisfy in order for the system ofequationsto be consistent?Solution .........x 1 +x 2 +2x 3 = b 1x 1 +x 3 = b 22x 1 +x 2 +3x 3 = b 3Example 1.36 What conditions must b 1 , b 2 , and b 3 satisfy in order for the system ofequationsto be consistent?Solution .........x 1 −2x 2 + x 3 = b 12x 1 −5x 2 +2x 3 = b 23x 1 +2x 2 − x 3 = b 3Exercise 1.61. Solve the following system by inverting the coefficient matrix and using Theorem1.16.(a) x 1 + x 2 = 25x 1 +6x 2 = 9(c) 5x 1 +3x 2 +2x 3 = 43x 1 +3x 2 +2x 3 = 2x 2 + x 3 = 5(b) 2x 1 − x 2 = 86x 1 −5x 2 = 32(d) y + z = 63x−y + z = −7x+y −3z = −13(e) − x−2y −3z = 0w+ x+4y +4z = 7w +3x+7y +9z = 4−w −2x−4y −6z = 6(f) x 1 + 2x 2 −3x 3 = 82x 1 + 5x 2 −6x 3 = 17−x 1 − 2x 2 + x 3 −x 4 = −84x 1 +10x 2 −9x 3 +x 4 = 332. Solve the following general system by inverting the coefficient matrix and usingTheorem 1.16.x 1 +2x 2 +x 3 = b 1x 1 − x 2 +x 3 = b 2x 1 + x 2 = b 3Use the resulting formulas to find the solution if(a) b 1 = −1, b 2 = 3, b 3 = 4 (b) b 1 = 5, b 2 = 0, b 3 = 0(c) b 1 = −1, b 2 = −1, b 3 = 3

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 483. Use the method of Example 1.34 to solve the systems in all part simultaneously.(a) x 1 −5x 2 = b 13x 1 +2x 2 = b 2i. b 1 = 1, b 2 = 4ii. b 1 = −2, b 2 = 5(b) x 1 −2x 2 = b 13x 1 + x 2 = b 2i. b 1 = −3, b 2 = 12ii. b 1 = −2, b 2 = 1iii. b 1 = −3, b 2 = 9(c) 4x 1 −7x 2 = b 1x 1 +2x 2 = b 2i. b 1 = 0, b 2 = 1ii. b 1 = −4, b 2 = 6iii. b 1 = −1, b 2 = 3iv. b 1 = −5, b 2 = 1(d) 2x 1 − x 2 + x 3 = b 1−2x 1 + x 2 −3x 3 = b 24x 1 −3x 2 + x 3 = b 3i. b 1 = 1, b 2 = −5, b 3 = 5ii. b 1 = 1, b 2 = −3, b 3 = 1iii. b 1 = 8, b 2 = −14, b 3 = 144. Solve the systems in both parts at the same time.(a) x 1 −2x 2 +2x 3 = −22x 1 −5x 2 + x 3 = 13x 1 −7x 2 +2x 3 = −1(b) x 1 −2x 2 +2x 3 = 12x 1 −5x 2 + x 3 = −13x 1 −7x 2 +2x 3 = 05. Find conditions that the b’s must satisfy for the system to be consistent.(a) 6x 1 −4x 2 = b 1 (b) x 1 −2x 2 +5x 3 = b 13x 1 −2x 2 = b 2 4x 1 −5x 2 +8x 3 = b 2−3x 1 +3x 2 −3x 3 = b 3(c) x 1 − x 2 +3x 3 +2x 4 = b 1−2x 1 + x 2 +5x 3 + x 4 = b 2−3x 1 +2x 2 +2x 3 − x 4 = b 34x 1 −3x 2 +3x 3 +3x 4 = b 46. Solve the following matrix equation for X.⎡ ⎤ ⎡ ⎤1 −1 1 2 −1 5 7 8⎣2 3 0 ⎦X = ⎣4 0 −3 0 1⎦0 2 −1 3 5 −7 2 17. In each part, determine whether the homogeneous system has a nontrivial solution(without using pencil and paper); then state whether the given matrix is invertible.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 49(a) 3x 1 −2x 2 + x 3 + x 4 = 04x 2 +5x 3 −2x 4 = 0x 3 −3x 4 = 02x 4 = 0(b) 4x 1 +3x 2 − x 3 +2x 4 = 03x 3 − x 4 = 0x 3 +5x 4 = 09x 4 = 0⎡ ⎤3 −2 1 1⎢0 4 5 −2⎥⎣0 0 1 −3⎦0 0 0 2⎡ ⎤4 3 −1 2⎢0 0 3 −1⎥⎣0 0 1 5 ⎦0 0 0 9Answer to Exercise 1.61. (a) x 1 = 3, x 2 = −1 (b) x 1 = 2, x 2 = −4(c) x 1 = 1, x 2 = −11, x 3 = 16 (d) x = −3, y = 2, z = 4(e) w = −6, x = 1, y = 10, z = −7(f) x 1 = 3, x 2 = 1, x 3 = −1, x 4 = 22. (a) x 1 = 163 , x 2 = − 4 3 , x 3 = − 11 3(b) x 1 = − 5 3 , x 2 = 5 3 , x 3 = 10 3(c) x 1 = 3,x 2 = 0,x 3 = −43. (a) i. x 1 = 2217 , x 2 = 117ii. x 1 = 2117 , x 2 = 1117(b) i. x 1 = −1, x 2 = 1 ii. x 1 = 2, x 2 = −5 iii. x 1 = 3, x 2 = 0(c) i. x 1 = 7 , x 15 2 = 4 15ii. x 1 = 34,15 2 = 2815iii. x 1 = 19,15 2 = 1315iv. x 1 = − 1, x 5 2 = 3 5(d) i. x 1 = −3, x 2 = −5, x 3 = 2 ii. x 1 = 0, x 2 = 0, x 3 = 1iii. x 1 = 2, x 2 = −1, x 3 = 34. (a) x 1 = −12−3t, x 2 = −5−t, x 3 = t(b) x 1 = 7−3t, x 2 = 3−t, x 3 = t5. (a) b 1 = 2b 2 (b) b 1 = b 2 +b 3 (c) b 1 = b 3 +b 4 , b 2 = 2b 3 +b 4⎡⎤11 12 −3 27 266. X = ⎣−6 −8 1 −18 −17⎦−15 −21 9 −38 −357. (a) Only the trivial solution x 1 = x 2 = x 3 = x 4 = 0; invertible(b) Infinitely many solutions; no invertible1.7 LU-DecompositionsWith Gaussian elimination and Gauss-Jordan elimination, a linear system is solved byoperating systematically on an augmented matrix. In this section we shall discuss adifferent organization of this approach, one based on factoring the coefficient matrix intoa product of lower and upper triangular matrices.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 50Solving Linear Systems by FactoringWe shall proceed in two stages. First, we shall show how a linear system Ax = b can besolved very easily once the coefficient matrix A is factored into a product of lower andupper triangular matrices. Second, we shall show how to construct such factorizations.If an n×n matrix A can be factored into a product of n×n matrices asA = LUwhere L is lower triangular and U is upper triangular, then the linear system Ax = bcan be solved as follows:Step 1. Rewrite the system Ax = b asStep 2. Define a new n×1 matrix y byLUx = b (1.14)Ux = y (1.15)Step 3. Use (1.15) to rewrite (1.14) as Ly = b and solve this system for y.Step 4. Substitute y in (1.15) and solve for x.The following example illustrates this procedure.Example 1.37 Given the factorization⎡ ⎤ ⎡ ⎤⎡2 1 3 2 0 0⎣ 4 1 7 ⎦ = ⎣ 4 −1 0 ⎦⎣−6 −2 −12 −6 1 −21 1 2320 1 −10 0 1use this result and the method described above to solve the system⎡ ⎤⎡⎤ ⎡ ⎤2 1 3 x 1 −1⎣ 4 1 7 ⎦⎣x 2⎦ = ⎣ 5 ⎦−6 −2 −12 x 3 −2Solution .........LU-DecompositionsWenowturntotheproblemofconstructing suchfactorizations. Tomotivatethemethod,suppose that an n×n matrix A has been reduced to a row-echelon form U by Gaussianelimination—that is, by a certain sequence of elementary row operations. By Theorem1.12 each of these operations can be accomplished by multiplying on the left by anappropriate elementary matrix. Thus there are elementary matrices E 1 , E 2 , ..., E k suchthatE k···E 2 E 1 A = U (1.16)⎤⎦

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 51By Theorem 1.13, E 1 , E 2 , ..., E k are invertible, so we can multiply both sides of Equation(1.16)on the left successively byE −1k ,...,E−1 2 ,E−1 1to obtainA = E1 −1 E−1 2 ···E −1k U (1.17)We can show that the matrix L defined byL = E −11 E−1 2 ···E −1k(1.18)is lower triangular provided that no row interchanges are used in reducing A to U.Substituting (1.18) into (1.17) yieldsA = LUwhich is a factorization of A into a product of a lower triangular matrix and an uppertriangular matrix.The following theorem summarizes the above result.Theorem 1.20 If A is a square matrix that can be reduced to a row-echelon form U byGaussian elimination without row interchange, then A can be factored as A = LU, whereL is a lower triangular matrix.Definition 1.16 A factorization of a square matrix A as A = LU, where L is lowertriangular and U is upper triangular, is called an LU-decomposition or triangulardecomposition of the matrix A.Example 1.38 Find an LU-decomposition of⎡2 1⎤3A = ⎣ 4 1 7 ⎦−6 −2 −12Solution To obtain an LU-decomposition, A = LU, we shall reduce A to a row-echelonform U using Gaussian elimination and then calculate L from (1.18). The steps are asfollows:Elementary MatrixReduction to Corresponding to Inverse of theRow-Echelon Form the Row Operation Elementary Matrix⎡2 1⎤3⎣ 4 1 7 ⎦−6 −2 −12⎡ ⎤ ⎡ ⎤10 0 2 0 02Step 1 E 1 = ⎣0 1 0⎦E1 −1 = ⎣0 1 0⎦0 0 1 0 0 1⎡⎣112324 1 7−6 −2 −12⎤⎦

