If the system of linear equations $2x + 2ay + az = 0$,$2x + 3by + bz = 0$,and $2x + 4cy + cz = 0$,where $a, b, c \in R$ are non-zero and distinct,has a non-zero solution,then:

Consider the following system of equations in matrix form $\begin{bmatrix} 1 \\ 2 \\ \lambda \end{bmatrix} \begin{bmatrix} 1 & 2 & \lambda \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 0$. Then which one of the following statements is true?

Consider the simultaneous linear equations $\beta x + \alpha y - z = -1$, $3x - \beta y + \alpha z = 0$, and $\alpha x + \beta y + z = 1$. In the usual notation used in Cramer's rule, given that $\frac{\Delta_1}{\Delta} = -1$, $\frac{\Delta_2}{\Delta} = 1$, and $\frac{\Delta_3}{\Delta} = 2$, then $(\alpha, \beta) = $

If $\begin{bmatrix} 1 & 1 & 1 \\ 1 & -2 & -2 \\ 1 & 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 3 \\ 4 \end{bmatrix}$,then $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is equal to

The values of $x, y, z$ in order for the system of equations $3x + y + 2z = 3,$ $2x - 3y - z = -3,$ and $x + 2y + z = 4$ are:

If for $AX = B,$ $B = \begin{bmatrix} 9 \\ 52 \\ 0 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 3 & -\frac{1}{2} & -\frac{1}{2} \\ -4 & \frac{3}{4} & \frac{5}{4} \\ 2 & -\frac{1}{4} & -\frac{3}{4} \end{bmatrix},$ then $X$ is equal to