Let $A = \begin{bmatrix} 1 & 3 \\ 4 & -3 \end{bmatrix}$. Let $S = \left\{ \begin{bmatrix} x \\ y \end{bmatrix} \in \mathbb{R}^2 \mid A \begin{bmatrix} x \\ y \end{bmatrix} = 3 \begin{bmatrix} x \\ y \end{bmatrix} \right\}$. What is the cardinality of $S$?

Statement $-1$: The system of linear equations
$x + (\sin \alpha)y + (\cos \alpha)z = 0$
$x + (\cos \alpha)y + (\sin \alpha)z = 0$
$x - (\sin \alpha)y - (\cos \alpha)z = 0$
has a non-trivial solution for only one value of $\alpha$ lying in the interval $(0, \frac{\pi}{2})$.
Statement $-2$: The equation in $\alpha$
$\left| \begin{matrix} \cos \alpha & \sin \alpha & \cos \alpha \\ \sin \alpha & \cos \alpha & \sin \alpha \\ \cos \alpha & -\sin \alpha & -\cos \alpha \end{matrix} \right| = 0$
has only one solution lying in the interval $(0, \frac{\pi}{2})$.

Solve the system of linear equations using the matrix method: $2x + 3y + 3z = 5$,$x - 2y + z = -4$,$3x - y - 2z = 3$.

Let $x = \alpha, y = \beta, z = \gamma$ be the unique solution of the system of simultaneous linear equations $2x + 3y - 2z + 4 = 0$, $3x - 4y + 3z + 5 = 0$, and $kx - 2y + z + 3 = 0$. If $\alpha = -2$, then $k =$

Given $A = \begin{bmatrix} 1 & 3 \\ 2 & 2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$. If $A - \lambda I$ is a singular matrix,then:

If the system of linear equations,$x+y+z = 6$,$x+2y+3z = 10$,and $3x+2y+\lambda z = \mu$ has more than two solutions,then $\mu-\lambda^{2}$ is equal to