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Math340Exam1Solutions

# Math340Exam1Solutions - Math 340 Lecture 3 1 Problem 1 Let...

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Math 340, Lecture 3 Solutions to Exam # 1 Problem 1: Let A = 1 5 - 3 0 2 1 - 3 0 2 and let B = 4 - 1 5 2 1 1 0 3 0 . Compute each of the following: (a) 3 A - B = 3 1 5 - 3 0 2 1 - 3 0 2 - 4 - 1 5 2 1 1 0 3 0 = 3 15 - 9 0 6 3 - 9 0 6 - 4 - 1 5 2 1 1 0 3 0 = - 1 16 - 14 - 2 5 2 - 9 - 3 6 . (b) AB = 1 5 - 3 0 2 1 - 3 0 2 4 - 1 5 2 1 1 0 3 0 = 14 - 5 10 4 5 2 - 12 9 - 15 . (c) Since ( AB ) T = B T A T , we have B T A T = ( AB ) T = 14 - 5 10 4 5 2 - 12 9 - 15 T = 14 4 - 12 - 5 5 9 10 2 - 15 . Problem 2: (a) What does it mean that an n × n matrix is invertible ? An n × n matrix A is invertible if and only if there is an n × n matrix B so that AB = I n and BA = I n where I n is the identity n × n matrix. (b) Suppose that A and B are n × n matrices, and that both A and B are invertible with inverses A - 1 and B - 1 . Prove that the product AB is also invertible, and find its inverse. We prove that AB is invertible by showing that the matrix B - 1 A - 1 acts as an inverse. We have: ( AB )( B - 1 A - 1 ) = A ( BB - 1 ) A - 1 = AI n A - 1 = AA - 1 = I n , and ( B - 1 A - 1 )( AB ) = B - 1 ( A - 1 A ) B = B - 1 I n B = B - 1 B = I n . Thus if we multiply ( AB ) on either side by the matrix B - 1 A - 1 , we get the identity matrix, and so ( AB ) is invertible with inverse B - 1 A - 1 . (c) Suppose that A , B 1 and B 2 are n × n matrices, and that B 1 A = I n and AB 2 = I n . Prove that B 1 = B 2 . (In this problem, I n denotes the n × n identity matrix).

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