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081st_tut2sol

# 081st_tut2sol - MATH1111/2008-09/Tutorial II Solution 1...

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MATH1111/2008-09/Tutorial II Solution 1 Tutorial II Suggested Solution 1. Let A = ( a ij ) n × n be a square matrix of order n . The trace of A , denoted by tr( A ), is defined as tr( A ) = n i =1 a ii = a 11 + a 22 + · · · + a nn . Given any two square matrices A = ( a ij ) n × n and B = ( b ij ) n × n of order n . Prove that (a) tr( A + B ) = tr( A ) + tr( B ), (b) tr( AB ) = tr( BA ), (c) tr( A T ) = tr( A ). Ans . Let A = ( a ij ) n × n and B = ( b ij ) n × n . Then, (a) tr( A + B ) = n i =1 ( a ii + b ii ) = n i =1 a ii + n i =1 b ii = tr( A ) + tr( B ) . (b) tr( AB ) = n i =1 n r =1 a ir b ri and tr( BA ) = n i =1 n r =1 b ir a ri . Renaming the indices i and r , it is clear that they are equal. (c) Write A T = ( b ij ) n × n , then b ij = a ij . Now tr( A T ) = n i =1 b ii = n i =1 a ii = tr( A ) .

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MATH1111/2008-09/Tutorial II Solution 2 2. (a) Apply the type II and type III elementary row operations (i.e. no interchange of rows) to reduce A = 2 - 6 - 2 - 1 3 3 - 1 3 7 into row echelon form. (b) Hence find a lower triangular matrix L and an upper triangular matrix U such that A = LU . [See p.67 of textbook for the definition of upper and lower triangular matrices.] Ans . (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 .
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081st_tut2sol - MATH1111/2008-09/Tutorial II Solution 1...

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