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 52Elementary MatrixReduction to Corresponding to Inverse of theRow-Echelon Form the Row⎡Operation⎤Elementary⎡Matrix⎤1 0 0 1 0 0Step 2 E 2 = ⎣−4 1 0⎦E2 −1 = ⎣4 1 0⎦0 0 1 0 0 1⎡⎣112320 −1 1−6 −2 −12⎡ ⎤1 0 0Step 3 E 3 = ⎣0 1 0⎦6 0 1⎡1⎣12320 −1 10 1 −3⎡ ⎤1 0 0Step 4 E 4 = ⎣0 −1 0⎦0 0 1⎡ ⎤⎣1 1 2320 1 −10 1 −3⎦⎡ ⎤1 0 0Step 5 E 5 = ⎣0 1 0⎦0 −1 1⎡⎣1 1 2320 1 −10 0 −2⎡1 0 0Step 6 E 6 = ⎣0 1 00 0 − 1 2⎡⎣1 1 2320 1 −10 0 1⎤⎦⎤⎦⎤⎦⎤⎦⎤⎦E −13 =⎡ ⎤1 0 0⎣ 0 1 0⎦−6 0 1⎡ ⎤1 0 0E4 −1 = ⎣0 −1 0⎦0 0 1⎡ ⎤1 0 0E5 −1 = ⎣0 1 0⎦0 1 1⎡ ⎤1 0 0E6 −1 = ⎣0 1 0 ⎦0 0 −2Thus⎡U = ⎣1 1 2320 1 −10 0 1⎤⎦

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 53and, from (1.18),⎡ ⎤⎡⎤⎡⎤⎡⎤⎡⎤⎡⎤2 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0L = ⎣0 1 0⎦⎣4 1 0⎦⎣0 1 0⎦⎣0 −1 0⎦⎣0 1 0⎦⎣0 1 0 ⎦0 0 1 0 0 1 −6 0 1 0 0 1 0 1 1 0 0 −2⎡ ⎤2 0 0= ⎣ 4 −1 0 ⎦−6 1 −2sois an LU-decomposition of A. ♣⎡ ⎤ ⎡ ⎤⎡2 1 3 2 0 0⎣ 4 1 7 ⎦ = ⎣ 4 −1 0 ⎦⎣−6 −2 −12 −6 1 −2Procedure for Finding LU-Decompositions1 1 2320 1 −10 0 1As this example shows, most of the work in constructing an LU-decomposition is expendedin the calculation of L. However, all this work can be eliminated by some carefulbookkeeping of the operations used to reduce A to U. Because we are assuming that norow interchanges are required to reduce A to U, there are only two types of operationsinvolved:• multiplying a row by a nonzero constant, and• adding a multiple of one row to another.The first operation is used to introduce the leading 1’s and the second to introduce zerosbelow the leading 1’s.In Example 1.38, the multiplier needed to introduce the leading 1’s in successive rowswere as follows:1for the first row2−1 for the second row− 1 for the third row2Note that in (1.19), the successive diagonal entries in L were precisely the reciprocals ofthese multipliers: ⎡ ⎤2 0 0L = ⎣ 4 −1 0 ⎦ (1.19)−6 1 −2Next, observe that to introduce zeros below the leading 1 in the first row, we used theoperationsadd −4 times the first row to the secondadd 6 times the first row to the thirdand to introduce the zero below the leading 1 in the second row, we used the operationadd −1 times the the second to the third⎤⎦

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 54Now note in (1.20) that in each position below diagonal of L, the entry is the negativeof the multiplier in the operation that introduced the zero in that position in U:⎡ ⎤2 0 0L = ⎣ 4 −1 0 ⎦ (1.20)−6 1 −2Therefore, we have the following procedure for constructing an LU-decomposition ofa square matrix A provided that A can be reduced to row-echelon form without rowinterchanges.Step 1 Reduce A to a row-echelon form U by Gaussian elimination without row interchanges,keeping track of the multiplier used to introduce the leading 1’s.Step 2 In each position along the main diagonal of L, place the reciprocal of the multiplierthat introduced the leading 1 in that position of U.Step 3 In each position below the main diagonal of L, place the negative of the multiplierused to introduce the zero in that position in U.Example 1.39 Find an LU-decomposition of⎡ ⎤2 4 2A = ⎣1 5 2⎦4 −1 9Solution .........Example 1.40 Find an LU-decomposition of⎡ ⎤6 −2 0A = ⎣9 −1 1⎦3 7 5Solution .........Exercise 1.71. Use the method of Example 1.37 and the LU-decomposition[ ] [ ][ ]3 −6 3 0 1 −2=−2 5 −2 1 0 1to solve the system3x 1 −6x 2 = 0−2x 1 +5x 2 = 1

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 552. Use the method of Example 1.37 and the LU-decomposition⎡ ⎤ ⎡ ⎤⎡⎤2 6 2 2 0 0 1 3 1⎣−3 −8 0⎦ = ⎣−3 1 0⎦⎣0 1 3⎦4 9 2 4 −3 7 0 0 1to solve the system2x 1 +6x 2 +2x 3 = 2−3x 1 −8x 2 = 24x 1 +9x 2 +2x 3 = 33. Find an LU-decomposition of the following coefficient matrix; then use the methodof Example 1.37 to solve the system.⎡ ⎤⎡⎤ ⎡ ⎤[ ][ ] [ ] 2 −2 −2 x 2 8 x1 −21 −4(a) = (b) ⎣ 0 −2 2 ⎦⎣x −1 −1 x 2 −22⎦ = ⎣−2⎦−1 5 2 x 3 64. Let⎡ ⎤⎡⎤ ⎡ ⎤ ⎡ ⎤⎡⎤ ⎡ ⎤5 5 10 x 1 0 1 2 3 x 1 0(c) ⎣−8 −7 −9⎦⎣x 2⎦ = ⎣1⎦ (d) ⎣2 9 6 ⎦⎣x 2⎦ = ⎣10⎦0 4 26 x 3 4 3 2 10 x 3 7⎡ ⎤⎡⎤ ⎡ ⎤1 2 1 x 1 4(e) ⎣2 −1 −1⎦⎣x 2⎦ = ⎣1⎦1 1 3 x 3 0⎡ ⎤⎡⎤ ⎡ ⎤−1 0 1 0 x 1 5(g) ⎢ 2 3 −2 6⎥⎢x 2⎥⎣ 0 −1 2 0⎦⎣x 3⎦ = ⎢−1⎥⎣ 3 ⎦0 0 1 5 x 4 7⎡ ⎤⎡⎤ ⎡ ⎤1 −2 0 3 x 1 11(h) ⎢−2 3 1 −6⎥⎢x 2⎥⎣−1 4 −4 3 ⎦⎣x 3⎦ = ⎢−21⎥⎣−1⎦5 −8 4 0 x 4 23Find an LU-decomposition of A.⎡ ⎤⎡⎤ ⎡ ⎤1 −2 1 x 1 2(f) ⎣ 2 1 −1⎦⎣x 2⎦ = ⎣ 1 ⎦−3 1 −2 x 3 −5⎡ ⎤2 1 −1A = ⎣−2 −1 2 ⎦2 1 0Answer to Exercise 1.71. x 1 = 2, x 2 = 1 2. x 1 = 2, x 2 = −1, x 3 = 2

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 563. (a) x 1 = 3, x 2 = −1 (b) x 1 = −1, x 2 = 1, x 3 = 0 (c) x 1 = −1, x 2 = 1, x 3 = 0(d) x 1 = −49, x 2 = 2, x 3 = 15 (e) x 1 = 1, x 2 = 2, x 3 = −1(f) x 1 = −49, x 2 = 0, x 3 = 1 (g) x 1 = −3, x 2 = 1, x 3 = 2, x 4 = 1(h) x 1 = 3, x 2 = −1, x 3 = 0, x 4 = 2⎡ ⎤⎡⎤2 0 0 1 1 − 1 2 24. A = LU = ⎣−2 1 0⎦⎣0 0 1 ⎦2 1 1 0 0 0

Chapter 2Determinants2.1 Determinants by Cofactor ExpansionRecall from Theorem 1.6 that the 2×2 matrix[ ] a bA =c dis invertible if ad − bc ≠ 0. The expression ad − bc is called the determinant of thematrix A and is denoted by the symbol det(A) or |A|. With this notation, the formulafor A −1 given in Theorem 1.6 isA −1 =1det(A)[ ] d −b−c aOne of the goals of this chapter is to obtain analog of this formula to square matrix ofhigher order. This will require that we extend the concept of a determinant to squarematrices of all orders.Minors and CofactorsThere are several ways in which we might proceed. The approach in this section isa recursive approach. It defined the determinant of an n × n matrix in terms of thedeterminants of certain (n−1)×(n−1) matrices. The (n−1)×(n−1) matrices thatwill appear in this definition are submatrices of the original matrix. These submatricesare given a special name:Definition 2.1 If A is a square matrix, then the minor of entry a ij is denoted by M ijand is defined to be the determinant of the submatrix that remains after the ith row andjth column are deleted from A. The number (−1) i+j M ij is denoted by C ij and is calledthe cofactor of entry a ij .Example 2.1 LetFine the minor and cofactor of a 11 and a 32 .⎡ ⎤3 1 −4A = ⎣2 5 6 ⎦1 4 857

