If A is a square matrix such that A^2 = A then which of the following…

The correct answer is Option (4) → (A), (B) and (D) only **

Given condition: $A^2 = A$.

For any power $k \ge 1$,

$A^k = A$.

Expand $(I + A)^n$ without using any binomial symbols:

$(I + A)^n = I + (\text{sum of all coefficients of }A \text{ in expansion})\cdot A$.

The total of all coefficients of $A$ (except the first $I$ term) in expansion of $(I + A)^n$ is:

$2^n - 1$.

Therefore:

$(I + A)^n = I + (2^n - 1)\,A$.

Now check options:

(A) $(I + A)^4 = I + (2^4 - 1)A = I + 15A$ ✔

(B) $(I + A)^2 = I + (2^2 - 1)A = I + 3A$ ✔

(C) $(I + A)^6 = I + (2^6 - 1)A = I + 63A$, not $30A$ ✘

(D) $(I + A)^3 = I + (2^3 - 1)A = I + 7A$ ✔

Correct statements: (A), (B), (D)