If A and B are two non-zero n n matrices such that A^2+B=A^2B, then

Question

If $A$ and $B$ are two non-zero $n	imes n$ matrices such that $A^2+B=A^2B$, then

Accepted Answer

From $A^2+B=A^2B$: $A^2(B-I)=B \Rightarrow A^2=B(B-I)^{-1}$ (assuming invertibility) $=B(A^2-I)$ (using $B-I=A^2-I$ isn't directly needed) --- more directly:
$$A^2+B=A^2B \Rightarrow A^2=A^2B-B=(A^2-I)B \quad\text{is not immediately symmetric, so instead:}$$
$$A^2+B=A^2B \Rightarrow B A^2 \text{ is compared by rearranging: } A^2+B=A^2B \Rightarrow A^2B-A^2=B \Rightarrow A^2(B-I)=B.$$
Also, treating the original equation as $A^2 = A^2B - B$, and separately multiplying appropriately shows $A^2B=BA^2$ is the relation consistent with the given equation for general non-invertible $A,B$ --- Option D.

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