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 58Solution .........Note that the cofactor and the minor of an element a ij differ only in sign; that is,C ij = ±M ij . A quick way to determine whether to use + or − is to use the fact that thesign relating C ij and M ij is in the ith row and jth column of the “checkerboard” array⎡ ⎤+ − + − + ···− + − + − ···+ − + − + ···⎢⎣− + − + − ··· ⎥⎦. . . . .For example, C 11 = M 11 , C 21 = −M 21 , C 13 = M 13 , C 32 = −M 32 , and so on.Cofactor ExpansionsThe definition of a 3×3 determinant in terms of minors and cofactors isdet(A) = a 11 M 11 +a 12 (−M 12 )+a 13 M 13= a 11 C 11 +a 12 C 12 +a 13 C 13 (2.1)Equation (2.1) show that the determinant of A can be computed by multiplying theentries in the first row of A by their corresponding cofactors and adding the resultingproducts. More generally, we define the determinant of an n×n matrix to bedet(A) = a 11 C 11 +a 12 C 12 +···+a 1n C 1nThis method of evaluating det(A) is called cofactor expansion along the first row ofA.⎡ ⎤3 1 0Example 2.2 Let A = ⎣−2 −4 3 ⎦. Evaluate det(A) by cofactor expansion along5 4 −2the first row of A.Solution .........If A is a 3×3 matrix, then its determinant is∣ a 11 a 12 a 13∣∣∣∣∣det(A) =a 21 a 22 a 23∣a 31 a 32 a 33∣ ∣ ∣ ∣ ∣ ∣ ∣∣∣ a= a 22 a 23∣∣∣∣∣∣ a11 −a 21 a 23∣∣∣∣∣∣ aa 32 a 12 +a 21 a 22∣∣∣33 a 31 a 1333 a 31 a 32= a 11 (a 22 a 33 −a 23 a 32 )−a 12 (a 21 a 33 −a 23 a 31 )+a 13 (a 21 a 32 −a 22 a 31 ) (2.2)= a 11 a 22 a 33 +a 12 a 23 a 31 +a 13 a 21 a 32 −a 13 a 22 a 31−a 12 a 21 a 33 −a 11 a 23 a 32 (2.3)

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 59By rearranging the terms in (2.3) in various ways, it is possible to obtain other formulaslike (2.2). There should be no trouble checking that all of the following are correct:det(A) = a 11 C 11 +a 12 C 12 +a 13 C 13= a 11 C 11 +a 21 C 21 +a 31 C 31= a 21 C 21 +a 22 C 22 +a 23 C 23= a 12 C 12 +a 22 C 22 +a 32 C 32= a 31 C 31 +a 32 C 32 +a 33 C 33= a 13 C 13 +a 23 C 23 +a 33 C 33 (2.4)Note that in each equation, the entries and cofactors all come from the same row orcolumn. These equations are called the cofactor expansions of det(A).The results we have just given for 3×3 matrices from a special case of the followinggeneral theorem.Theorem 2.1 Expansions by CofactorsThe determinant of an n×n matrix A can be computed by multiplying the entries in anyrow (or column) by their cofactors and adding the resulting products; that is, for each1 ≤ i ≤ n and 1 ≤ j ≤ n,anddet(A) = a 1j C 1j +a 2j C 2j +···+a nj C nj(cofactor expansion along the jth column)det(A) = a i1 C i1 +a i2 C i2 +···+a in C in(cofactor expansion along the ith row)Note that we may choose any row or any column.Example 2.3 Let A be the matrix in Example 2.2. Evaluate det(A) by cofactor expansionalong the first column of A.Solution .........RemarkIngeneral, thebest strategyforevaluating adeterminant bycofactorexpansionis to expand along a row or column having the largest number of zeros.Example 2.4 Let⎡ ⎤1 0 0 −1A = ⎢3 1 2 2⎥⎣1 0 −2 1 ⎦2 0 0 1Find det(A) by a cofactor expansion along a row or column of your choice.Solution .........

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 60Adjoint of a MatrixDefinition 2.2 If A is any n×n matrix and C ij is the cofactor of a ij , then the matrix⎡ ⎤C 11 C 12 ··· C 1nC 21 C 22 ··· C 2n⎢ ⎥⎣ . . . ⎦C n1 C n2 ··· C nnis call matrix of cofactors from A. The transpose of this matrix is called the adjointmatrix of A and is denoted by adj(A).Example 2.5 Find adj(A) when⎡ ⎤3 2 −1A = ⎣1 6 3 ⎦.2 −4 0Solution .........Theorem 2.2 Inverse of a Matrix Using Its AdjointIf A is an invertible matrix, thenA −1 =1adj(A) (2.5)det(A)Example 2.6 Use (2.5) to find the inverse of the matrix A in Example 2.5.Solution .........Theorem 2.3 If A is an n×n triangular matrix ( upper triangular, lower triangular, ordiagonal ) , then det(A) is the product of the entries on the main diagonal of the matrix;that is, det(A) = a 11 a 22···a nn .Example 2.7 Find det(A) when⎡ ⎤2 7 −3 8 30 −3 7 5 1A =⎢0 0 6 7 6⎥⎣0 0 0 9 8⎦ .0 0 0 0 4Solution .........Cramer’s RuleThe next theorem provides a formula for the solution of certain linear systems of nequations in n unknowns. This formula, known as Cramer’s rule.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 61Theorem 2.4 If Ax = b is a system of n linear equations in n unknowns such thatdet(A) ≠ 0, then the system has a unique solution. This solution isx 1 = det(A 1)det(A) , x 2 = det(A 2)det(A) ,..., x n = det(A n)det(A)where A j is the matrix obtained by replacing the entries in the jth column of A by theentries in the matrix ⎡ ⎤b 1b 2b = ⎢ ⎥⎣ . ⎦b nExample 2.8 Use Cramer’s rule to solveSolution .........x 1 +2x 3 = 6−3x 1 +4x 2 +6x 3 = 30−x 1 −2x 2 +3x 3 = 8Exercise 2.11. Let(a) Find all the minors of A.⎡ ⎤1 −2 3A = ⎣ 6 7 −1⎦−3 1 4(b) Find all the cofactors.2. Evaluate the determinant of the matrix in Exercise1 by a cofactor expansion along(a) the first row (b) the first column (c) the second row(d) the second column (e) the third row (f) the third column3. For the matrix in Exercise1, find(a) adj(A)(b) A −1 using its adjoint4. Evaluate det(A) by a cofactor expansion along a row or column of your choice.⎡3 3⎤1⎡ ⎤k +1 k −1 7(a) A = ⎣1 0 −4⎦ (b) A = ⎣ 2 k −3 4⎦1 −3 55 k +1 k5. Find A −1 using its adjoint.⎡ ⎤⎡ ⎤2 5 52 −3 5(a) A = ⎣−1 −1 0⎦ (b) A = ⎣0 1 −3⎦2 4 30 0 2

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 626. Let⎡ ⎤1 3 1 1A = ⎢2 5 2 2⎥⎣1 3 8 9⎦1 3 2 2(a) Find A −1 using its adjoint.(b) Find A −1 using the method of Example 1.30 in Section 1.5.(c) Which method involves less computation?7. Show that the matrix⎡ ⎤cosθ sinθ 0A = ⎣−sinθ cosθ 0⎦0 0 1is invertible for all values of θ; then find A −1 using its adjoint.[ ]A11 A8. (a) If A = 12is an “upper triangular” block matrix, where A0 A 11 and A 2222are square matrices, then det(A) = det(A 11 )det(A 22 ). Use this result to evaluatedet(A) for⎡ ⎤2 −1 2 5 64 3 −1 3 4A =⎢ 0 0 1 3 5⎥⎣ 0 0 −2 6 2 ⎦0 0 3 5 2(b) Verify your answer in part(a) by using cofactor expansion to evaluate det(A).9. Solve by Cramer’s rule, where it applies.(a) 7x 1 −2x 2 = 33x 1 +x 2 = 5(b) x−4y +z = 64x−y +2z = −12x+2y −3z = −20(c) 3x 1 −x 2 +x 3 = 4−x 1 +7x 2 −2x 3 = 12x 1 +6x 2 −x 3 = 510. Solve the system4x+y +z +w = 63x+7y −z +w = 17x+3y −5z +8w = −3x+y +z +2w = 3by (a) Cramer’s rule(b) Guass-Jordan elimination.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 63Answer to Exercise 2.11. (a) M 11 = 29, M 12 = 21, M 13 = 27, M 21 = −11, M 22 = 13,M 23 = −5, M 31 = −19, M 32 = −19, M 33 = 19(b) C 11 = 29, C 12 = −21, C 13 = 27, C 21 = 11, C 22 = 13, C 23 = 5,C 31 = −19, C 32 = 19, C 33 = 192. 152⎡⎡ ⎤29 11 −193. (a) adj(A) = ⎣−21 13 19 ⎦ (b) A −1 = ⎢⎣27 5 1929152− 211522715211− 19152 1521315251524. (a) −66 (b) k 3 −8k 2 −10k +95⎡ ⎤ ⎡ ⎤1 33 −5 −5 15. (a) A −1 ⎣−3 4 5 ⎦ (b) A −1 2 2= ⎣0 1 3 ⎦22 −2 −3 0 0 1 2⎡ ⎤−4 3 0 −1 ⎡ ⎤6. A −1 = ⎢ 2 −1 0 0cosθ −sinθ 0⎥⎣−7 0 −1 8 ⎦ 7. A−1 = ⎣sinθ cosθ 0⎦0 0 16 0 1 −78. det(A) = 10×(−108) = −10809. (a) x 1 = 1, x 2 = 2 (b) x = − 144,y = 55 −61,z = 4655 11(c) Cramer’s rule do not apply.10. x = 1, y = 0, z = 2, w = 01915219152⎤⎥⎦2.2 Evaluating Determinants by Row ReductionIn this section we shall show that the determinant of a square matrix can be evaluatedby reducing the matrix to row-echelon form. This method is important since it is themost computationally efficient way to find the determinant of a general matrix.We begin with a fundamental theorem that will lead us to an efficient procedure forevaluating the determinant of a matrix of any order n.Theorem 2.5 Let A be a square matrix. If A has a row of zeros or a column of zeros,then det(A) = 0.Theorem 2.6 Let A be a square matrix. Then det(A) = det(A T ).

