08bAss2sol

# 08bAss2sol - AS2sol/MATH1111/YKL/08-09 THE UNIVERSITY OF...

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AS2sol/MATH1111/YKL/08-09 THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH1111: Linear Algebra Assignment 2 Suggested Solution 1. (a) Apply the elementary row operations 1 2 R 1 , R 1 + R 2 , R 1 + R 3 , 1 2 R 2 , - 6 R 2 + R 3 in order, A is reduced into U = 1 - 3 - 1 0 0 1 0 0 0 . (b) Observe that U is upper triangular. Let E 1 = 1 2 0 0 0 1 0 0 0 1 , E 2 = 1 0 0 1 1 0 0 0 1 , E 3 = 1 0 0 0 1 0 1 0 1 , E 4 = 1 0 0 0 1 2 0 0 0 1 , E 5 = 1 0 0 0 1 0 0 - 6 1 . Then, U = E 5 E 4 · · · E 1 A . Take L = E - 1 1 E - 1 2 · · · E - 1 5 , then L = 2 0 0 0 1 0 0 0 1 1 0 0 - 1 1 0 0 0 1 1 0 0 0 1 0 - 1 0 1 1 0 0 0 2 0 0 0 1 1 0 0 0 1 0 0 6 1 is lower triangular; indeed L = 2 0 0 - 1 2 0 - 1 6 1 . [If you don’t know that the product of lower triangular matrices is lower triangular , prove it now. Hint: Use induction on the order.] 2. Let A = ( a ij ) n × n and B = ( b ij ) n × n . Then, (a) tr( A + B ) = n X i =1 ( a ii + b ii ) = n X i =1 a ii + n X i =1 b ii = tr( A ) + tr( B ) . (b) Straightforward. Skipped. 1

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(c) tr( AB ) = n X i =1 n X r =1 a ir b ri and tr( BA ) = n X i =1 n X r =1 b ir a ri . Renaming the indices i and r , it is clear that they are equal. (d) Write A T = ( b ij ) n × n , then b ij = a ij . Now tr( A T ) = n X i =1 b ii = n X i =1 a ii = tr( A ) . 3. Here we use the fact that det( EA ) = det( AE ) = det( A ) if E is the type I or type III elementary matrix. In case of type II elementary matrix where E scales a row by α , then det( EA ) = det( AE ) = α det( A ).
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