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 the system of equations,$a^2 x - ay = 1 - a$ and $bx + (3 - 2b) y = 3 + a$ possess a unique solution $x = 1, y = 1$,then:

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

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 $X=$

The system $2x + 3y + z = 5$,$3x + y + 5z = 7$ and $x + 4y - 2z = 3$ has

If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$,then the system of three equations $\begin{aligned} & 2x + 3|y| + 5[z] = 0, \\ & x + |y| - 2[z] = 4, \\ & x + |y| + [z] = 1 \end{aligned}$ has