129a_ans5 - 4 1 0 8-5 6 7 1 3 0 0 4 2 3 2 0    ...

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Spring 2010 Answer of HW5 2.3 #6 not invertible 2.3 #7 invertible 2.3 #15 No. Invertible matrix has linearly independent columns. 2.3 #20 If EF = I then E and F are inverses of each other, hence EF = I = FE . 2.3 #26 The existence of A 2 means A is a square matrix, and so A 2 is a square matrix. Now if the columns of A are linearly independent then, by IMT, A is invertible and so is A 2 . By IMT again, the columns of A 2 span R n . 3.1 #3 det 2 - 4 3 3 1 2 1 4 - 2 = 2 det " 1 2 4 - 2 # - ( - 4) det " 3 2 1 - 2 # +3 det " 3 1 1 4 # = 2( - 10)+4( - 8)+3(11) = - 19 det 2 - 4 3 3 1 2 1 4 - 2 = - ( - 4) det " 3 2 1 - 2 # +1 det " 2 3 1 - 2 # - 4 det " 2 3 3 2 # = 4( - 8)+( - 7) - 4( - 5) = - 19 3.1 #14 det 6 3 2 4 0 9 0 -
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Unformatted text preview: 4 1 0 8-5 6 7 1 3 0 0 4 2 3 2 0         = (-3) det      3 2 4 0-4 1 0-5 6 7 1 2 3 2 0      = (-3)(-1) det    3 2 4-4 1 2 3 2    = (-3)(-1) ± 3 det "-4 1 3 2 # + 2 det " 2 4-4 1 #! = 9 3.1 #22 the row operation is kR 2 + R 1 → R 1 , the determinant is unchanged 3.1 #26 det = 1 3.1 #31 the determinant is 1 because the matrix is a lower triangular matrix with all diagonal entries equal to 1...
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This note was uploaded on 09/08/2010 for the course MATH 129A at San Jose State University .

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