For the system $x-y+z=4, 2x+y-3z=0, x+y+z=2$,the values of $x, y, z$ respectively are given by

For the system $S$ of linear equations $x+y+z=3, 2x+2y-z=3, x+y+\lambda z=1$, the incorrect option among the following statements is:

The system of equations $\lambda x - y + (\cos\theta) z = 0$,$3x + y + 2z = 0$,and $(\cos\theta) x + y + 2z = 0$ for $0 < \theta < 2\pi$ has non-trivial solution$(s)$:

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

Let $AX=D$ be a system of three linear non-homogeneous equations. If $|A|=0$ and $\operatorname{rank}(A)=\operatorname{rank}([AD])=\alpha$, then

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