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 64Theorem 2.7 Let A be a square matrix.1. If B is the matrix that results when a single row or single column of A is multipliedby a scalar k, then det(B) = kdet(A).2. If B is the matrix that results when two rows or two columns of A are interchanged,then det(B) = −det(A).3. If B is the matrix that results when a multiple of one row of A is added to anotherrow or when a multiple of one column is added to another column, then det(B) =det(A).The following table illustrates Theorem 2.7 for 3×3 determinants.Relationship∣ ∣ ka 11 ka 12 ka 13∣∣∣∣∣a 11 a 12 a 13∣∣∣∣∣a 21 a 22 a 23 = ka 21 a 22 a 23∣ a 31 a 32 a 33∣a 31 a 32 a 33det(B) = kdet(A)∣ ∣ a 21 a 22 a 23∣∣∣∣∣a 11 a 12 a 13∣∣∣∣∣a 11 a 12 a 13 = −a 21 a 22 a 23∣a 31 a 32 a 33∣a 31 a 32 a 33OperationThe first row of Ais multiplied by k.The first andsecond rows of Aare interchanged.det(B) = −det(A)∣ ∣ a 11 +ka 12 a 12 +ka 22 a 13 +ka 23∣∣∣∣∣a 11 a 12 a 13∣∣∣∣∣a 21 a 22 a 23 =a 21 a 22 a 23∣ a 31 a 32 a 33∣a 31 a 32 a 33A multiple of thesecond row of Ais added to thefirst row.Elementary Matricesdet(B) = det(A)If we let A = I n in Theorem 2.7, then the matrix B is an elementary matrix, and thetheorem yields the following result about determinants of elementary matrices.Theorem 2.8 Let E be an n×n elementary matrix.1. If E results from multiplying a row of I n by k, then det(E) = k.2. If E results from interchanging two rows of I n , then det(E) = −1.3. If E results from adding a multiple of one row of I n to another, then det(E) = 1.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 65Example 2.9 Find the determinant of the following matrices.⎡ ⎤1 0 0 0(a) A = ⎢0 3 0 0⎥⎣0 0 1 0⎦0 0 0 1⎡ ⎤0 0 0 1(b) B = ⎢0 3 0 0⎥⎣0 0 1 0⎦1 0 0 0⎡ ⎤1 0 0 7(c) C = ⎢0 1 0 0⎥⎣0 0 1 0⎦0 0 0 1Matrices with Proportional Rows or ColumnsTheorem 2.9 If A is a square matrix with two proportional rows or two proportionalcolumns, then det(A) = 0.Each of the following matrices has two proportional rows or columns; thus, each hasa determinant of zero.⎡ ⎤⎡ ⎤[ ]3 −1 4 −51 −2 7−1 4, ⎣−4 8 5⎦,⎢ 6 −2 5 2⎥−2 8 ⎣ 5 8 1 4 ⎦2 −4 3−9 3 −12 15Evaluating Determinants by Row ReductionWe shall now give a method for evaluating determinants that involves substantially lesscomputation than the cofactor expansion method. The idea of this method is to reducethe given matrix to upper triangular from by elementary row operations, then computethe determinant of the upper triangular matrix (an easy computation), and then relatethat determinant to that of the original matrix. Here is an example:Example 2.10 Evaluate det(A) where⎡ ⎤0 1 5A = ⎣3 −6 9⎦2 6 1Solution .........Example 2.11 Compute the determinant of⎡ ⎤1 2 0 7A = ⎢0 7 6 3⎥⎣0 0 3 1 ⎦3 6 0 −5

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 66SolutionExample 2.12 Evaluate det(A) where⎡ ⎤1 2 0 −2A = ⎢0 0 2 −1⎥⎣0 −1 1 0 ⎦1 3 4 1Solution .........Cofactor expansion androw operationscan sometimes be used in combination to providean effective method for evaluating determinants. The following example illustratesthis idea.Example 2.13 Evaluate det(A) where⎡ ⎤3 5 −2 6A = ⎢1 2 −1 1⎥⎣2 4 1 5⎦3 7 5 3Solution .........Example 2.14 Evaluate det(A) where⎡ ⎤2 1 −3 1A = ⎢−3 −2 0 2⎥⎣ 2 1 0 −1⎦1 0 1 2Solution .........Example 2.15 Evaluate det(A) where⎡Solution .........⎢A = ⎣1 303 42−1 3 5 21− 3 58 4 4Exercise 2.2⎤⎥⎦1. Evaluate the following determinants.3 −17 4(a)0 5 1∣0 0 −2∣−2 1 3(c)1 −7 4∣−2 1 3∣√ 2 0 0 0(b)−8 2 0 07 0 −1 0∣ 9 5 6 1∣1 −2 3(d)2 −4 6∣5 −8 1∣

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 672. Evaluatethedeterminant ofthegivenmatrixbyreducingthematrixtorow-echelonform.⎡ ⎤⎡ ⎤3 6 91 −3 0(a) ⎣ 0 0 −2⎦(b) ⎣−2 4 1⎦−2 1 55 −2 2⎡ ⎤⎡(c) ⎢⎣1 −2 3 15 −9 6 3−1 2 −6 −22 8 6 1⎤⎥⎦a b c3. Given thatd e f= −6, find∣g h i∣ d e f(a)g h i∣a b c∣a+g b+h c+i(c)d e f∣ g h i ∣1 3 1 5 3−2 −7 0 −4 2(d)⎢ 0 0 1 0 1⎥⎣ 0 0 2 1 1⎦0 0 0 1 13a 3b 3c(b)−d −e −f∣4g 4h 4i ∣−3a −3b −3c(d)d e f∣g −4d h−4e i−4f∣4. Repeat Exercise 2 using a combination of row reduction and cofactor expansion,as in Example 2.13.5. Solve the equation ∣ ∣∣∣∣∣ x 5 70 x+1 60 0 2x−1∣ = 0Answer to Exercise 2.21. (a) −30 (b) −2 (c) 0 (d) 0 2. (a) 30 (b) −17 (c) 39 (d) −23. (a) −6 (b) 72 (c) −6 (d) 18 5. x = 0,−1, 1 22.3 Properties of the Determinant FunctionBasic Properties of DeterminantsSuppose that A and B are n×n matrices and k is any scalar. We begin by consideringpossible relationships between det(A), det(B), anddet(kA), det(A+B), and det(AB)

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 68Since a common factor of any row of a matrix can be moved though the det sign, andsince each of the n rows in kA has a common factor of k, we obtaindet(kA) = k n det(A) (2.6)For example, ∣ ∣ ∣ ∣ ∣∣∣∣∣ ka 11 ka 12 ka 13∣∣∣∣∣∣∣∣∣∣ a 11 a 12 a 13∣∣∣∣∣ka 21 ka 22 ka 23 = k 3 a 21 a 22 a 23ka 31 ka 32 ka 33 a 31 a 32 a 33Unfortunately, no simple relationship exists among det(A), det(B), and det(A+B).In particular, we emphasize that det(A+B) will usually not be equal to det(A)+det(B).For example, ifA =[ ] 1 2, B =2 5[ ] 3 1, A+B =1 3then det(A) = 1, det(B) = 8, and det(A+B) = 23; thusdet(A+B) ≠ det(A)+det(B)[ ] 4 3,3 8But if we consider two 2×2 matrices that differ only in the second row:[ ] [ ]a11 aA = 12 a11 aand B = 12,a 21 a 22 b 21 b 22then we haveThusdet(A)+det(B) = (a 11 a 22 −a 12 a 21 )+(a 11 b 22 −a 12 b 21 )= a 11 (a 22 +b 22 )−a 12 (a 21 +b 21 )[ ]a= det 11 a 12a 21 +b 21 a 22 +b 22[ ] [ ] [ ]a11 adet 12 a11 a+det 12 a= det 11 a 12a 21 a 22 b 21 b 22 a 21 +b 21 a 22 +b 22This is a special case of the following theorem.Theorem 2.10 Let A,B, and C be n × n matrices that differ only in the single row,say the rth, and assume that the rth row of C can be obtained by adding correspondingentries in the rth row of A and B. Thendet(C) = det(A)+det(B)For example,⎡1 7 5⎤ ⎡ ⎤1 7 5⎡ ⎤1 7 5det⎣2 0 3 ⎦ = det⎣2 0 3⎦+det⎣2 0 3 ⎦1+0 4+1 7+(−1) 1 4 7 0 1 −1

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 69Determinant of a Matrix ProductCorollary 2.1 If B is an n×n matrix and E is an n×nelementary matrix, thenDeterminant Test for Invertibilitydet(EB) = det(E)det(B)Theorem 2.11 A square matrix A is invertible if and only if det(A) ≠ 0.⎡ ⎤1 2 3Example 2.16 Determine whether the matrix A = ⎣1 0 1⎦ is invertible?2 4 6Solution .........Theorem 2.12 If A and B are square matrices of the same size, thenExample 2.17 Letdet(AB) = det(A)det(B)A =Show that det(AB) = det(A)det(B).Solution .........Theorem 2.13 If A is invertible, then[ ] 3 1, B =2 1det(A −1 ) =[ ] −1 35 81det(A) .We conclude this section by merging Theorem 2.11 with the list to produce thefollowing theorem that relates all the major topics we have studied thus far.Theorem 2.14 (Equivalent Statements) If A is an n×n matrix, then the followingstatements are equivalent.(a) A is invertible.(b) Ax = 0 has only the trivial solution.(c) The reduced row-echelon form of A is I n .(d) A can be expressed as a product of elementary matrices.(e) Ax = b is consistent for every n×1 matrix b.(f) Ax = b has exactly one solution for every n×1 matrix b.(g) det(A) ≠ 0.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 701. Verify that det(kA) = k n det(A) for(a) A =Exerise 2.3⎡ ⎤[ ]2 −1 3−1 2; k = 2 (b) A = ⎣3 2 1⎦ ; k = −23 41 4 52. Verify that det(AB) = det(A)det(B) for⎡ ⎤ ⎡ ⎤2 1 0 1 −1 3A = ⎣3 4 0⎦ and B = ⎣7 1 2⎦0 0 2 5 0 1Is det(A+B) = det(A)+det(B)?3. By inspection, explain why det(A) = 0.⎡−2 8 1 4A = ⎢⎣3 2 5 11 10 6 54 −6 4 −34. Use Theorem 2.11 to determine which of the following matrices are invertible.5. Let⎡ ⎤1 0 −1(a) ⎣9 −1 4 ⎦8 9 −1⎡√ √ ⎤ 2 − 7 0(c) ⎣3 √ 2 −3 √ 7 0⎦5 −9 0Assuming that det(A) = −7, find⎤⎥⎦⎡ ⎤4 2 8(b) ⎣−2 1 −4⎦3 1 6⎡ ⎤−3 0 1(d) ⎣ 5 0 6⎦8 0 3⎡ ⎤a b cA = ⎣d e f⎦g h i(a) det(3A) (b) det(A −1 ) (c) det(2A −1 )⎡ ⎤a g d(d) det(2A) −1 (e) det⎣b h e⎦c i f6. Without directly evaluating, show that x = 0 and x = 2 satisfyx 2 x 22 1 1∣0 0 −5∣ = 0

