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By Kenneth Kuttler

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14 is entirely similar. 15, the associative law of multiplication. Before reading this, review the definition of matrix multiplication in terms of entries of the matrices. (A (BC))ij Aik (BC)kj = k = Aik k = Bkl Clj l (AB)il Clj l = ((AB) C)ij . 15. Another important operation on matrices is that of taking the transpose. The following example shows what is meant by this operation, denoted by placing a T as an exponent on the matrix.  T 1 1 + 2i 1 3 2  3  = 1 1 + 2i 1 6 2 6 What happened? The first column became the first row and the second column became the second row.

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. 1 2 3 4 Write the matrix associated with this digraph and find the number of ways to go from 3 to 4 in three steps. 16. ♠ Show that if A−1 exists for an n × n matrix, then it is unique. That is, if BA = I and AB = I, then B = A−1 . −1 17. Show (AB) = B −1 A−1 . 18. ♠ Show that if A is an invertible n × n matrix, then so is AT and AT −1 = A−1 T .

First, here is an interesting lemma. 10 Suppose v ∈ / span (u1 , · · · , uk ) and {u1 , · · · , uk } is linearly independent. Then {u1 , · · · , uk , v} is also linearly independent. k Proof: Suppose i=1 ci ui + dv = 0. It is required to verify that each ci = 0 and that d = 0. But if d = 0, then you can solve for v as a linear combination of the vectors, {u1 , · · · , uk }, k v=− i=1 ci ui d k contrary to assumption. Therefore, d = 0. But then i=1 ci ui = 0 and the linear independence of {u1 , · · · , uk } implies each ci = 0 also.

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