If A and B are two non-zero n times n matrice | vedclass

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(D) Given the equation: $A^2 + B = A^2 B$Rearrange the terms to group $B$ on one side:$A^2 = A^2 B - B$$A^2 = (A^2 - I)B$Alternatively,rearrange to fa

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$A^2 B = B A^2$

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(D) Given the equation: $A^2 + B = A^2 B$Rearrange the terms to group $B$ on one side:$A^2 = A^2 B - B$$A^2 = (A^2 - I)B$Alternatively,rearrange to factorize:$A^2 B - B = A^2$$B(A^2 - I) = A^2$Consider the expression $(A^2 - I)(B - I) = A^2 B - A^2 - B + I$Substitute $A^2 B = A^2 + B$ into the expression:$(A^2 - I)(B - I) = (A^2 + B) - A^2 - B + I = I$Since $(A^2 - I)(B - I) = I$,it implies that the matrices $(A^2 - I)$ and $(B - I)$ commute.Therefore,$(A^2 - I)(B - I) = (B - I)(A^2 - I) = I$Expanding $(B - I)(A^2 - I) = I$:$B A^2 - B - A^2 + I = I$$B A^2 = A^2 + B$Since $A^2 + B = A^2 B$,we conclude that:$A^2 B = B A^2$

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

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