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(B) Given that $A^2 = A$. We need to evaluate $(I + A)^3 - 8A$. Using the binomial expansion for matrices,$(I + A)^3 = I^3 + 3I^2A + 3IA^2 + A^3$. Sin

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$I - A$

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(B) Given that $A^2 = A$. We need to evaluate $(I + A)^3 - 8A$. Using the binomial expansion for matrices,$(I + A)^3 = I^3 + 3I^2A + 3IA^2 + A^3$. Since $I^n = I$ and $I 	imes A = A$,this simplifies to $I + 3A + 3A^2 + A^3$. Given $A^2 = A$,we can substitute $A^2$ with $A$: $A^3 = A^2 	imes A = A 	imes A = A^2 = A$. Substituting these into the expression: $(I + A)^3 = I + 3A + 3(A) + A = I + 7A$. Now,subtract $8A$: $(I + 7A) - 8A = I - A$.

If $A$ is a square matrix such that $A^2 = A$,then $(I + A)^3 - 8A =$ . . . . . . .

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