Question

If \( A = \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1 \end{bmatrix} \), then \( A^{-1} \) is:

• A. \( A^T \)
• B. \( A^2 \)
• C. \( A \)
• D. \( I \)
• E. \( 0 \)

Hint:

For matrices with only \( 0, 1, \text{ or } -1 \), check for orthogonality first. If each row and each column has exactly one non-zero entry, and that entry's square is 1, the inverse is always the transpose.

Answer 1

The Correct Option is A

Concept: This question tests whether the matrix \( A \) is orthogonal. For any orthogonal matrix \( A \), the inverse is simply the transpose (\( A^{-1} = A^T \)). This holds true if every row is a unit vector and is orthogonal to every other row.

Step 1: Identify the transpose \( A^T \).
By flipping rows into columns: \[ A^T = \begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \end{bmatrix} \]

Step 2: Test the identity \( A \cdot A^T = I \).
Perform the matrix multiplication: \[ A \cdot A^T = \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \end{bmatrix} \] Multiplying the first row by the first column:
\( (0)(0) + (-1)(-1) + (0)(0) = 1 \).
Multiplying the second row by the second column:
\( (1)(1) + (0)(0) + (0)(0) = 1 \).
Multiplying the third row by the third column:
\( (0)(0) + (0)(0) + (-1)(-1) = 1 \).
Since all diagonal elements are 1 and off-diagonals are 0, we have \( A \cdot A^T = I \). Therefore, \( A^{-1} = A^T \).