Examine the consistency of the system of equations: $2x - y = 5$ and $x + y = 4$.

For a matrix $A=\begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$, if $U_{1}, U_{2}$, and $U_{3}$ are $3 \times 1$ column matrices satisfying $A U_{1}=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$, $A U_{2}=\begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}$, $A U_{3}=\begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}$, and $U$ is a $3 \times 3$ matrix whose columns are $U_{1}, U_{2}$, and $U_{3}$, then the sum of the elements of $U^{-1}$ is:

Let $S$ be the set of all $\lambda \in \mathbb{R}$ for which the system of linear equations
$2x - y + 2z = 2$
$x - 2y + \lambda z = -4$
$x + \lambda y + z = 4$
has no solution. Then the set $S$

The system of equations $kx + 2y - z = 1$,$(k - 1)y - 2z = 2$,and $(k + 2)z = 3$ has a unique solution if $k$ is equal to:

If the system of linear equations $8x + y + 4z = -2$,$x + y + z = 0$,and $\lambda x - 3y = \mu$ has infinitely many solutions,then the distance of the point $\left(\lambda, \mu, -\frac{1}{2}\right)$ from the plane $8x + y + 4z + 2 = 0$ is:

Consider the simultaneous linear equations $\beta x + \alpha y - z = -1$, $3x - \beta y + \alpha z = 0$, and $\alpha x + \beta y + z = 1$. In the usual notation used in Cramer's rule, given that $\frac{\Delta_1}{\Delta} = -1$, $\frac{\Delta_2}{\Delta} = 1$, and $\frac{\Delta_3}{\Delta} = 2$, then $(\alpha, \beta) = $