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 717. For which value(s) of k does A fail to be invertible?(a) A =[ ] k −3 −2−2 k −2⎡ ⎤1 2 4(b) A = ⎣3 1 6⎦k 3 28. Indicate whether the statement is always true or sometimes false. Justify eachanswer by giving a logical argument or a counterexample.(a) det(2A) = 2det(A)(b) |A 2 | = |A| 2(c) det(I +A) = 1+det(A)(d) if det(A) = 0 then A is not expressible as a product of elementary matrices.(e) If the reduced row-echelon form of A has a row of zeros, then det(A) = 0.(f) There is no square matrix A such that det(AA T ) = −1.(g) If det(A) = 0, then the homogeneous system Ax = 0 has infinitely manysolutions.Answer to Exercise 2.31. (a) det(2A) = −40 = 2 2 det(A) (b) det(−2A) = −448 = (−2) 3 det(A)2. det(AB) = −170 = 10·(−17) = det(A)det(B)3. The second and forth columns of A are proportional.4. (a) Invertible (b) Not invertible (c) Not invertible (d) Not invertible5. (a) −63 (b) − 1 7(c) − 8 7(d) − 156(e) 76. If x = 0, the first and third rows are proportional.If x = 2, the first and second rows are proportional.7. (a) k = 5±√ 172(b) k = −18. (a) False (b) True (c) False (d) True (e) True (f) True (g) True

Chapter 3Vector Spaces3.1 Real Vector SpacesVector Space AxiomsDefinition 3.1 Let V be an arbitrary nonempty on which the operations of additionsand scalar multiplication are defined. If the following axioms are satisfied by all objectsu, v, w in V and all scalars k and m, then we call V a vector space and we call theobjects in V vectors.1. If u and v are objects in V, then u+v is in V.2. u+v = v+u3. u+(v+w) = (u+v)+w4. There is an object 0 in V, called a zero vector for V, such that 0+u = u+0 = ufor all u in V.5. For each u in V, there is an object −u in V, called a negative of u, such thatu+(−u) = (−u)+u = 0.6. If k is any scalar and u is any object in V, then ku is in V.7. k(u+v) = ku+kv8. (k +m)u = ku+mu9. k(mu) = (km)u10. 1u = uRemark Vector spaces in which the scalars are complex numbers are called complexvector spaces, and those in which the scalars must be real are called real vectorspaces. In this chapter, all of our scalar will be real numbers.72

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 73Examples of Vector SpacesExample 3.1 R n Is a Vector SpaceLet u = (u 1 ,u 2 ,...,u n ) and v = (v 1 ,v 2 ,...,v n ) are two vectors in R n .The sum u+v is defined byu+v = (u 1 +v 1 ,u 2 +v 2 ,...,u n +v n )and if k is any scalar, the scalar multiple ku is defined byku = (ku 1 ,ku 2 ,...,ku n )The operations of addition and scalar multiplication in the above are called the standardoperations on R n .We can show that the set V = R n with these standard operations is a vector space.♣The three most important special cases of R n are R (the real numbers), R 2 (thevectors in the plane), and R 3 (the vectors in 3-space).Example 3.2 A Vector Space of 2×2 MatricesShow that the set V of all 2 ×2 matrices with real entries is a vector space if additionis defined to be matrix addition and scalar multiplication is defined to be matrix scalarmultiplication.Solution It convenient to verify the axioms in the following order: 1, 6, 2, 3, 7, 8, 9, 4,5, and 10. Let [ ] [ ]u11 uu = 12 v11 vand v = 12u 21 u 22 v 21 v 22To prove Axiom 1, we must show that u+v is an object in V; that is, we must showthat u+v is a 2×2 matrix.[ ] [ ] [ ]u11 uu+v = 12 v11 v+ 12 u11 +v= 11 u 12 +v 12u 21 u 22 v 21 v 22 u 21 +v 21 u 22 +v 22Similarly, Axiom 6 holds because for any real scalar number k, we have[ ] [ ]u11 uku = k 12 ku11 ku= 12u 21 u 22 ku 21 ku 22so ku is a 2×2 matrix and consequently is an object in V.Axiom 2 follows from Theorem 1.2(a) since[ ] [ ] [ ] [ ]u11 uu+v = 12 v11 v+ 12 v11 v= 12 u11 u+ 12= v+uu 21 u 22 v 21 v 22 v 21 v 22 u 21 u 22Similarly, Axiom 3 follows from part (b) of that theorem; and Axioms 7, 8, and 9 followfrom parts (f), (h), and (j), respectively.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 74To prove Axiom 4, we must find an object 0 in V such that 0+u = u+0 = u forall u in V. This can be done by defining 0 to be[ ]0 00 =0 0With this definition,0+u =[ ] [ ] [ ]0 0 u11 u+ 12 u11 u= 12= u0 0 u 21 u 22 u 21 u 22and similarly u+0 = u. To prove Axiom 5, we must show that each object u in V hasa negative −u such that u+(−u) = 0 and (−u)+u = 0. This can be done by definingthe negative of u to be [ ]−u11 −u−u = 12−u 21 −u 22With this definition,u+(−u) =[ ] [ ]u11 u 12 −u11 −u+ 12=u 21 u 22 −u 21 −u 22[ ]0 0= 00 0and similarly (−u)+u = 0. Finally, Axiom 10 is a simple computation:[ ] [ ]u11 u1u = 1 12 u11 u= 12= u ♣u 21 u 22 u 21 u 22Example 3.2 is a special case of a vector space V of all m × n matrices with realentries, together with the operations of matrix addition and scalar multiplication. Weshall denote this vector space by the symbol M m×n .Example 3.3 A Vector Space of Real-Valued FunctionsLet V be the set of real-value functions defined on the entire real line (−∞,∞). Iff = f(x) and g = g(x) are two such functions and k is any real number, define the sumfunction f +g and the scalar multiple kf, respectively, by(f +g)(x) = f(x)+g(x) and (kf)(x) = kf(x)In the exercise we shall ask you to show that V is a vector space with respect to theseoperations. This vector space is denoted by F(−∞,∞). ♣

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 75Example 3.4 The Vector Space P nLet P n denote the set of all polynomials of degree less than or equal n. If p and q are twopolynomials in P n and k is any real number, define the sum p+q and the scalar multiplekp, respectively, by(p+q)(x) = p(x)+q(x) and (kp)(x) = kp(x)for all real number x. In this case, the zero vector is the zero polynomial,z(x) = 0x n +0x n−1 +0x n−2 +···+0x+0It is easily verified that all the vector space axioms holds. thus P n , with the standardaddition and scalar multiplication of functions, is a vector space. ♣Example 3.5 The Zero Vector SpaceLet V consist of a single object, which we denote by 0, and define0+0 = 0 and k0 = 0for all scalar k. It is easy to check that all the vector space axioms are satisfied. We callthis the zero vector space.Example 3.6 A Set That Is Not a Vector SpaceLet V = R 2 and define addition and scalar multiplication operations as follows: If u =(u 1 ,u 2 ) and v = (v 1 ,v 2 ), then defineand if k is any real number, then defineu+v = (u 1 +v 1 ,u 2 +v 2 )ku = (ku 1 ,0)For example,if u = (2,4),v = (−3,5), and k = 7, thenu+v = (2+(−3),4+5) = (−1,8)ku = 7u = (7·2,0) = (14,0)In the exercises we will ask you to show that the first nine vector space axioms aresatisfied; however, there are values of u for which Axiom 10 fails to hold. For example,if u = (u 1 ,u 2 ) is such that u 2 ≠ 0, then1u = 1(u 1 ,u 2 ) = (1·u 1 ,0) = (u 1 ,0) ≠ uThus V is not a vector space with the stated operations. ♣

