For what values of $a$ will the system of equations $x+y+z=1$, $2x+3y+2z=2$, and $ax+ay+2az=4$ have a unique solution?

Solve the system of linear equations using the matrix method: $2x + y + z = 1$,$x - 2y - z = \frac{3}{2}$,and $3y - 5z = 9$.

If the system of equations $2x + 3y - 3z = 3$,$x + 2y + \alpha z = 1$,and $2x - y + z = \beta$ has infinitely many solutions,then $\frac{\alpha}{\beta} - \frac{\beta}{\alpha} =$

Let $M = (a_{ij})$,$i, j \in \{1, 2, 3\}$,be a $3 \times 3$ matrix such that $a_{ij} = 1$ if $j+1$ is divisible by $i$,otherwise $a_{ij} = 0$. Then which of the following statements is (are) true?
$(A)$ $M$ is invertible
$(B)$ There exists a nonzero column matrix $\begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}$ such that $M \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} = \begin{bmatrix} -a_1 \\ -a_2 \\ -a_3 \end{bmatrix}$
$(C)$ The set $\{X \in \mathbb{R}^3 : MX = 0, X \neq 0\}$ is non-empty,where $0 = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$
$(D)$ The matrix $(M - 2I)$ is invertible,where $I$ is the $3 \times 3$ identity matrix

For a real number $\alpha$,if the system of linear equations $\begin{bmatrix} 1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}$ has infinitely many solutions,then $1+\alpha+\alpha^2=$

If the following system of linear equations
$2x + y + z = 5$
$x - y + z = 3$
$x + y + az = b$
has no solution,then :