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8 Let A = (Aij ) be an m × n matrix and let B = (Bij ) be an n × p matrix. Then AB is an m × p matrix and (AB)ij = n ∑ k=1 Aik Bkj . 1. MATRICES 45 Two matrices, A and B are said to be conformable in a particular order if they can be multiplied in that order. Thus if A is an r × s matrix and B is a s × p then A and B are conformable in the order AB. The above formula for (AB)ij says that it equals the ith row of A times the j th column of B.   ( ) 1 2  2 3 1  . 9 Multiply if possible  3 1  7 6 2 2 6 First check to see if this is possible.

Then there exists a unique m × n matrix A such that Ax = Lx for all x ∈ Fn . 22) Stated in another way, the k th column of A equals Lek . Proof: By the lemma, (Lx)i = eTi Lx = eTi ∑ xk Lek = k ∑( ) eTi Lek xk . k Let Aik = eTi Lek , to prove the existence part of the theorem. To verify uniqueness, suppose Bx = Ax = Lx for all x ∈ Fn . Then in particular, this is true for x = ej and then multiply on the left by eTi to obtain Bij = eTi Bej = eTi Aej = Aij showing A = B. 5 A linear transformation, L : Fn → Fm is completely determined by the vectors {Le1 , · · · , Len } .

Actually, it produces the right inverse. 25 Suppose A is an n × n matrix. 18) if possible. When this has been done, B = A−1 . 18). As described above, the following is a description of what you have just done. A I Rq Rq−1 ···R1 → I → B Rq Rq−1 ···R1 52 CHAPTER 2. LINEAR TRANSFORMATIONS where those Ri sympolize row operations. It follows that you could undo what you did by doing the inverse of these row operations in the opposite order. Thus I B −1 R1−1 ···Rq−1 Rq−1 → −1 R1−1 ···Rq−1 Rq−1 → A I Here R−1 is the row operation which undoes the row operation R.