solution6 - M341 Homework Solution 6 Section 2.4: 2(d) :...

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Unformatted text preview: M341 Homework Solution 6 Section 2.4: 2(d) : The matrix is row equivalent to I 4 . Therefore its rank is 4 and it is nonsingular. 3(b): = 20 20 = 0 . Therefore the matrix is singular. 3(d): = 3 8 = 11 . Therefore the matrix is nonsingular and its inverse is [ 3 / 11 2 / 11 4 / 11 1 / 11 ] 4(b): The augmented matrix 5 7 6 3 1 2 1 5 2 1 1 1 reduces to 1 7 / 5 6 / 5 1 1 / 2 1 / 5 3 / 16 5 / 16 1 2 1 . Therefore the matrix is singular. 7(b): First we compute the inverse of the coe cient matrix. The augmented matrix 5 3 6 3 1 7 2 1 2 1 1 1 reduces to 1 1 1 1 3 8 / 5 2 / 5 17 / 5 1 / 5 1 / 5 4 / 5 . Therefore the inverse of the coe cient matrix is 1 3 8 / 5 2 / 5 17 / 5 1 / 5 1 / 5 4 / 5 and the solution of the system is 1 3 8 / 5 2 / 5 17 / 5 1 / 5 1 / 5 4 / 5 4 11 2 = 2 4 3 That is x 1 = 2 , x 2 = 4 , x 3 = 3 . 9(a): Let A = I 2 and B = I 2 . Then they are both nonsingular but A + B = O 2 which is singular. 9(b): Consider an example as following: A = [ 1 ] and B = [ 1 ] 1 Then they are both singular since rank ( A ) = 1 < 2 and rank ( B ) = 1 < 2 , but A + B = I 2 which is nonsingular....
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solution6 - M341 Homework Solution 6 Section 2.4: 2(d) :...

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