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Unformatted text preview: 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 a 1 1 1 2 a + 2 1 2 3 a 3 , 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 1 1 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 nonzero for all a, the matrix A is invertible for all a....
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This note was uploaded on 03/30/2008 for the course MATH 141B taught by Professor Marcfabbri during the Spring '08 term at Pennsylvania State University, University Park.
 Spring '08
 MARCFABBRI
 Math, Logic

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