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If $A$ is a non-singular matrix such that $A \cdot A^T=A^T \cdot A$ and $B=A^{-1} \cdot A^T$, then
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The correct answer is:
$B \cdot B^T=I$
We have, $A \cdot A^T=A^T \cdot A$ and $B=A^{-1} A^T$
Now, $B B^T=\left(A^{-1} A^T\right)\left(A^{-1} A^T\right)^T$
$=A^{-1} A^T\left(A^T\right)^T\left(A^{-1}\right)^T \quad\left[\because(A B)^T=B^T A^T\right]$
$=A^{-1} A^T A\left(A^T\right)^{-1}\left[\because\left(A^T\right)^T=A,\left(A^{-1}\right)^T=\left(A^T\right)^{-1}\right]$
$=A^{-1} A A^T\left(A^T\right)^{-1}$
$\begin{aligned} B B^T & =I \cdot I=I \\ \therefore B B^T & =I\end{aligned}$
Now, $B B^T=\left(A^{-1} A^T\right)\left(A^{-1} A^T\right)^T$
$=A^{-1} A^T\left(A^T\right)^T\left(A^{-1}\right)^T \quad\left[\because(A B)^T=B^T A^T\right]$
$=A^{-1} A^T A\left(A^T\right)^{-1}\left[\because\left(A^T\right)^T=A,\left(A^{-1}\right)^T=\left(A^T\right)^{-1}\right]$
$=A^{-1} A A^T\left(A^T\right)^{-1}$
$\begin{aligned} B B^T & =I \cdot I=I \\ \therefore B B^T & =I\end{aligned}$
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