Question

Let $A$ and $B$ be $3\times 3$ matrices of real numbers, where $A$ is symmetric, $B$ is skew-symmetric and $\left(A+B\right)\left(A-B\right)=\left(A-B\right)\left(A+B\right).$ If ${\left(AB\right)}^T={\left(-1\right)}^{k}AB,$ where ${\left(AB\right)}^T$ is the transpose of the matrix $AB,$ then the value of $k$ is:

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

$1$

$A=A^T[A$ is symmetric matrix$]$

$B=-B^T[B$ is skew-symmetric matrix$]$

$(A+B)(A-B)=(A-B)(A+B)$

$\Rightarrow A^2-AB+BA-B^2=A^2+AB-BA-B^2$

$\Rightarrow -AB+BA=AB-BA$

$\Rightarrow 2BA=2AB$

$\Rightarrow BA=AB$

Now

$(AB{)}^T=(-1{)}^{k}AB$

L.H.S

$(AB{)}^T$

$=(BA{)}^T\therefore k=1$

$=A^T{B}^T$

$=-AB$