If \(A = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}\), then \(A^2\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(-4I\)
Multiplying row by column:
\( \displaystyle A^2 = \begin{bmatrix} 0 \times 0 + 2 \times (-2) & 0 \times 2 + 2 \times 0 \\ -2 \times 0 + 0 \times (-2) & -2 \times 2 + 0 \times 0 \end{bmatrix} \)
\( \displaystyle A^2 = \begin{bmatrix} -4 & 0 \\ 0 & -4 \end{bmatrix} = -4I \)
So the square of this skew-symmetric matrix is the scalar matrix \(-4I\).