The inverse of a matrixSeptember 10, 2007Theorem 1 Let A be an n × n matrix of rank n. Then TAis invertible,and its inverse is TA−1, where A−1can be computed as follows. Let Inbe then× n identity matrix and let ( A In) be the n×2n matrix formed by addingthe columns of Into the right of An. Then rref ( A In) = ( InA−1).Proof: We have already proved that TAis invertible. Recall also that sinceA has rank n, all the rows of rref(A) are nonzero, hence every row andevery column has a leading index, hence rref(A) = In. Let A−1be thematrix described above. To show that TA−1is the inverse of TA, it is enoughto show that for any Y ∈ Rn, X := TA−1Y satisfies the equation TAX = Y .Let us first check this when Y = ej, where (e1, . . . en) be the standard framefor Rn. To solve the equation TA(X) = ej, one puts the augmented matrix( A ej) in reduced row eche lon form. This will look like ( InCj), whereCjis some column vector, and in fact Cjis the solution: TA(Cj) = ej. Nowyou can see easily that all the Cj’s can be calculated together by using themethod of the theorem: Cjis just the jth column of the matrix A−1describedabove.To deduce the general case we use the principle of superposition. For anyY , Y =Pjyjej, where the yj’s are the entries of Y . Hence the principle ofsuperposition tells us thatTA−1(Y ) =XjyjTA−1(ej).Recall that f or any matrix B, TB(ej) = Cj(B), the jth column of B. ThusTA−1(Y ) =XjyjCj(A−1).1Now by the principle of superposition again,TA(TA−1(Y )) = TA(XjyjCj(A−1)) =XjyjTA(CjA−1).By what we saw above, TA(CjA−1) = ej, soTA(TA−1(Y )) =XjyjTA(Cj(A−1)) =Xjyjej=
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