If A is a square matrix such that A2 = A and \( I \) is identity matrix of same order, then \( (2I+A)^3-19A \) is equal to:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 8I \)
I and A commute (IA = AI = A), so the binomial expansion can be used:
\( \displaystyle (2I+A)^3=8I^3+3(4I^2)A+3(2I)A^2+A^3 \)
\( \displaystyle =8I+12A+6A^2+A^3 \)
From \( A^2=A \): \( A^3=A^2\cdot A=A\cdot A=A \). So
\( \displaystyle (2I+A)^3=8I+12A+6A+A \)
\( \displaystyle =8I+19A \)
\( \displaystyle (2I+A)^3-19A=8I \)
Why the other options are incorrect
Option (2) A — Incorrect. The A terms cancel completely (12A + 6A + A = 19A), so no A is left.
Option (3) O — Incorrect. The constant term \( (2I)^3=8I \) remains; the result is not the zero matrix.
Option (4) 8 — Incorrect. The result of matrix operations is a matrix: \( (2I)^3 \) is 8I, not the number 8.