The system of equations $kx + 2y - z = 1$,$(k - 1)y - 2z = 2$,and $(k + 2)z = 3$ has a unique solution if $k$ 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

If the solution of the system of simultaneous equations $\frac{1}{x}+\frac{2}{y}-\frac{3}{z}-1=0$,$\frac{2}{x}-\frac{4}{y}+\frac{3}{z}-1=0$ and $\frac{3}{x}+\frac{6}{y}-\frac{6}{z}-4=0$ is $x=\alpha, y=\beta, z=\gamma$,then $\alpha^2+\gamma^2=$

For the system of linear equations:
$x - 2y = 1, x - y + kz = -2, ky + 4z = 6, k \in R$
Consider the following statements:
$(A)$ The system has a unique solution if $k \neq 2, k \neq -2$.
$(B)$ The system has a unique solution if $k = -2$.
$(C)$ The system has a unique solution if $k = 2$.
$(D)$ The system has no solution if $k = 2$.
$(E)$ The system has an infinite number of solutions if $k \neq -2$.
Which of the following statements are correct?

The system of equations $\begin{cases} \alpha x + y + z = \alpha - 1 \\ x + \alpha y + z = \alpha - 1 \\ x + y + \alpha z = \alpha - 1 \end{cases}$ has no solution,if $\alpha$ is

If $A$ is a matrix such that $\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] A \left[\begin{array}{ll} 1 & 1 \end{array}\right] = \left[\begin{array}{ll} 1 & 1 \\ 0 & 0 \end{array}\right]$,then $A$ is equal to