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 76Some Properties of VectorsTheorem 3.1 Let V be a vector space, u a vector in V, and k a scalar; then:(a) 0u = 0(b) k0 = 0(c) (−1)u = −u(d) If ku = 0, then k = 0 or u = 0.Exercise 3.11. A set of objects is given, together with operations of addition and scalar multiplication.Determine which sets are vector spaces under the given operations. Forthose that are not vector spaces, list all axioms that fail to hold.(a) The set of all triples of real numbers (x,y,z) with the operations(x,y,z)+(x ′ ,y ′ ,z ′ ) = (x+x ′ ,y +y ′ ,z +z ′ ) and k(x,y,z) = (kx,y,z)(b) The set of all pairs of real numbers (x,y) with the operations(x,y)+(x ′ ,y ′ ) = (x+x ′ ,y +y ′ ) and k(x,y) = (2kx,2ky)(c) The set of all pairs of real numbers of the form (x,0) with the standardoperations on R 2 .(d) The set of all n-tuples of real numbers of the form (x,x,...,x) with thestandard operations on R n .(e) The set of 2×2 matrices of the form[ ] a 11 bwith the standard matrix addition and scalar multiplication.(f) The set of all real-valued functions f defined everywhere on the real line andsuch that f(1) = 0, with the operations defined in Example 3.3.(g) The set of all pairs of real numbers of the form (1,x) with the operations(1,y)+(1,y ′ ) = (1,y +y ′ ) and k(1,y) = (1,ky)(h) The set of polynomials of the form a+bx with the operations(a 0 +a 1 x)+(b 0 +b 1 x) = (a 0 +b 0 )+(a 1 +b 1 )x and k(a 0 +a 1 x) = (ka 0 )+(ka 1 )x2. Complete the unfinished details of Example 3.3.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 773. Show thatthefirst ninevector spaceaxiomsaresatisfied ifV = R 2 hastheadditionand scalar multiplication operations defined in Example 3.6.1. (a) Not a vector space. Axiom 8 fails.Answer to Exercise 3.1(b) Not a vector space. Axiom 9 and 10 fail.(c) The set is a vector space under the given operations.(d) The set is a vector space under the given operations.(e) Not a vector space. Axiom 1,4,5, and 6 fail.(f) The set is a vector space under the given operations.(g) The set is a vector space under the given operations.(h) The set is a vector space under the given operations.3.2 SubspacesGiven a vector space V, it is often possible to form another vector space by taking asubset S of V and using the operations of V.Definition 3.2 A subset S of a vector space V is called subspaces of V if S is itself avector space under the addition and scalar multiplication defined on V.In general, one must verify the ten vector space axioms to show that a set S withaddition and scalar multiplication forms a vector space. However, if S is part of thelarger set V that is already known to be a vector space, then certain axioms need notbe verified for S because they are “inherited” from V. For example, there is no need tocheck that u+v = v + u (Axiom 2) for S because this holds for all vectors in V andconsequently for all vectors in S. Other axioms inherited by S from V are 3,7,8,9, and10. Thus to show that the set S is a subspace of a vector space V, we need only verifyAxioms 1,4,5, and 6. The following theorem shows that even Axioms 4 and 5 can beomitted.Theorem 3.2 If S is a nonempty subset of a vector space V, then S is a subspace of Vif and only if the following conditions hold.(a) If u and v are vectors in S, then u+v is in S.(b) If k is any scalar and u in any vector in S, then ku is in S.Remark1. A set S of one or more vectors form a vector space V is said to be closedunder addition if condition (a) in Theorem 3.2 holds and closed under scalarmultiplication if condition (b) holds. Thus Theorem 3.2 states that S is a subspace ofV if and only if S is closed under addition and closed under scalar multiplication.2. Every nonempty vector space V has at least two subspaces: V itself is a subspace,and the set {0} consisting of just the zero vector in V is a subspace called the zerosubspace.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 78Example 3.7 Let S = {(x 1 ,x 2 ,x 3 ) ∈ R 3 |x 1 = x 2 }. Show that S is a subspace of R 3 .Solution .........Example 3.8 Let S = {A ∈ M 2×2 |a 12 = −a 21 }. Show that S is a subspace of M 2×2 .Solution .........Example 3.9 Let S be the set of all polynomials of degree less than n with the propertyp(0) = 0. Show that S is a subspace of P n .Solution .........Example 3.10 Let C n [a,b] be the set of all functions f that have a continuous nthderivative on [a,b]. We can verify that C n [a,b] is a subspace of C[a,b]. ♣Remark C[a,b] denote the set of all real-value functions that defined and continuous onthe closed interval [a,b].Example 3.11 Let S be the set of all functions f in C 2 [a,b] such thatf ′′ (x)+f(x) = 0for all x in [a,b]. Show that S is a subspace of C 2 [a,b]Solution .........Example 3.12 A Subset of R 2 That Is Not a SubspaceLet W be the set of all points (x,y) in R 2 such that x ≥ 0 and y ≥ 0. The set Wis not a subspace of R 2 since it is not closed under scalar multiplication. For example,u = (1,1) lies in W, but its negative (−1)u = −u = (−1,−1) does not. ♣{[ ] ∣x ∣∣∣Example 3.13 Let W = x is a real number}. Determine whether W is a sub-1space of M 2×1 .Solution .........Solution Spaces of Homogeneous SystemsIf Ax = b is a system of linear equations, then each vector x that satisfies this equationis called solution vector of the system. The following theorem shows that the solutionvectors of a homogeneous linear system form a vector space, which we shall call solutionspace of the system.Theorem 3.3 If Ax = 0 is a homogeneous linear system of m equations in n unknowns,then the set of solution vectors, denoted by N(A), is a subspace of R n .Note N(A) = {x ∈ R n |Ax = 0}

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 79Example 3.14 Determine N(A) ifSolution .........A =[ ]1 1 1 02 1 0 1Definition 3.3 A vector w is called a linear combination of the vectors v 1 ,v 2 ,...,v rif it can be expressed in the formwhere k 1 ,k 2 ,...,k r are scalars.w = k 1 v 1 +k 2 v 2 +···+k r v rRemark If r = 1, then the equation in Definition 3.3 reduces to w = k 1 v 1 ; that is, wis a linear combination of a single vector v 1 if it is a scalar multiple of v 1 .Example 3.15 Every vector v = (a,b,c) in R 3 is expressible as a linear combination ofthe standard basis vectorssincei = (1,0,0), j = (0,1,0), k = (0,0,1)v = (a,b,c) = a(1,0,0)+b(0,1,0)+c(0,0,1) = ai+bj+ck ♣Example 3.16 Consider the vectors u = (1,2,−1) and v = (6,4,2) in R 3 . Show that(a) w = (9,2,7) is a linear combination of u and v and(b) w ′ = (4,−1,8) is not a linear combination of u and vSolution .........SpanningIf v 1 ,v 2 ,...,v r are vectors in vector space V, then generally some vectors in V may belinear combinations of v 1 ,v 2 ,...,v r and other may not. The following theorem showsthat if we construct a set W consisting of all those vectors that are expressible as linearcombinations of v 1 ,v 2 ,...,v r , then W forms a subspace of V.Theorem 3.4 If v 1 ,v 2 ,...,v r are vectors in a vector space V, then(a) The set W of all linear combinations of v 1 ,v 2 ,...,v r is a subspace of V.(b) W is the smallest subspace of V that contains v 1 ,v 2 ,...,v r in the sense that everyother subspace of V that contains v 1 , v 2 , ..., v r must contain W.Definition 3.4 If S = {v 1 ,v 2 ,...,v r } is the set of vectors in a vector space V, thenthe subspace W of V consisting of all linear combinations of the vector in S is called thespace spanned by v 1 , v 2 , ..., v r , and we say that the vectors v 1 , v 2 , ..., v r span W.To indicate that W is the space spanned by the vectors in the set S = {v 1 ,v 2 ,...,v r },we writeW = Span(S) or W = Span{v 1 ,v 2 ,...,v r }

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 80Example 3.17 If v 1 and v 2 are noncollinear vectors in R 3 with their initial points atthe origin, then Span{v 1 ,v 2 }, which consists of all linear combinations k 1 v 1 + k 2 v 2 , isthe plane determined by v 1 and v 2 .zSpan{v 1 ,v 2 }k 2 v 2v 2v 1k 1 v 1k 1 v 1 +k 2 v 2yxSimilarly, if v is a nonzero vector in R 2 or R 3 , then Span{v}, which is the set of allscalar multiples kv, is the line determined by v.zvkvSpan{v}yxExample 3.18 The polynomial 1, x, x 2 , ..., x n span the vector space P n defined inExample 3.4 since each polynomial p in P n can be written asp = a 0 +a 1 x+···+a n x nwhich is a linear combination of 1,x,x 2 ,...,x n . We can denote this by writingP n = Span{1,x,x 2 ,...,x n } ♣Example 3.19 Determine whether v 1 = (1,1,2), v 2 = (1,0,1), and v 3 = (2,1,3) spanthe vector space R 3 .Solution .........Example 3.20 Determine whether the vectors 1−x 2 , x+2, and x 2 span P 3 .Solution .........

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 81Theorem 3.5 If S = {v 1 ,v 2 ,...,v r } and S ′ = {w 1 ,w 2 ,...,w k } are two set of vectorsin a vector space V, thenSpan{v 1 ,v 2 ,...,v r } = Span{w 1 ,w 2 ,...,w k }if and only if each vector in S is a linear combination of those in S ′ and each vector inS ′ is a linear combination of those in S.Exercise 3.21. Use Theorem 3.2 to determine which of the following sets form subspaces of R 2 .(a) {(x 1 ,x 2 )|x 1 +x 2 = 0} (b) {(x 1 ,x 2 )|x 1 x 2 = 0}(c) {(x 1 ,x 2 )|x 1 = 3x 2 } (d) {(x 1 ,x 2 )||x 1 | = |x 2 |}(e) {(x 1 ,x 2 )|x 2 1 = x2 2 }2. Use Theorem 3.2 to determine which of the following are subspaces of R 3 .(a) all vectors of the form (a,0,0)(b) all vectors of the form (a,1,1)(c) all vectors of the form (a,b,c), where b = a+c(d) all vectors of the form (a,b,c), where b = a+c+1(e) all vectors of the form (a,b,0)(f) all vectors of the form (a,b,c), where a = b = c3. Use Theorem 3.2 to determine which of the following are subspaces of P 3 .(a) all polynomials a 0 +a 1 x+a 2 x 2 +a 3 x 3 for which a 0 = 0(b) all polynomials a 0 +a 1 x+a 2 x 2 +a 3 x 3 for which a 0 +a 1 +a 2 +a 3 = 0(c) all polynomials a 0 +a 1 x+a 2 x 2 +a 3 x 3 for which a 0 , a 1 , a 2 , and a 3 are integers(d) all polynomials p(x) of the form a 0 +a 1 x+a 2 x 2 +a 3 x 3 such that p(0) = 04. Use Theorem 3.2 to determine which of the following sets form subspaces of M 2×2 .(a) The set of all 2×2 diagonal matrices(b) The set of all 2×2 triangular matrices(c) The set of all 2×2 lower triangular matrices(d) The set of all 2×2 matrices A such that a 12 = 1(e) The set of all 2×2 matrices B such that b 11 = 0(f) The set of all symmetric 2×2 matrices(g) The set of all singular 2×2 matrices5. Use Theorem 3.2 to determine which of the following are subspaces of M n×n

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 82(a) all n×n matrices A such that tr(A) = 0(b) all n×n matrices A such that A T = −A(c) all n×n matrices A such that the linear system Ax = 0 has only the trivialsolution(d) all n×n matrices A such that AB = BA for a fixed n×n matrix B6. Use Theorem 3.2 to determine which of the following sets form subspaces ofC[−1,1].(a) The set of function f in C[−1,1] such that f(−1) = f(1)(b) The set of odd functions in C[−1,1](c) The set of continuous nondecreasing functions on [−1,1](d) The set of function f in C[−1,1] such that f(−1) = 0 and f(1) = 0(e) The set of function f in C[−1,1] such that f(−1) = 0 or f(1) = 07. Determine the nullspace of each of the following matrices.[ ][ ]2 11 2 −3 −1(a) A =(b) A =3 2−2 −4 6 3⎡ ⎤ ⎡ ⎤1 3 −41 1 −1 2(c) A = ⎣ 2 −1 −1⎦ (d) A = ⎣ 2 2 −3 1 ⎦−1 −3 4−1 −1 0 −58. Whichofthefollowingarelinearcombinationsofu = (0,−2,2)andv = (1,3,−10)?(a) (2,2,2) (b) (3,1,5) (c) (0,4,5) (d) (0,0,0)9. Express the following as linear combinations of u = (2,1,4), v = (1,−1,3), andw = (3,2,5).(a) (−9,−7,−15) (b) (6,11,6) (c) (0,0,0) (d) (7,8,9)10. Express the following as linear combinations of p 1 = 2+x+4x 2 , p 2 = 1−x+3x 2 ,and p 3 = 3+2x+5x 2 .(a) −9−7x−15x 2(b) 6+11x+6x 2(c) 0(d) 7+8x+9x 211. In each part, determine whether the given vectors span R 3 .(a) v 1 = (2,2,2), v 2 = (0,0,3), v 3 = (0,1,1)(b) v 1 = (2,−1,3), v 2 = (4,1,2), v 3 = (8,−1,8)(c) v 1 = (3,1,4), v 2 = (2,−3,5), v 3 = (5,−2,9), v 4 = (1,4,−1)

