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

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
  1. $A \cdot B^T=I$
  2. $B \cdot B^T=I$
  3. $A^T \cdot B^T=I$
  4. $B^{-1} \cdot B^T=I$

Solution

We have, $A \cdot A^T=A^T \cdot A$ and $B=A^{-1} A^T$ Now, $B B^T=\left(A^{-1} A^T\right)\left(A^{-1} A^T\right)^T$ $=A^{-1} A^T\left(A^T\right)^T\left(A^{-1}\right)^T \quad\left[\because(A B)^T=B^T A^T\right]$ $=A^{-1} A^T A\left(A^T\right)^{-1}\left[\because\left(A^T\right)^T=A,\left(A^{-1}\right)^T=\left(A^T\right)^{-1}\right]$ $=A^{-1} A A^T\left(A^T\right)^{-1}$ $\begin{aligned} B B^T & =I \cdot I=I \\ \therefore B B^T & =I\end{aligned}$

Asked in: AP EAMCET 2021 (24 Aug Shift 2)

Practice more Matrices questions on Aicharya