If A is a square matrix and $I$ is an identity matrix of same order such that $A^2 = A$, then $(I+A)^3 - 8I$ is equal to

$7(A-I)$

The correct answer is Option (3) → $7(A-I)$

Given: $A^{2} = A$ and $I$ is the identity matrix of the same order.

We need to find $(I + A)^{3} - 8I$.

Expand using the binomial theorem:

$(I + A)^{3} = I^{3} + 3I^{2}A + 3IA^{2} + A^{3}$

Since $I^{2} = I$ and $IA = A$, this simplifies to:

$(I + A)^{3} = I + 3A + 3A^{2} + A^{3}$

Given $A^{2} = A \;\Rightarrow\; A^{3} = A^{2}A = A \cdot A = A$.

Substitute these values:

$(I + A)^{3} = I + 3A + 3A + A = I + 7A$

Now, $(I + A)^{3} - 8I = (I + 7A) - 8I = 7A - 7I = 7(A - I)$

Final Answer: $(I + A)^{3} - 8I = 7(A - I)$