Linear Algebra Exercises-n-Answers.pdf

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4 Linear Algebra, by Hefferon Topic: Cramer’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 Topic: Speed of Calculating Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . 175 Topic: Projective Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 Chapter Five: Similarity 178 Subsection Five.II.1: Definition and Examples . . . . . . . . . . . . . . . . . . . . . . . . 179 Subsection Five.II.2: Diagonalizability . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 Subsection Five.II.3: Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . 186 Subsection Five.III.1: Self-Composition . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 Subsection Five.III.2: Strings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192 Subsection Five.IV.1: Polynomials of Maps and Matrices . . . . . . . . . . . . . . . . . . 196 Subsection Five.IV.2: Jordan Canonical Form . . . . . . . . . . . . . . . . . . . . . . . . 203 Topic: Method of Powers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 Topic: Stable Populations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 Topic: Linear Recurrences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
Chapter One: Linear Systems Subsection One.I.1: Gauss’ Method One.I.1.16 Gauss’ method can be performed in different ways, so these simply exhibit one possible way to get the answer. (a) Gauss’ method −(1/2)ρ 1+ρ 2 2x + 3y = 7 −→ − (5/2)y = −15/2 gives that the solution is y = 3 and x = 2. (b) Gauss’ method here −3ρ 1 +ρ 2 −→ gives x = −1, y = 4, and z = −1. One.I.1.17 (a) Gaussian reduction 1 3 y = 4 ρ +ρ x − z = 0 y + 3z = 1 x − z = 0 −ρ 2 +ρ 3 −→ y + 3z = 1 −3z = 3 −(1/2)ρ 1 +ρ 2 −→ 2x + 2y = 5 −5y = −5/2 shows that y = 1/2 and x = 2 is the unique solution. (b) Gauss’ method gives y = 3/2 and x = 1/2 as the only solution. (c) Row reduction ρ 1+ρ 2 −→ −x + y = 1 2y = 3 −ρ 1 +ρ 2 −→ x − 3y + z = 1 4y + z = 13 shows, because the variable z is not a leading variable in any row, that there are many solutions. (d) Row reduction shows that there is no solution. (e) Gauss’ method −3ρ 1+ρ 2 −→ −x − y = 1 0 = −1 ρ 1 ↔ρ 4 2x − 2y + z = 0 −2ρ 1 +ρ 2 −4y + 3z = −20 −→ −→ x + z = 5 −ρ 1 +ρ 3 −y + 2z = −5 x + y − z = 10 x + y − z = 10 4y + z = 20 4y + z = 20 gives the unique solution (x, y, z) = (5, 5, 0). (f) Here Gauss’ method gives −2ρ 1+ρ 4 − (5/2)z − (5/2)w = −15/2 2x + z + w = 5 y − w = −1 y − w = −1 −(3/2)ρ 1 +ρ 3 −→ which shows that there are many solutions. −ρ 2 +ρ 4 ρ 2 +ρ 4 (5/4)z = 0 x + y − z = 10 −4y + 3z = −20 4z = 0 −(1/4)ρ 2 +ρ 3 −→ −→ y − w = −1 − (5/2)z − (5/2)w = −15/2 2x + z + w = 5 0 = 0 One.I.1.18 (a) From x = 1 − 3y we get that 2(1 − 3y) = −3, giving y = 1. (b) From x = 1 − 3y we get that 2(1 − 3y) + 2y = 0, leading to the conclusion that y = 1/2. Users of this method must check any potential solutions by substituting back into all the equations.
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