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MA 131: Homework 2

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Semester 1/2012<strong>MA</strong> <strong>131</strong>: <strong>Homework</strong> 2 1Name. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Student ID. . . . . . . . . . . . . . . . . .1. In each of the following fill the correct answer in the space provided.⎡ ⎤1 0 0 00 2 0 0(a) Let A = ⎢ ⎥⎣0 0 3 0⎦ . Evaluate ( 1A) −1 .20 0 0 4Solution⎡ ⎤ ⎡ ⎤1 0 0 0 0 1⎢ ⎥ ⎢ ⎥(b) Given A = ⎣0 1 −2⎦ and B = ⎣0 1 0⎦. Evaluate (AB) −1 .0 0 1 1 0 0Solution(c) Let A be 3×3 matrix such that A 3 = I. Find the value of det(2A) −1 .Solution(d) Solve for x.∣ ∣∣∣∣∣∣ x −11 0 −3∣3 1−x∣ = 3 x −61 2 x−5∣Solution(e) Find the value of cosα by using Cramer’s rule.bcosγ +ccosβ = accosα+acosγ = bbcosα +acosβ = cSolution


Semester 1/2012<strong>MA</strong> <strong>131</strong>: <strong>Homework</strong> 2 2Name. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Student ID. . . . . . . . . . . . . . . . . .In problems 2 - 11, show all your work.2. Solve the following systems using matrix inversion.(a) x 1 +x 3 = 2x 1 − x 2 = −12x 2 +x 3 = 1Solution(b) x 1 +x 3 = 1x 1 − x 2 = 32x 2 +x 3 = −2


Semester 1/2012<strong>MA</strong> <strong>131</strong>: <strong>Homework</strong> 2 3Name. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Student ID. . . . . . . . . . . . . . . . . .3. Solve the following systems using matrix inversion.(i) x 1 −x 2 = −2−x 1 +x 2 +x 3 = 1−x 2 +x 3 = −1Solution(ii) x 1 −x 2 = 1−x 1 +x 2 +x 3 = −1−x 2 +x 3 = 0


Semester 1/2012<strong>MA</strong> <strong>131</strong>: <strong>Homework</strong> 2 4Name. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Student ID. . . . . . . . . . . . . . . . . .4. Solve the following two systems of linear equations by applying the method of Gauss-Jordan elimination.(a) x 1 − x 2 +3x 3 = 82x 1 − x 2 +4x 3 = 11−x 1 +2x 2 −4x 3 = −11(b) x 1 − x 2 +3x 3 = 02x 1 − x 2 +4x 3 = 1−x 1 +2x 2 −4x 3 = 2Solution


Semester 1/2012<strong>MA</strong> <strong>131</strong>: <strong>Homework</strong> 2 5Name. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Student ID. . . . . . . . . . . . . . . . . .5. Find an LU-decomposition of⎡ ⎤1 2 3 4−1 0 1 2A = ⎢ ⎥⎣ 1 0 0 0⎦ .−1 0 0 2Solution


Semester 1/2012<strong>MA</strong> <strong>131</strong>: <strong>Homework</strong> 2 6Name. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Student ID. . . . . . . . . . . . . . . . . .6. Find an LU-decomposition of⎡ ⎤1 1 1 11 2 2 2A = ⎢ ⎥⎣1 2 3 3⎦ .1 2 3 4Solution


Semester 1/2012<strong>MA</strong> <strong>131</strong>: <strong>Homework</strong> 2 7Name. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .⎡ ⎤2 0 3 −11 0 2 27. Evaluate the determinant of A = ⎢ ⎥⎣0 −1 1 4 ⎦2 0 1 −3any row or column that seem convenient.SolutionStudent ID. . . . . . . . . . . . . . . . . .by using cofactor expansion along


Semester 1/2012<strong>MA</strong> <strong>131</strong>: <strong>Homework</strong> 2 8Name. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Student ID. . . . . . . . . . . . . . . . . .8. Use the adjoint method to compute the inverse of⎡ ⎤1 2 −1⎢ ⎥A = ⎣2 2 4 ⎦.1 3 −3Solution


Semester 1/2012<strong>MA</strong> <strong>131</strong>: <strong>Homework</strong> 2 9Name. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Student ID. . . . . . . . . . . . . . . . . .9. Evaluate det(A) where⎡ ⎤1 1 1 11 1 2 2A = ⎢ ⎥⎣1 2 2 1⎦1 2 1 2by reducing the matrix to upper triangular form.Solution


Semester 1/2012<strong>MA</strong> <strong>131</strong>: <strong>Homework</strong> 2 10Name. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .10. Evaluate the determinant of A =reduction and cofactor expansion.Solution⎡ ⎤0 4 −1 1−3 1 1 2⎢ ⎥⎣ 1 0 −2 3⎦ 2 3 0 1Student ID. . . . . . . . . . . . . . . . . .by using a combination of row


Semester 1/2012<strong>MA</strong> <strong>131</strong>: <strong>Homework</strong> 2 11Name. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Student ID. . . . . . . . . . . . . . . . . .11. Let⎡ ⎤ ⎡ ⎤ ⎡ ⎤a 1 a 2 a 3 x 1 b 1⎢ ⎥ ⎢ ⎥ ⎢ ⎥A = ⎣a 4 a 5 a 6 ⎦, X = ⎣x 2 ⎦, B = ⎣b 2 ⎦.a 7 a 8 a 9 x 3 b 3Suppose that det(A) = −2 and M = 2A. If∣ ∣ ∣∣∣∣∣∣where k = det(M) a 1 a 4 a 7 ∣∣∣∣∣∣det(A −1 ) , then find b 1 b 2 b 3 .a 3 a 6 a 9Solution⎡⎢⎣−kkk +1⎤⎥⎦ is a solution to the system AX = B

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