19_hwc1SolnsODDA

19_hwc1SolnsODDA - 3.5 Let A = 0 0 0 1 0 0 0 1 Compute A2...

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Unformatted text preview: 3.5 Let A = 0 0 0 1 0 0 0 1 . Compute A2 and A3 . 0 SOLUTION One can just do the computations, using the definition of matrix–matrix multiplication, and the result is: 0 2 0 A= 0 0 0 0 1 0 0 0 3 0 A= 0 0 0 0 0 0 . 0 SECOND SOLUTION Although the computations are not hard, with all of the entries being either zero or one, there is a more insightful way to solve the problem: By Theorem 2, for any matrix B , the j th column of B is B ej . Then since A = [0, e1 , e2 ], we see Ae1 = 0 Then from (∗) we see A2 e1 = A(Ae1 ) = A0 = 0 A2 e2 = A(Ae2 ) = Ae1 = 0 A2 e3 = A(Ae3 ) = Ae2 = e1 . That is, A2 e1 = 0 Next, from (∗) and (∗∗) we see A3 e1 = A(A2 e1 ) = A0 = 0 A3 e2 = A(A2 e2 ) = A0 = 0 A3 e3 = A(A2 e3 ) = Ae1 = 0 . That is, A2 e1 = 0 A2 e2 = 0 and A2 e3 = e1 . (∗ ∗ ∗) Finally, once more using the fact that for any matrix B , the j th column of B is B ej , we can use (∗∗) and (∗ ∗ ∗) to write down the matrices A2 and A3 . Notice in conclusion that for any k > 3, Ak = Ak−3 A3 = Ak−3 0 = 0. 3.7 For any three numbers a, b and c, let A = (a) Compute A2 . 21/september/2005; 22:09 Ae2 = e1 and Ae3 = e2 . (∗) A2 e2 = 0 and A2 e3 = e1 . (∗∗) a 0 b . c 20 ...
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