As the title says, if I have a $n$-by-$n$ matrix $M$ which is invertible how do I show that the transposed matrix is invertible with ${({M^{ - 1}})^t} = {({M^t})^{ - 1}}$. I found this in a book but there isn't a solution provided, any assistance would be great.
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2$MM^{-1}=I$, take transpose on both sides, what do you get? – user160738 Dec 13 '15 at 18:12
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Since $M$ is invertible, $MM^{-1}=I$. Transposing both sides produces $(MM^{-1})^t=(M^{-1})^tM^t=I^t=I$, so $M^t$ is invertible with inverse $(M^{-1})^t$. That is, $(M^t)^{-1}=(M^{-1})^t$.
Mb123
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$M$ and its transpose have the same determinant. If you accept a nonzero determinant as a criterion of invertibility then that does the trick.
Justpassingby
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