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08aAss1sol

08aAss1sol - AS1sol/MATH1111/YKL/08-09 THE UNIVERSITY OF...

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AS1sol/MATH1111/YKL/08-09 THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH1111: Linear Algebra Assignment 1 Suggested Solution 1. Substituting x = 3, y = - 1, z = 2 in turn into the equations, we obtain a system of linear equations in unknowns a, b, c : 3 - a + 2 c = 0 3 b - c - 6 = 1 3 a - 2 + 2 b = 5 Its augmented matrix is - 1 0 2 - 3 0 3 - 1 7 3 2 0 7 3 R 1 + R 3 -----------→ - 1 0 2 - 3 0 3 - 1 7 0 2 6 - 2 - 2 3 R 2 + R 3 -----------→ - 1 0 2 - 3 0 3 - 1 7 0 0 20 3 - 20 3 - R 1 , 1 3 R 2 , 3 20 R 3 -----------→ 1 0 - 2 3 0 1 - 1 3 7 3 0 0 1 - 1 . Hence the solution is a = 1 , b = 2 , c = - 1. 2. ”if part”: Let AB be symmetric, i.e. ( AB ) T = AB . The L.H.S. is ( AB ) T = B T A T = BA, as A and B are symmetric. So AB = BA . ”only if part”: Assume AB = BA . Consider the transpose of AB , ( AB ) T = B T A T = BA as A and B are symmetric. This follows ( AB ) T = BA = AB by our condition. Hence AB is symmetric. 3. Let A be a square matrix of order n . Prove the following: (a) As A is invertible, its inverse A - 1 exists. Thus, from AB = 0, we have A - 1 AB = A - 1 0 IB = 0 B = 0 . 1

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(b) By Theorem 1.4.2, if A is not invertible, then there is a nontrivial solution for A x = 0 .
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