MCQ
Let $A$ and $B$ be two $3 \times 3$ real matrices such that $\left(A^{2}-B^{2}\right)$ 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:
  • $0$
  • B
    $2$
  • C
    $1$
  • D
    $4$

Answer

Correct option: A.
$0$
a
$C=A^{2}-B^{2} ;|C| \neq 0$

$A^{5}=B^{5}$ and $A^{3} B^{2}=A^{2} B^{3}$

Now, $A^{5}-A^{3} B^{2}=B^{5}-A^{2} B^{3}$

$\Rightarrow A^{3}\left(A^{2}-B^{2}\right)+B^{3}\left(A^{2}-B^{2}\right)=0$

$\Rightarrow\left(A^{3}+B^{3}\right)\left(A^{2}-B^{2}\right)=0$

Post multiplying inverse of $A^{2}-B^{2}$ : $A^{3}+B^{3}=0$

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