Question

Let
\[ A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}. \]
Which of the following is true?

• A. \( A \) is a null matrix
• B. \( A \) is skew-symmetric
• C. \( A^{-1} \) does not exist
• D. \( A^2 = I \)

Hint:

When verifying properties of matrices, compute powers or transposes explicitly to check conditions.

Answer 1

The Correct Option is D

1. Calculate \( A^2 \):

\[ A^2 = A \cdot A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \cdot \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}. \]

2. Multiply the matrices:

\[ A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I, \]

where \( I \) represents the identity matrix.

3. As \( A^2 = I \), the property \( A^2 = I \) is confirmed. The other choices are not correct:

• \( A \) is not a null matrix.
• \( A \) is not skew-symmetric (because \( A \ne -A^T \)).
• \( A^{-1} \) exists due to \( A^2 = I \implies A^{-1} = A \).