Solve the system of linear equations using the matrix method: $5x + 2y = 4$ and $7x + 3y = 5$.

The system of equations $x + y + z = 2$,$3x - y + 2z = 6$ and $3x + y + z = -18$ has

The system of linear equations $\lambda x + 2y + 2z = 5$,$2\lambda x + 3y + 5z = 8$,and $4x + \lambda y + 6z = 10$ has:

If $\begin{bmatrix} 1 & 1 & 1 \\ 1 & -2 & -2 \\ 1 & 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 3 \\ 4 \end{bmatrix}$,then $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is equal to

Let $\alpha$ be a solution of $x^2+x+1=0$,and for some $a$ and $b$ in $\mathbb{R}$,$\begin{bmatrix} 4 & a & b \end{bmatrix} \begin{bmatrix} 1 & 16 & 13 \\ -1 & -1 & 2 \\ -2 & -14 & -8 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}$. If $\frac{4}{\alpha^4} + \frac{m}{\alpha^a} + \frac{n}{\alpha^b} = 3$,then $m + n$ is equal to

Let the system of linear equations $x+y+\alpha z=2$,$3x+y+z=4$,and $x+2z=1$ have a unique solution $(x^{*}, y^{*}, z^{*})$. If $(\alpha, x^{*}), (y^{*}, \alpha)$ and $(x^{*}, -y^{*})$ are collinear points,then the sum of absolute values of all possible values of $\alpha$ is