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Unformatted text preview: AS2sol/MATH1111/YKL/08-09 THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH1111: Linear Algebra Assignment 2 Suggested Solution 1. (a) True. adj A T = ( ( A T ) ij ) T = ( A ji ) T = ( adj A ) T . (b) True. ”Only if” part: r ( A ) = n ⇒ A = E k ··· E 1 where E i ’s are elementary matrices ⇒ det A 6 = 0 ⇒ det(adj A ) 6 = 0 ⇒ adj A is invertible ⇒ adj A = F k ··· F 1 I where F i ’s are elementary matrices ⇒ r (adj A ) = n. ”If” part: Check that the last three implications can obviously go backward. r (adj A ) = n ⇒ adj A is invertible . Claim : adj A is invertible ⇒ A is invertible. Proof . Assume A is not invertible. Then det A = 0, so A (adj A ) = (det A ) I = 0 I = 0 (a zero matrix). So adj A is invertible ⇒ A = 0(adj A ))- 1 = 0. When A = 0, it is found from definition that adj A = 0. So adj A is not invertible. Contradiction arises. Hence, r (adj A ) = n ⇒ A is invertible ⇒ A is row equivalent to I ....
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