JEE MAIN - Mathematics (2023 - 24th January Morning Shift - No. 10)
If A and B are two non-zero n $$\times$$ n matrices such that $$\mathrm{A^2+B=A^2B}$$, then :
$$\mathrm{A^2B=I}$$
$$\mathrm{A^2=I}$$ or $$\mathrm{B=I}$$
$$\mathrm{A^2B=BA^2}$$
$$\mathrm{AB=I}$$
Explanation
Given : $A^{2}+B=A^{2} B\quad...(i)$
$\Rightarrow A^{2}+B-I=A^{2} B-I$
$\Rightarrow A^{2} B-A^{2}-B+I=I$
$\Rightarrow A^{2}(B-I)-I(B-I)=I$
$\Rightarrow\left(A^{2}-I\right)(B-I)=I$
$\therefore A^{2}-I$ is the inverse matrix of $B-I$ and vice versa.
So, $(B-I)\left(A^{2}-I\right)=I$
$\Rightarrow B A^{2}-B-A^{2}+I=I$
$\therefore A^{2}+B=B A^{2} \quad...(ii)$
So, by (i) and (ii)
$A^{2} B=B A^{2}$
$\Rightarrow A^{2}+B-I=A^{2} B-I$
$\Rightarrow A^{2} B-A^{2}-B+I=I$
$\Rightarrow A^{2}(B-I)-I(B-I)=I$
$\Rightarrow\left(A^{2}-I\right)(B-I)=I$
$\therefore A^{2}-I$ is the inverse matrix of $B-I$ and vice versa.
So, $(B-I)\left(A^{2}-I\right)=I$
$\Rightarrow B A^{2}-B-A^{2}+I=I$
$\therefore A^{2}+B=B A^{2} \quad...(ii)$
So, by (i) and (ii)
$A^{2} B=B A^{2}$
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