Step 1: Write the given matrices.
\[ A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, B = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \]
Step 2: Compute $BA$. \\ \[ BA = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \] Performing multiplication: \[ BA = \begin{bmatrix} (1 \cdot 0 + 0 \cdot 1) & (1 \cdot 1 + 0 \cdot 0) \\ (0 \cdot 0 + (-1) \cdot 1) & (0 \cdot 1 + (-1) \cdot 0) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \]
Step 3: Compare with given options.
This matches option (B).
Step 4: Conclusion.
The correct answer is (B) $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$.
Find the values of \( x, y, z \) if the matrix \( A \) satisfies the equation \( A^T A = I \), where
\[ A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} \]
If \[ A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \] prove that \[ A^n = \begin{bmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{bmatrix}, \] where \( n \in \mathbb{N} \).
If matrix \[ A = \begin{bmatrix} 1 & 1 & 3 \\ 1 & 3 & -3 \\ -2 & -4 & -4 \end{bmatrix}, \] then find \( A^{-1} \).