Stitz-Zeager_College_Algebra_e-book

11 and the right hand side then px 0 for all x in the

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Unformatted text preview: iscussion on page 494 shows the solution to AX = B to be X = A−1 B , and since A−1 is unique, so is A−1 B . Conversely, if AX = B has a unique solution for every n × r matrix B , then, in particular, there is a unique solution X0 to the equation AX = In . The solution matrix X0 is our candidate for A−1 . We have AX0 = In by definition, but we need to also show X0 A = In . To that end, we note that A (X0 A) = (AX0 ) A = In A = A. In other words, the matrix X0 A is a solution to the equation AX = A. Clearly, X = In is also a solution to the equation AX = A, and since we are assuming every such equation as a unique solution, we must have X0 A = In . Hence, we have X0 A = AX0 = In , so that X0 = A−1 and A is invertible. The foregoing discussion justifies our quest to find A−1 using our super-sized augmented matrix approach A In Gauss Jordan Elimination −− − − − − − −→ −−−−−−−− I n A− 1 We are, in essence, trying to find the unique solution to the equation AX = In u...
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