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

If $S$ is the set of distinct values of $b$ for which the following system of linear equations $x + y + z = 1$,$x + ay + z = 1$,and $ax + by + z = 0$ has no solution,then $S$ is:

Let the system of linear equations $x+y+kz=2$; $2x+3y-z=1$; $3x+4y+2z=k$ have infinitely many solutions. Then the system $(k+1)x+(2k-1)y=7$; $(2k+1)x+(k+5)y=10$ has:

Let the system of equations $x+5y-z=1$,$4x+3y-3z=7$,$24x+y+\lambda z=\mu$,where $\lambda, \mu \in R$,have infinitely many solutions. Then the number of solutions of this system,if $x, y, z$ are integers and satisfy $7 \leq x+y+z \leq 77$,is

For how many different values of $a$ does the following system of linear equations have at least two distinct solutions?
$ax + y = 0$
$x + (a + 10)y = 0$

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