Question

If \[ A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \] then

• A. \(A\) is not invertible
• B. \(A = A^{-1}\)
• C. \(A^{-1} = 2A\)
• D. \(A^{-1} = I\)

Hint:

If a matrix satisfies \(A^2 = I\), then the matrix is called involutory and is its own inverse.

Answer 1

The Correct Option is B

Step 1: Compute \(A^2\).
Multiply the matrix \(A\) by itself: \[ A^2 = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \] Step 2: Perform matrix multiplication.
\[ A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I \] Step 3: Use the inverse definition.
If \(A^2 = I\), then \[ A^{-1} = A \] Step 4: Final conclusion.
Hence, the correct statement is \[ A = A^{-1} \]