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 linear equations $2x - 3y = \gamma + 5$ and $\alpha x + 5y = \beta + 1$,where $\alpha, \beta, \gamma \in R$,has infinitely many solutions,then the value of $|9\alpha + 3\beta + 5\gamma|$ is equal to

Consider the system of equations
$\begin{cases} x+y+z = 0 \\ \alpha x+\beta y+\gamma z = 0 \\ \alpha^{2} x+\beta^{2} y+\gamma^{2} z = 0 \end{cases}$
Then the system of equations has

The system of equations $-k x+3 y-14 z=25$,$-15 x+4 y-k z=3$,and $-4 x+y+3 z=4$ is consistent for all $k$ in the set

In a legislative assembly election,a political group hired a public relations firm to promote its candidate in three ways: telephone,house calls,and letters. The cost per contact (in paise) is given in matrix $A$ as $A = \begin{bmatrix} 40 \\ 100 \\ 50 \end{bmatrix} \begin{matrix} \text{Telephone} \\ \text{Housecall} \\ \text{Letter} \end{matrix}$. The number of contacts of each type made in two cities $X$ and $Y$ is given by $B = \begin{bmatrix} 1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{bmatrix} \begin{matrix} \text{Telephone} & \text{Housecall} & \text{Letter} \\ \to X \\ \to Y \end{matrix}$. Find the total amount spent by the group in the two cities $X$ and $Y$.

The system of linear equations $\lambda x + y + z = 3$,$x - y - 2z = 6$,and $-x + y + z = \mu$ has: