The product of all real values of $b$ such that there is no solution to the system of equations $2x + 5y + z = 19$,$-4x + by + 6z = -42$,and $-3y - bz = 81$ is:

If the system of equations $(\lambda-1) x+(\lambda-4) y+\lambda z=5$,$\lambda x+(\lambda-1) y+(\lambda-4) z=7$,and $(\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9$ has infinitely many solutions,then $\lambda^2+\lambda$ is equal to

For a real number $\alpha$,if the system of linear equations $\begin{bmatrix} 1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}$ has infinitely many solutions,then $1+\alpha+\alpha^2=$

The system of equations $x+y+z=6$,$x+2y+5z=9$,$x+5y+\lambda z=\mu$ has no solution if

Let $AX=D$ be a system of three linear non-homogeneous equations. If $|A|=0$ and $\operatorname{rank}(A)=\operatorname{rank}([AD])=\alpha$, then

If $(x, y, z)=(\alpha, \beta, \gamma)$ is the unique solution of the system of simultaneous linear equations $3x - 4y + z + 7 = 0$, $2x + 3y - z = 10$, and $x - 2y - 3z = 3$, then $\alpha = $