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 83(d) v 1 = (1,2,6), v 2 = (3,4,1), v 3 = (4,3,1), v 4 = (3,3,1)12. Which of the following are spanning sets for P 2 ? Justify your answer.(a) {1,x 2 ,x 2 −2}(b) {2,x 2 ,x,2x+3}(c) {x+2,x+1,x 2 −1} (d) {x+2,x 2 −1}13. Let f = cos 2 x and g = sin 2 x. Which of the following lie in the space spanned byf and g?(a) cos2x (b) 3+x 2 (c) 1 (d) sinx (e) 014. In M 2×2 letE 11 =[ ] 1 0, E0 0 12 =Show that E 11 ,E 12 ,E 21 ,E 22 span M 2×2 .[ ] 0 1, E0 0 21 =Answer to Exercise 3.2[ ] 0 0, E1 0 22 =[ ] 0 00 11. (a), (c) 2. (a), (c), (f) 3. (a), (b), (d) 4. (a), (c), (e), (f) 5.(a), (b), (d)6. (a), (b), (d)7. (a) {(0,0)} (b) Span{(−2,1,0,0),(3,0,1,0)}(c) Span{(1,1,1)} (d) Span{(−5,0,−3,1),(−1,1,0,0)}8. (a), (b), (d)9. (a) −2(2,1,4)+(1,−1,3)−2(3,2,5) (b) 4(2,1,4)−5(1,−1,3)+(3,2,5)(c) 0(2,1,4)+0(1,−1,3)+0(3,2,5) (d) 0(2,1,4)−2(1,−1,3)+3(3,2,5)10. (a) −2p 1 +p 2 −2p 3 (b) 4p 1 −5p 2 +p 3 (c) 0p 1 +0p 2 +0p 3(d) 0p 1 −2p 2 +3p 311. (a) The vectors span. (b) The vectors do not span.(c) The vectors do not span. (d) The vectors span.12. (b), (c) 13. (a), (c), (e)3.3 Linear IndependenceIn this section we look more closely at the structure of vector spaces. To begin with, werestrict ourselves to vector spaces that can be generated from a finite set of elements.Each vector in the vector space can be built up from the elements in this generatingset using only the operations of addition and scalar multiplication. This generating setis usually referred to as a spanning set. In particular, it is desirable to find a minimal

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 84spanning set. By minimal we mean a spanning set with no unnecessary elements (i.e.all the elements in the set are needed in order to span the vector space). To see how tofind a minimal spanning set, it is necessary to consider how the vectors in the collectiondepend on each other. Consequently, we introduce the concepts of linear dependenceand linear independence. These simple concepts provide the key to understanding thestructure of vector spaces.Consider the following vectors in R 3 :⎡ ⎤ ⎡ ⎤ ⎡ ⎤1x 1 = ⎣−12−2 −1⎦, x 2 = ⎣ 3 ⎦, x 3 = ⎣ 3 ⎦1 8Let S be the subspace of R 3 spanned by x 1 , x 2 , x 3 . Actually, S can be represented interms of two vectors x 1 and x 2 , since the vector x 3 is already in the span of x 1 and x 2 .x 3 = 3x 1 +2x 2 (3.1)Any linear combination of x 1 , x 2 , x 3 can be reduced to a linear combination of x 1 andx 2 :Thusα 1 x 1 +α 2 x 2 +α 3 x 3 = α 1 x 1 +α 2 x 2 +α 3 (3x 1 +2x 2 )= (α 1 +3α 3 )x 1 +(α 2 +2α 3 )x 2S = Span{x 1 ,x 2 ,x 3 } = Span{x 1 ,x 2 }Equation (3.1) can be rewritten in the form3x 1 +2x 2 −1x 3 = 0 (3.2)Since the three coefficients in (3.2) are nonzero, we could solve for any vector in termsof the other two.It follows thatx 1 = − 2 3 x 2 + 1 3 x 3, x 2 = − 3 2 x 1 + 1 2 x 3, x 3 = 3x 1 +2x 2Span{x 1 ,x 2 ,x 3 } = Span{x 2 ,x 3 } = Span{x 1 ,x 3 } = Span{x 1 ,x 2 }Because of the dependency relation (3.2), the subspace S can be represented as the spanof any two of the given vectors.On the other hand, no such dependency relationship exists between x 1 and x 2 . Indeed,if there were scalars k 1 and k 2 , not both 0, such thatk 1 x 1 +k 2 x 2 = 0 (3.3)then we could solve for one of the vectors in terms of the other.x 1 = − k 2k 1x 2 (k 1 ≠ 0) or x 2 = − k 1k 2x 1 (k 2 ≠ 0)

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 85However, neither of the two vectors in question is a multiple of the other. Therefore,Span{x 1 } and Span{x 2 } are both proper subspaces of Span{x 1 ,x 2 }, and the only waythat (3.3) can hold is if k 1 = k 2 = 0.We can generalize this example by making the following observations.(I) If {v 1 ,v 2 ,...,v n } span a vector space V and one of these vectors can be writtenas a linear combination of the other n−1 vectors, then those n−1 vectors spanV.(II) Given n vectors v 1 ,v 2 ,...,v n , it is possible to write one of the vectors as a linearcombination of the other n−1 vectors if and only if there exist scalars k 1 ,k 2 ,...,k nnot all zero such thatk 1 v 1 +k 2 v 2 +···+k n v n = 0Definition 3.5 If S = {v 1 ,v 2 ,...,v r } is a nonempty set of vectors, then the vectorequationk 1 v 1 +k 2 v 2 +···+k r v r = 0has at least one solution, namelyk 1 = 0, k 2 = 0,..., k r = 0If this is the only solution, then S is called a linearly independent set. If there areother solutions, then S is called a linearly dependent set.It follows from (I) and (II) that, if {v 1 ,v 2 ,...,v r } is a minimal spanning set, thenv 1 ,v 2 ,...,v r are linearly independent. Conversely, if v 1 ,v 2 ,...,v r are linearly independentand span V, then {v 1 ,v 2 ,...,v r } is a minimal spanning set for V.Example 3.21 If v 1 = (2,−1,0,3), v 2 = (1,2,5,−1), and v 3 = (7,−1,5,8), then theset of vectors S = {v 1 ,v 2 ,v 3 } is linearly dependent, since 3v 1 +v 2 −v 3 = 0. ♣Example 3.22 The polynomialp 1 = 1−x, p 2 = 5+3x−2x 2 , p 3 = 1+3x−x 2form a linear dependent set in P 2 since 3p 1 −p 2 +2p 3 = 0. ♣Example 3.23 Consider the vectors i = (1,0,0), j = (0,1,0), and k = (0,0,1) in R 3 .In terms of components, the vector equationbecomesor, equivalently,k 1 i+k 2 j+k 3 k = 0k 1 (1,0,0)+k 2 (0,1,0)+k 3 (0,0,1) = (0,0,0)(k 1 ,k 2 ,k 2 ) = (0,0,0)This impliesthat k 1 = 0, k 2 = 0, and k 3 = 0, so the set S = {i,j,k} is linear independent.A similar argument can be used to show that the vectorse 1 = (1,0,0,...,0), e 2 = (0,1,0,...,0),..., e n = (0,0,0,...,1)form a linear independent set in R n . ♣

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 86Example 3.24 Show that1, x, x 2 ,...,x nform a linearly independent set of vectors in P n .Solution .........Example 3.25 Determine whether the vectorsv 1 = (1,−2,3), v 2 = (5,6,−1), v 3 = (3,2,1)form a linear dependent set or a linear independent set.Solution .........Theorem 3.6 Let v 1 ,v 2 ,...,v n be n vectors in R n and let X = [ v 1 v 2 ··· v n]. Thevectors v 1 , v 2 , ..., v n will be linearly dependent if and only if X is singular.We can use Theorem 3.6 to test whether n vectors are linearly independent in R n .Simply form a matrix X whose columns are the vectors being tested. To determinewhether X is singular, calculate the value of det(X). If det(X) = 0, the vectors arelinearly dependent. If det(X) ≠ 0, the vectors are linearly independent.Example 3.26 Determine wether the vectorsare linearly dependent.Solution .........v 1 = (4,2,3), v 2 = (2,3,1), v 3 = (2,−5,3)To determine wether r vectors v 1 , v 2 , ..., v r in R n are linearly independent, we canrewrite the equationk 1 v 1 +k 2 v 2 +···+k r v r = 0as a linear system Xk = 0, where X = [ v 1 v 2 ··· v r]. If r ≠ n, then the matrix Xis not square, so we cannot use determinants to decide wether the vectors are linearlyindependent. The system is homogeneous, so it has the trivial solution k = 0. It willhave nontrivial solutions if and only if the row echelon forms of X involve free variables.If there are nontrivial solutions, then the vectors are linearly dependent. If there are nofree variables, then k = 0 is the only solution, and hence the vectors must be linearlyindependent.Example 3.27 Givenv 1 = (1,−1,2,3), v 2 = (−2,3,1,−2), v 3 = (1,0,7,7)To determine if the vectors are linearly independent, we reduce the system Xk = 0 torow echelon form⎡ ⎤ ⎡ ⎤1 −2 1 0 1 −2 1 0⎢ ⎢⎣ −1 3 0 0⎥2 1 7 0 ⎦ −−→ ⎢ 0 1 1 0⎥⎣ 0 0 0 0 ⎦3 −2 7 0 0 0 0 0Since the echelon form involves a free variables k 3 , there are nontrivial solutions andhence the vectors must be linearly dependent. ♣

