By Kenneth Kuttler

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Putting in the variables, this says x = 2 and y = −1. 19 to find z and w. Take (−1) times the first row and add to the second to get 1 1 0 . 0 1 1 Now take (−1) times the second row and add to the first to get 1 0 0 1 −1 1 . Putting in the variables, this says z = −1 and w = 1. Therefore, the inverse is 2 −1 −1 1 . Didn’t the above seem rather repetitive? Note that exactly the same row operations were used in both systems. In each case, the end result was something of the form (I|v) x where I is the identity and v gave a column of the inverse.

To check this, multiply 1 1 1 2 and 2 −1 2 −1 −1 1 −1 1 = 1 0 0 1 1 2 = 1 0 0 1 1 1 showing that this matrix is indeed the inverse of A. In the last example, how would you find A−1 ? You wish to find a matrix, such that 1 1 1 2 x z y w = 1 0 0 1 . x z y w 46 MATRICES AND LINEAR TRANSFORMATIONS This requires the solution of the systems of equations, x + y = 1, x + 2y = 0 and z + w = 0, z + 2w = 1. 19) for the first system and for the second. Lets solve the first system. Take (−1) times the first row and add to the second to get 1 1 1 0 1 −1 Now take (−1) times the second row and add to the first to get 1 0 0 1 2 −1 .

Use the result of Problem 10 to verify directly that (AB) = B T AT without making any reference to subscripts. 12. Let x = (−1, −1, 1) and y = (0, 1, 2) . Find xT y and xyT if possible. 13. ♠ Give an example of matrices, A, B, C such that B = C, A = 0, and yet AB = AC. 1 1 1 1 −3 1 −1 −2 0 . Find 14. Let A = −2 −1 , B = , and C = −1 2 2 1 −2 1 2 −3 −1 0 if possible. (a) AB (b) BA (c) AC (d) CA (e) CB (f) BC 50 MATRICES AND LINEAR TRANSFORMATIONS 15. ♠ Consider the following digraph.

### An Introduction To Linear Algebra by Kenneth Kuttler

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