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

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

If ${a_1}x + {b_1}y + {c_1}z = 0, {a_2}x + {b_2}y + {c_2}z = 0, {a_3}x + {b_3}y + {c_3}z = 0$ and $\left| \begin{matrix} {a_1} & {b_1} & {c_1} \\ {a_2} & {b_2} & {c_2} \\ {a_3} & {b_3} & {c_3} \end{matrix} \right| = 0$,then the given system has

The solution of the linear system of equations $\begin{bmatrix} 2 & 2 & 3 \\ 7 & 1 & 1 \\ 0 & 6 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 y + 11 \\ 6 z - 1 \\ 5 y + 11 \end{bmatrix} + \begin{bmatrix} x \\ x \\ 4 z \end{bmatrix} + \begin{bmatrix} z \\ 3 x \\ 4 y \end{bmatrix}$ is

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

If $\begin{bmatrix} \alpha & \beta & \gamma \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & -5 \\ 1 & 2 & 5 \end{bmatrix} = \begin{bmatrix} 3 & 5 & 2 \end{bmatrix}$,then $\alpha^3 + \beta^3 + \gamma^3 = $