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3 and 4 .Determinants and Matrices
hard
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:
A
$0$
B
$2$
C
$1$
D
$4$
(JEE MAIN-2021)
Solution
$C=A^{2}-B^{2} ;|C| \neq 0$
$A^{5}=B^{5}$ 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 $A^{2}-B^{2}$ : $A^{3}+B^{3}=0$
Standard 12
Mathematics
