If the system of linear equations $2x + y - z = 7$,$x - 3y + 2z = 1$,and $x + 4y + \delta z = k$,where $\delta, k \in R$,has infinitely many solutions,then $\delta + k$ is equal to

If $x^a y^b=e^m, x^c y^d=e^n, \Delta_1=\left|\begin{array}{ll}m & b \\ n & d\end{array}\right|, \Delta_2=\left|\begin{array}{ll}a & m \\ c & n\end{array}\right|, \Delta_3=\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|$,then the values of $x$ and $y$ are respectively ($e$ is the base of natural logarithm).

Let $A$ be a $3 \times 3$ real matrix such that $A\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = 2\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$,$A\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} = 4\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$,and $A\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = 2\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$. Then,the system $(A-3I)\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ has

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 $S$

Let $A = \begin{bmatrix} 2 & -1 & 3 \\ 1 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix}$ and $D = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$. The system $AX = D$ has

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