$\begingroup$I know that $(AB)^T$ = $B^TA^T$ and that $(A^T)^{-1}= (A^{-1})^T$ but couldn't reach any convincing answer. Can someone demonstrate the expression.
$\endgroup$2 Answers
$\begingroup$you have the next property too $$(AB)^{-1}=B^{-1}A^{-1}$$ then the inverse of $(AB)^{T}$ is $\left[(AB)^{T}\right]^{-1}$ $$\left[(AB)^{T}\right]^{-1}=\left[B^{T}A^{T}\right]^{-1}=\left[A^{T}\right]^{-1}\left[B^{T}\right]^{-1}=\left[A^{-1}\right]^{T}\left[B^{-1}\right]^{T}$$
$\endgroup$$\begingroup$One way to see, you can compare entries of two matrices $(AB)^t$ and $B^tA^t$.
Firstly, we have $$AB_{ij} = \sum_{k} a_{ik}b_{kj}.$$
When to transpose a matrix, we switch the rows and the columns. $$(AB)^t_{ij} = AB_{ji} = \sum_{k} a_{jk}b_{ki}.$$
Now, let see the entry ij of $B^tA^t$:
$$B^tA^t_{ij} = \sum_{k} B^t_{ik}A^t_{kj} = \sum_{k} b_{ki}a_{jk}.$$
Finally, you can apply this to see $(A^t)^{-1} = (A^{-1})^t$.
$$I = I^t = (AA^{-1})^t = (A^{-1})^tA^t.$$
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