Theorem 5 (Exercise):

5

I’m sure this has popped up somewhere on here before, anyway:

\begin{align*}(AB)(B^{-1}A^{-1}) &= A(BB^{-1})A^{-1} \\ & = AI_nA^{-1} \\ & = AA^{-1} \\ & = I_n\end{align*}

using associativity and BB^{-1} = AA^{-1} = I. Similarly:

\begin{align*}(B^{-1} A^{-1})(AB) & = B^{-1} (A^{-1}A)B \\ & = B^{-1} I_n B \\ & = B^{-1}B \\ & = I_n\end{align*}

so AB is invertible with inverse B^{-1}A^{-1}.

We could spot that AB was invertible from the fact that \det (AB) = \det A \det B, and since A,B are invertible this product is non-zero, so AB is invertible also.

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Think we might need to refactor all these once everything is up