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141Bquiz3csols

# 141Bquiz3csols - 1 MATH 141B Instructor Dr M.Fabbri QUIZ#3...

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1 MATH 141B QUIZ#3 Solutions Date: Tues Feb 5, 2008 Instructor: Dr. M.Fabbri 1. i) False . Adding the first two columns as the 4th and 5th columns we get the matrix: a - 1 1 0 a - 1 1 1 - 2 a + 2 1 - 2 3 0 a 3 0 , Now using the method of diagonals to compute the determinant, we get det A = [( - 2 a ( a - 1)+3( a +2)+0] - (0 - 0+ a ) = - 2 a 2 +2 a +3 a +6 - a = - 2 a 2 +4 a +6 = - 2( a +1)( a - 3) . ii) True . A is not invertible if a = - 1 or a = 3 . iii) True . If a = 2 then A is invertible, and so the system has a solution for all values b 1 ,b 2 and b 3 . In fact, the solution is x y z = A - 1 b 1 b 2 b 3 . Therefore, C) is the answer. 2. i) False . Adding the first two columns as the 4th and 5th columns we get the matrix: 1 0 1 1 0 0 - 1 0 0 - 1 1 a 0 1 a , Again, using the method of diagonals to compute the determinant, we get det A = (0 + 0 + 0) - ( - 1 - 0 + 0) = 1 . ii) True . Since the determinant is non-zero for all a, the matrix A is invertible for all a. iii) False . If you reduce the matrix A to I , then the matrix I is reduced to A - 1 = 0 a 1 0 - 1 0 1 - a - 1 and hence ( A ) 32 = - a.

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