Question

Let $A$ and $B$ are square matrices of order $3$ such that $A{B}^2=BA$ and $B{A}^2=AB.$ If ${\left(AB\right)}^3=A^3{B}^m$ , then $m$ is equal to

• A. $3$
• B. $4$
• C. $5$
• D. $7$

Answer 1

Answer: (D)

${\left(AB\right)}^3=\left(AB\right)\left(AB\right)\left(AB\right)$
$=A\left(BA\right)\left(BA\right)B$
$=A\left(A{B}^2\right)\left(A{B}^2\right)B=A^{2}B\left(BA\right)B^3$
$=A^{2}B\left(A{B}^2\right)B^3$
$=A^2\left(BA\right)B^5=A^2\left(A{B}^2\right)B^5$
$=A^3{B}^7$