The number of solutions of the equations $x + 4y - z = 0,$ $3x - 4y - z = 0,$ and $x - 3y + z = 0$ is

The equations $x+y+z=3$, $x+2y+2z=6$ and $x+ay+3z=b$ have

Let $A$ be a $3 \times 3$ real matrix such that $A\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = 2\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$,$A\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} = 4\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$,and $A\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = 2\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$. Then,the system $(A-3I)\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ has

Solve the following system of equations by matrix method: $3x - 2y + 3z = 8$,$2x + y - z = 1$,$4x - 3y + 2z = 4$.

The number of $3 \times 3$ matrices $A$,whose entries are either $1$ or $-1$ and for which the system $A\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}$ has exactly three distinct solutions,is

If the system of linear equations $x - 2y + kz = 1$,$2x + y + z = 2$,and $3x - y - kz = 3$ has a non-zero solution $(x, y, z) \neq 0$,then $(x, y)$ lies on the straight line whose equation is