Let A be a non-zero square matrix such that \(A^2 = A\), and let I be the identity matrix of the same order.
(A) \((I - A)^2 = I - A\)
(B) \((I + A)^2 = I + 2A\)
(C) \(A(I - A) = O\)
(D) \(A^5 = A\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (C) and (D) only
Since A and I commute, powers of \(I \pm A\) expand like ordinary binomials, and every power of A reduces to A because \(A^2 = A\):
\(A^3 = A^2A = A^2 = A\), and in the same way \(A^5 = A\)
\((I - A)^2 = I - 2A + A^2 = I - A\)
\((I + A)^2 = I + 2A + A^2 = I + 3A\)
\(A(I - A) = A - A^2 = A - A = O\)
So \((I - A)^2 = I - A\), \((I + A)^2 = I + 3A\), \(A(I - A) = O\) and \(A^5 = A\).
Checking the statements
(A) \((I - A)^2 = I - A\) — Correct. \(I - 2A + A^2 = I - A\).
(B) \((I + A)^2 = I + 2A\) — Incorrect. It equals \(I + 3A\), which differs from \(I + 2A\) because \(A \ne O\).
(C) \(A(I - A) = O\) — Correct. \(A - A^2 = A - A = O\).
(D) \(A^5 = A\) — Correct. Each extra factor A leaves \(A^2 = A\) unchanged.