Question

If $A$ is an $3\times 3$ non-singular matrix such that $A{A}^T=A^{T}A$ and $B=A^{-1}{A}^T$, then $B{B}^T$ equals

• A. $B^{-1}$
• B. ${\left(B^{-1}\right)}^T$
• C. $I+B$
• D. $I$

Answer 1

Answer: (D)

$I$

Given $B=A^{-1}{A}^T$
Multiplying both side by A, we get $BA=A^{-1}{A}^{T}A$

$\Rightarrow BA=A^{-1}A{A}^T$ ,Since $A^{T}A=A{A}^T$

$\Rightarrow BA=A^T$

Taking transpose both side, we get $(BA{)}^T=A$

$\Rightarrow A^T{B}^T=A$

$\Rightarrow B^T=\frac{A}{A^T}$

$\Rightarrow B^T=\frac{A^{-1}A}{A^{-1}{A}^T}$

$\Rightarrow B^T=\frac{I}{B}$

$\therefore B{B}^T=I$