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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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