The number of solutions of the following system of linear homogeneous equations $x-y+z=0$,$x+2y-z=0$,and $2x+y+3z=0$ is

The system of linear equations $x + y + z = 2$,$2x + y - z = 3$,and $3x + 2y + kz = 4$ has a unique solution if

Let $A$ be a $3 \times 3$ real matrix such that $A\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = 2\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$,$A\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} = 4\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$,and $A\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = 2\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$. Then,the system $(A-3I)\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ has

Let $S_1$ and $S_2$ be respectively the sets of all $a \in R - \{0\}$ for which the system of linear equations
$a x + 2 a y - 3 a z = 1$
$(2 a + 1) x + (2 a + 3) y + (a + 1) z = 2$
$(3 a + 5) x + (a + 5) y + (a + 2) z = 3$
has a unique solution and infinitely many solutions,respectively. Then:

If the system of linear equations
$2x + y - z = 3$
$x - y - z = \alpha$
$3x + 3y + \beta z = 3$
has infinitely many solutions,then $\alpha + \beta - \alpha \beta$ is equal to .... .

Solve the system of linear equations using the matrix method:
$5x + 2y = 3$
$3x + 2y = 5$