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

The solution of equations $x + y = 10$,$2x + y = 18$,and $4x - 3y = 26$ is:

Let $\lambda$ be a real number for which the system of linear equations $x + y + z = 6$,$4x + \lambda y - \lambda z = \lambda - 2$,and $3x + 2y - 4z = -5$ has infinitely many solutions. Then $\lambda$ is a root of the quadratic equation:

Let $A = \begin{bmatrix} 1 & 3 \\ 4 & -3 \end{bmatrix}$. Let $S = \left\{ \begin{bmatrix} x \\ y \end{bmatrix} \in \mathbb{R}^2 \mid A \begin{bmatrix} x \\ y \end{bmatrix} = 3 \begin{bmatrix} x \\ y \end{bmatrix} \right\}$. What is the cardinality of $S$?

The solution of the equation $\left[\begin{array}{rrr}1 & 0 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]$ is $(x, y, z)=$

If $\begin{bmatrix} 3 & 1 \\ 4 & 1 \end{bmatrix} X = \begin{bmatrix} 5 & -1 \\ 2 & 3 \end{bmatrix}$,then $X =$