Let $A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$ such that $AX = B$,then $X =$

If the system of equations $\begin{bmatrix} \alpha & -1 & -1 \\ 1 & -\alpha & -1 \\ 1 & -1 & -\alpha \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} \alpha-1 \\ \alpha-1 \\ \alpha-1 \end{bmatrix}$ is inconsistent, then $\alpha=$

For $\alpha, \beta \in [0, 2\pi]$ and $\gamma \in [0, \pi)$, consider the system of equations:
$2 \sin \alpha - \cos \beta + 3 \tan \gamma = 3$
$4 \sin \alpha + 2 \cos \beta - 2 \tan \gamma = 2$
$6 \sin \alpha - 3 \cos \beta + \tan \gamma = 9$
Then, which one of the following is true?

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 $1+\alpha+\alpha^2=$

Let $A$ be a $3 \times 3$ real matrix such that $A \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$,$A \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$,and $A \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$. If $X = (x_1, x_2, x_3)^T$ and $I$ is an identity matrix of order $3$,then the system $(A - 2I)X = \begin{bmatrix} 4 \\ 1 \\ 1 \end{bmatrix}$ has:

In matrix notation, if the system of equations $\begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 5 \\ -5 \\ 10 \end{bmatrix}$ has an infinite number of solutions, then all these solutions lie on