Consider the system of equations: $x + ay = 0$,$y + az = 0$,and $z + ax = 0$. The set of all real values of $a$ for which the system has a unique solution is:

Let $A=\begin{bmatrix} 1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & -6 \end{bmatrix}$ and $B=[b_{ij}]_{3 \times 3}$ with $b_{11}=2, b_{13}=-2, b_{12}=0$ such that $AB=\begin{bmatrix} 2 & 14 & -4 \\ 4 & 1 & -8 \\ -6 & 15 & 12 \end{bmatrix}$. Then $|B|+\operatorname{trace}(B)=$

The system of equations $3x + 2y + z = 6$, $3x + 4y + 3z = 14$ and $6x + 10y + 8z = a$ has an infinite number of solutions if $a$ is equal to

The solution of equations $x + y = 10$,$2x + y = 18$,and $4x - 3y = 26$ is:

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

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