Linear Algebra Solutions 17

# Linear Algebra Solutions 17 - AX = B where A = 1 1-1 1 2 1...

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(ii) Let B = 0 r s 0 0 t 0 0 0 . Then B 2 = 0 0 rt 0 0 0 0 0 0 and B 3 = 0. Hence from the preceding question ( I 3 - B ) - 1 = I 3 + B + B 2 = 1 0 0 0 1 0 0 0 1 + 0 r s 0 0 t 0 0 0 + 0 0 rt 0 0 0 0 0 0 = 1 r s + rt 0 1 t 0 0 1 . 9. (i) Suppose that A 2 = 0. Then if A - 1 exists, we deduce that A - 1 ( AA ) = A - 1 0, which gives A = 0 and this is a contradiction, as the zero matrix is singular. We conclude that A does not have an inverse. (ii). Suppose that A 2 = A and that A - 1 exists. Then A - 1 ( AA ) = A - 1 A, which gives A = I n . Equivalently, if A 2 = A and A 6 = I n , then A does not have an inverse. 10. The system of linear equations x + y - z = a z = b 2 x + y + 2 z = c is equivalent to the matrix equation
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Unformatted text preview: AX = B , where A = 1 1-1 1 2 1 2 , X = x y z , B = a b c . By Question 7, A-1 exists and hence the system has the unique solution X = -1-3 1 2 4-1 1 a b c = -a-3 b + c 2 a + 4 b-c b . Hence x =-a-3 b + c, y = 2 a + 4 b-c, z = b . 20...
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