Q20 / 50 · Manipal (MET) 2022 · Mathematics
If $A$ and $n \times n$ non-singular matrix such that $A A^T=A^{-1} A$ and $B=A^{-1} A^T$, then $B B^{\prime}$ is equal to
- A$1+B$
- Correct answerB$1$
- C$B^{-1}$
- D$\left(B^{-1}\right)^{\prime}$
Solution
If $A$ is non-singular matrix, then $|A| \neq 0$
$\begin{aligned}
A A^T & =A^{-1} A \text { and } B=A^{-1} A^T \\
\Rightarrow \quad B B^{\prime} & =\left(A^{-1} A^T\right)\left(A^{-1} A^T\right)^T \\
& =A^{-1} A^T \cdot A\left(A^{-1}\right)^T \quad\left[\because(A I)^{\prime}=A\right] \\
& =A^{-1} A A^T\left(A^{-1}\right)^T \\
& =I A^T\left(A^{-1}\right)^T \\
& =A^T\left(A^{-1}\right)^T \\
& =\left(A^{-1} A\right)^T \quad\left[\because(A B)^T=B^T A^T\right] \\
& =I^T=1
\end{aligned}$
Hence, the correct answer is option: 'B'.