If A^2 = A,then (I + A)^3 - 7A = . . . . . . | vedclass

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(B) Given that $A^2 = A$. We need to evaluate $(I + A)^3 - 7A$. Using the binomial expansion for matrices,$(I + A)^3 = I^3 + 3I^2A + 3IA^2 + A^3$. Sin

Vedclass · Accepted Answer

$I$

Vedclass · Answer

(B) Given that $A^2 = A$. We need to evaluate $(I + A)^3 - 7A$. Using the binomial expansion for matrices,$(I + A)^3 = I^3 + 3I^2A + 3IA^2 + A^3$. Since $I^n = I$ and $IA = AI = A$,we have: $(I + A)^3 = I + 3A + 3A^2 + A^3$. Given $A^2 = A$,it follows that $A^3 = A^2 \cdot A = A \cdot A = A^2 = A$. Substituting these into the expression: $(I + A)^3 = I + 3A + 3A + A = I + 7A$. Now,subtract $7A$: $(I + A)^3 - 7A = (I + 7A) - 7A = I$. Therefore,the correct option is $B$.

If $A^2 = A$,then $(I + A)^3 - 7A =$ . . . . . . ,where $A$ is a square matrix.

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