The solution of the equation $\begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$ is $(x, y, z) = $

Let $A = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 1 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 2 \\ 1 \\ 7 \end{bmatrix}$. For the equation $AX = B$, find the matrix $X$.

If the system of linear equations $(\sin \theta) x - y + z = 0$, $x - (\cos \theta) y + z = 0$, and $x + y + (\sin \theta) z = 0$ has a non-trivial solution, then the least positive value of $\theta$ is

The existence of a unique solution for the system of equations $x+y+z=\beta$,$5x-y+\alpha z=10$,and $2x+3y-z=6$ depends on:

The system of equations $x + y + z = 6$,$x + 2y + 3z = 10$,and $x + 2y + \lambda z = \mu$ has no solution for:

In solving a system of linear equations $AX=B$ by Cramer's rule,in the usual notation,if $\Delta_1=\left|\begin{array}{ccc}-11 & 1 & -7 \\ -4 & 1 & -2 \\ 5 & 1 & 1\end{array}\right|$ and $\Delta_3=\left|\begin{array}{ccc}4 & 1 & -11 \\ 1 & 1 & -4 \\ 4 & 1 & 5\end{array}\right|$,then $X=$