For an invertible $n$-by-$n$ matrix $M$ show the transpose is also invertible. 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.

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, $(M^t)^{-1}=(M^{-1})^t$.