Question

Let $A$ and $B$ be two $3\times 3$ real matrices such that $(A^2–B^2)$ is invertible matrix. If $A^5=B^5$ and $A^3{B}^2=A^2{B}^3$, then the value of the determinant of the matrix $A^3+B^3$ is equal to

• A. $1$
• B. $2$
• C. $4$
• D. $0$

Answer 1

The correct option is D $0$

$(A^2–B^2)$ is invertible matrix
$\Rightarrow |A^2-B^2|\ne 0\dots \left(1\right)$
$A^5=B^5\dots \left(2\right)$
$A^3{B}^2=A^2{B}^3\dots \left(3\right)$

Now,
$(A^3+B^3)(A^2-B^2)=A^5-A^3{B}^2+B^3{A}^2-B^5$
$\Rightarrow (A^3+B^3)(A^2-B^2)=[O{]}_{3\times 3}$
Taking determinant both sides, we get
$|(A^3+B^3)(A^2-B^2)|=|[O{]}_{3\times 3}|$
$\Rightarrow |A^3+B^3||A^2-B^2|=0$
$\Rightarrow |A^3+B^3|=0(\because |A^2-B^2|\ne 0)$