Perform the matrix multiplication:
\[ \begin{bmatrix} 1 & 3 \\ 4 & 5 \end{bmatrix} \cdot \begin{bmatrix} x \\ 2 \end{bmatrix} = \begin{bmatrix} 1 \cdot x + 3 \cdot 2 \\ 4 \cdot x + 5 \cdot 2 \end{bmatrix} = \begin{bmatrix} x + 6 \\ 4x + 10 \end{bmatrix}. \]
Equating this with \(\begin{bmatrix} 5 \\ 6 \end{bmatrix}\), solve:
\[ x + 6 = 5 \implies x = -1, \]
\[ 4x + 10 = 6 \implies 4x = -4 \implies x = -1. \]
Both equations confirm \(x = -1\).
Let $ A = \begin{bmatrix} \alpha & -1 \\6 & \beta \end{bmatrix},\ \alpha > 0 $, such that $ \det(A) = 0 $ and $ \alpha + \beta = 1 $. If $ I $ denotes the $ 2 \times 2 $ identity matrix, then the matrix $ (1 + A)^5 $ is:
Let $A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}$. If for some $\theta \in (0, \pi)$, $A^2 = A^T$, then the sum of the diagonal elements of the matrix $(A + I)^3 + (A - I)^3 - 6A$ is equal to