Step 1: Find $(A + B)$. \[ A + B = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} (2+1) & (4+3) \\ (3+(-2)) & (2+5) \end{bmatrix} = \begin{bmatrix} 3 & 7 \\ 1 & 7 \end{bmatrix} \] Step 2: Find $(A - B)$. \[ A - B = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} (2-1) & (4-3) \\ (3-(-2)) & (2-5) \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 5 & -3 \end{bmatrix} \]
Final Answer: \[ A + B = \begin{bmatrix} 3 & 7 \\ 1 & 7 \end{bmatrix}, A - B = \begin{bmatrix} 1 & 1 \\ 5 & -3 \end{bmatrix} \]
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 \( A^{-1} \).