# Answers Homework Week 2 - Part 2.pdf - Matrix Algebra 2...

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Matrix Algebra 2 Answers Homework Week 2 Part 2 Section 4.2. Exercise 60. We solve the system of equations 0 1 1 1 1 1 1 1 1 z y x = 3 2 1 using Cramer’s rule . We first note that 0 1 1 1 1 1 1 1 1 2 3 1 2 R R R R = 1 2 0 2 0 0 1 1 1 3 2 R R = 2 0 0 1 2 0 1 1 1 = 4 (Theorem 4.2). Then we apply Cramer’s rule: , 4 9 4 ) 9 ( 4 1 3 2 3 ) 1 ( ) 1 ( 4 0 1 3 0 2 3 1 1 1 4 0 1 3 1 1 2 1 1 1 3 1 1 2 R R x , 4 3 4 3 1 3 2 ) 1 ( ) 1 ( 4 0 3 1 0 3 2 1 1 1 4 0 3 1 1 2 1 1 1 1 3 1 1 2 R R y . 2 1 4 2 4 2 1 1 1 2 ) 1 ( 4 3 1 2 2 1 0 1 1 0 4 3 1 1 2 1 1 1 1 1 1 3 2 1 K K z Note: the solution method developed in Chapter two gives the same solution. Section 4.3. Exercise 12. (a) Block decomposition gives: 2 0 0 1 1 0 0 1 0 1 4 0 1 1 1 = 1 4 1 1 2 1 1 = [(1 λ ) 2 4 ][ ( 1 λ )( λ ) 2] = [ λ 2 2 λ + 1 4]( 1)[ λ 2 + λ 2] = ( 1)[ λ 2 2 λ 3] [ λ 2 + λ 2] =

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= –(λ – 3)(λ + 1)(λ + 2)(λ – 1). So the characteristic polynomial of A is ( λ 3)( λ + 1)( λ + 2)( λ 1). (b) The eigenvalues of A are the roots of the characteristic polynomial, so λ = 1, λ = 3, λ = 1 and λ = 2. (c) First the eigenvalue λ = 1. 0 0 0 0 1 2 0 0 1 2 0 0 1 0 0 4 0 1 1 0 3 4 2 1 R R R R 0 0 0 0 0 0 0 0 1 2 0 0 0 1 1 0 1 0 0 4 .

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