If \( A=\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix} \), then \( A^2 \) is equal to :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix} \)
Each entry of \( A^2=A\cdot A \) is a row of A times a column of A.
Row 1 (2, −3, −5):
\( \displaystyle 4+3-5=2 \)
\( \displaystyle -6-12+15=-3 \)
\( \displaystyle -10-15+20=-5 \)
Row 2 (−1, 4, 5):
\( \displaystyle -2-4+5=-1 \)
\( \displaystyle 3+16-15=4 \)
\( \displaystyle 5+20-20=5 \)
Row 3 (1, −3, −4):
\( \displaystyle 2+3-4=1 \)
\( \displaystyle -3-12+12=-3 \)
\( \displaystyle -5-15+16=-4 \)
\( \displaystyle A^2=\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}=A \)
So A is an idempotent matrix.
Why the other options are incorrect
Option (2) \( \begin{bmatrix}2 & -1 & 1\\-3 & 4 & -3\\-5 & 5 & -4\end{bmatrix} \) — Incorrect. This is \( A^{T} \), the transpose of A, not its square.
Option (3) \( \begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix} \) — Incorrect. The first entry of \( A^2 \) is 2, not 1, so \( A^2\ne\mathrm{I} \).
Option (4) \( \begin{bmatrix}4 & 9 & 25\\1 & 16 & 25\\1 & 9 & 16\end{bmatrix} \) — Incorrect. Squaring each entry separately is not matrix multiplication; the first entry of \( A^2 \) is 2, not 4.