Examine the consistency of the system of equations: $5x - y + 4z = 5$,$2x + 3y + 5z = 2$,and $5x - 2y + 6z = -1$.

The equations $x + 2y + 3z = 1,$ $2x + y + 3z = 2,$ and $5x + 5y + 9z = 4$ have:

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

Let $A=\left[\begin{array}{lll}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right]$,$B=\left[B_1, B_2, B_3\right]$,where $B_1, B_2, B_3$ are column matrices,and $AB_1=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$,$AB_2=\left[\begin{array}{l}2 \\ 3 \\ 0\end{array}\right]$,$AB_3=\left[\begin{array}{l}3 \\ 2 \\ 1\end{array}\right]$. If $\alpha=|B|$ and $\beta$ is the sum of all the diagonal elements of $B$,then $\alpha^3+\beta^3$ is equal to

If the system of linear equations given by $x+y+z=3$,$2x+2y-z=3$,and $x+y-z=1$ is consistent and if $(x_0, y_0, z_0)$ is a solution,then $2x_0+2y_0+z_0=$

If the solution for the system of equations $x+2y-z=3$,$3x-y+2z=1$ and $2x-2y+3z=2$ is $(\alpha, \beta, \gamma)$,then $\alpha^2+\beta^2+\gamma^2=$