# The Inverse Matrix of the Transpose is the Transpose of the Inverse Matrix ## Problem 506

Let $A$ be an $n\times n$ invertible matrix. Then prove the transpose $A^{\trans}$ is also invertible and that the inverse matrix of the transpose $A^{\trans}$ is the transpose of the inverse matrix $A^{-1}$.
Namely, show that
$(A^{\trans})^{-1}=(A^{-1})^{\trans}.$ Add to solve later

## Proof.

By definition of inverse matrices, if there is an $n\times n$ matrix $B$ such that
$A^{\trans}B=I \text{ and } B A^{\trans}=I,$ where $I$ is the $n\times n$ identity matrix, then $A^{\trans}$ is invertible and its inverse is $B$, that is, $B=(A^{\trans})^{-1}$.

We claim that we can take $(A^{-1})^{\trans}$ for this $B$.
In fact, we have
\begin{align*}
A^{\trans}(A^{-1})^{\trans}=(A^{-1}A)^{\trans}=I^{\trans}=I.
\end{align*}

Here in the first equality, we used the fact about transpose matrices that
$(CD)^{\trans}=D^{\trans}C^{\trans}$ for any matrices $C, D$ such that the matrix product $CD$ is well-defined.

Similarly, we have
$(A^{-1})^{\trans}A^{\trans}=(AA^{-1})^{\trans}=I^{\trans}=I.$

This proves the transpose $A^{\trans}$ is invertible and that $(A^{-1})^{\trans}$ is the inverse matrix of $A^{\trans}$, that is,
$(A^{\trans})^{-1}=(A^{-1})^{\trans}$ as required. Add to solve later

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