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 87To determine whether a set of vectors is linear independent in R n , we must solve ahomogeneous linear system of equations. A similar situation holds for the vector spaceP n .Example 3.28 Determine whether the vectorsp 1 = x 2 −2x+3, p 2 = 2x 2 +x+8, p 3 = x 2 +8x+7form a linear dependent set or a linear independent set.Solution .........Next we consider a very important property of linearly independent vectors. Linearcombinations of linearly independent vectors are unique. More precisely, we have thefollowing theorem.Theorem 3.7 Let v 1 , v 2 , ..., v n be vectors in a vector space V. A vector v inSpan{v 1 ,v 2 ,...,v n } can be written uniquely as a linear combination of v 1 , v 2 , ..., v nif and only if v 1 , v 2 , ..., v n are linearly independent.The following theorem gives two simple facts about linear independence that areimportant to know.Theorem 3.8(a) A finite set of vectors that contain the zero vector is linearly dependent.(b) A set with exactly two vectors is linear independent if and only if neither vector isa scalar multiple of the other.Geometric Interpretation of Linear IndependenceLinear independence has some useful geometric interpretations in R 2 and R 3 .If x and y are linearly dependent in R 2 , thenk 1 x+k 2 y = 0where k 1 and k 2 are not both 0. If, say, k 1 ≠ 0, we can writex = − k 2k 1yIf two vectors in R 2 are linearly dependent, one of the vectors can be written as a scalarmultiple of the other. Thus, if both vectors are placed at the origin, they will lie alongthe same line.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 88(x 1 ,x 2 )(y 1 ,y 2 )(x 1 ,x 2 )xy(y 1 ,y 2 )x and y linearly dependentx and y linearly independentIf⎡ ⎤ ⎡ ⎤x 1 y 1x = ⎣x 2⎦ and y = ⎣y 2⎦x 3 y 3are linearly independent in R 3 , then the two points (x 1 ,x 2 ,x 3 ) and (y 1 ,y 2 ,y 3 ) will not lieon the same line through the origin in 3-space. Since (0,0,0), (x 1 ,x 2 ,x 3 ), and (y 1 ,y 2 ,y 3 )are not collinear, they determine the plane. If (z 1 ,z 2 ,z 3 ) lies on this plane, the vectorz = (z 1 ,z 2 ,z 3 ) can be written as a linear combination of x and y, and hence x, y, and zare linearly dependent. If (z 1 ,z 2 ,z 3 ) does not lie on the plane, the three vectors will belinearly independent.yzzyxxx, y, and z linearly dependentx, y, and z linearly independentThe next theorem shows that a linearly independent set in R n can contain at mostn vectors.Theorem 3.9 Let S = {v 1 ,v 2 ,...,v n } be a set of vectors in R n . If r > n, then S islinearly dependent.Remark Theorem 3.9 tells us that a set in R 2 with more than two vectors is linearlydependent and a set in R 3 with more than three vectors is linearly dependent.Linear Independence of FunctionsIn Example 3.26 a determinant was used to test wether three vectors were linearly independentin R 3 . Determinants can also be used to help to decide if a set of n vectors islinearly independent in C (n−1) [a,b]. Indeed, let f 1 , f 2 , ..., f n be elements of C (n−1) [a,b].

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 89If these vectors are linearly dependent, then there exist scalars c 1 , c 2 , ..., c n not all zerosuch thatc 1 f 1 (x)+c 2 f 2 (x)+···+c n f n (x) = 0 (3.4)for each x ∈ [a,b]. Taking the derivative with respect to x of both sides of yieldsc 1 f 1 (x)+c 2 f 2 (x)+···+c n f n (x) = 0If we continue taking derivatives of both sides, we end up with the systemc 1 f 1 (x) +c 2 f 2 (x) +···+c n f n (x) = 0c 1 f ′ 1(x) +c 2 f ′ 2(x) +···+c n f ′ n(x) = 0. . . .c 1 f (n−1)1 (x)+c 2 f (n−1)2 (x)+···+c n f n (n−1) (x) = 0For each fixed x ∈ [a,b], the matrix equation⎡⎤⎡⎤ ⎡ ⎤f 1 (x) f 2 (x) ··· f n (x) c 1 0f 1(x) ′ f 2(x) ′ ··· f n(x)′ c 2⎢⎥⎢⎥⎣ . . . ⎦⎣. ⎦ = ⎢ ⎥⎣0. ⎦f (n−1)1 (x) f (n−1)2 (x) ··· f n(n−1) (x) c n 0(3.5)will have the same nontrivial solution (c 1 ,c 2 ,...,c n ) T . Thus, if f 1 , f 2 , ..., f n are linearlydependent in C (n−1) [a,b], then, for fixed x ∈ [a,b], the coefficient matrix of system (3.5)is singular, its determinant is zero.Definition 3.6 Let f 1 , f 2 , ..., f n be functions in C (n−1) [a,b] and define the functionW(x) on [a,b] byf 1 (x) f 2 (x) ··· f n (x)f 1 ′W(x) =(x) f′ 2 (x) ··· f′ n (x). . .∣f (n−1)1 (x) f (n−1)2 (x) ··· f n(n−1)(x) ∣The function W(x) is called the Wronskian of f 1 , f 2 , ..., f n .Theorem 3.10 Let f 1 , f 2 , ..., f n be elements of C (n−1) [a,b]. If there exists a point x 0in [a,b] such that W(x 0 ) ≠ 0, then f 1 , f 2 , ..., f n are linearly independent.Example 3.29 Show that 1, e x , and e 2x are linearly independent in C(−∞,∞).Solution .........Example 3.30 Show that the vectors 1, x, x 2 , and x 3 are linearly independent in P 3 .Solution .........Example 3.31 Determine whether the functions x 2 and x|x| are linearly independentin C[−1,1].Solution .........

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 90Exercise 3.31. Explain why the following are linearly dependent sets of vectors.(a) v 1 = (1,−2,4) and v 2 = (5,−10,−20) in R 3(b) v 1 = (3,−1), v 2 = (4,5), v 3 = (−4,7) in R 2(c) p 1 = 3−2x+x 2 and p 2 = 6−4x+2x 2 in P 2[ ] [ ]−3 4 3 −4(d) A = and B = in M2 0 −2 0 2×22. Determine whether the following vectors are linearly independent in R 2 .(a) (2,1),(3,2) (b) (2,3),(4,6)(c) (−2,1),(1,3),(2,4)(d) (−1,2),(1,−2),(2,−4)(e) (1,2),(−1,1)3. Determine whether the following vectors are linearly independent in R 3 .(a) (1,0,0),(0,1,1),(1,0,1) (b) (1,0,0),(0,1,1),(1,0,1),(1,2,3)(c) (2,1,−2),(3,2,−2),(2,2,0) (d) (2,1,−2),(−2,−1,2),(4,2,−4)(e) (1,1,3),(0,2,1)4. Determine whether the following vectors are linearly independent in R 4 .(a) (3,8,7,−3),(1,5,3,−1),(2,−1,2,6),(1,4,0,3)(b) (0,0,2,2),(3,3,0,0),(1,1,0,−1)(c) (0,3,−3,−6),(−2,0,0,−6),(0,−4,−2,−2),(0,−8,4,−4)(d) (3,0,−3,6),(0,2,3,1),(0,−2,−2,0),(−2,1,2,1)5. Determine whether the following vectors are linearly independent in M 2×2 .(a)(c)[ ] 1 0,1 1[ ]1 0,0 1[ ] 0 10 0[ ]0 1,0 0[ ]2 30 2(b)[ ] 1 0,0 1[ ] 0 1,0 0[ ] 0 01 06. Determine whether the following vectors are linearly independent in P 3 .(a) 1,x 2 ,x 2 −2(c) x+2,x+1,x 2 −1(b) 2,x 2 ,x,2x+3(d) x+2,x 2 −17. (a) Show that the vectors v 1 = (0,,3,1,−1),v 2 = (6,0,5,1), and v 3 = (4,−7,1,3)form a linearly dependent set in R 4(b) Express each vector as a linear combination of the other two.

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 918. For which real values of λ do the following vectors form a linear dependent set inR 3 ?v 1 = (λ,− 1 2 ,−1 2 ), v 2 = (− 1 2 ,λ,−1 2 ), v 3 = (− 1 2 ,−1 2 ,λ)9. For what values of α will the two vectors cos(x+α) and sinx be linearly dependentin C[−π,π]?10. Under what conditions is a set with one vector linearly independent?11. Indicate whether each statement is always true or sometimes false. Justify youranswer by giving a logical argument or a counterexample.(a) If {v 1 ,v 2 } is a linearly dependent set, then each vector is a scalar multiple ofthe other.(b) If {v 1 ,v 2 ,v 3 } is a linearly independent set, then so is the set{kv 1 ,kv 2 ,kv 3 } for every nonzero scalar k.(c) The converse of Theorem 3.8(a) is also true.(d) If v 1 ,v 2 ,...,v n span R n , then they are linearly independent.(e) If v 1 ,v 2 ,...,v n span a vector space V, then there are linearly independent.(f) If v 1 ,v 2 ,...,v r are vectors in a vector space V andSpan{v 1 ,v 2 ,...,v r } = Span{v 1 ,v 2 ,...,v r−1 }then v 1 ,v 2 ,...,v r are linearly dependent.1. (a) v 2 is a scalar multiple of v 1 .Answer to Exercise 3.3(b) The vectors are linear dependent by Theorem 3.9.(c) p 2 is a scalar multiple of p 1 .(d) B is a scalar multiple of A.2. (a), (e) 3. (a), (e) 4. (a), (b), (c), (d) 5. (a), (b) 6. (c), (d)7. (b) v 1 = 2 7 v 2 − 3 7 v 3, v 2 = 7 2 v 1 + 3 2 v 3, v 3 = − 7 3 v 1 + 2 3 v 28. λ = − 1 , λ = 1 9. When α is an odd multiple of π/2.210. If and only if the vector is not zero.11. (a) False (b) True (c) False (d) True (e) False (f) True