The number of real values $\lambda$,such that the system of linear equations $2x - 3y + 5z = 9$,$x + 3y - z = -18$,and $3x - y + (\lambda^2 - |\lambda|)z = 16$ has no solution,is :-

  • A
    $0$
  • B
    $1$
  • C
    $2$
  • D
    $4$

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Let $\alpha, \beta$ and $\gamma$ be real numbers. Consider the following system of linear equations:
$x+2y+z=7$
$x+\alpha z=11$
$2x-3y+\beta z=\gamma$
Match each entry in List-$I$ to the correct entries in List-$II$:
List-$I$ List-$II$
$(P)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma=28$,then the system has $(1)$ a unique solution
$(Q)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma \neq 28$,then the system has $(2)$ no solution
$(R)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$,then the system has $(3)$ infinitely many solutions
$(S)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma=28$,then the system has $(4)$ $x=11, y=-2$ and $z=0$ as a solution
$(5)$ $x=-15, y=4$ and $z=0$ as a solution

If the system of equations $x+y+z=2$,$2x+4y-z=6$,and $3x+2y+\lambda z=\mu$ has infinitely many solutions,then:

Let $A$ be a $3 \times 3$ real matrix such that $A \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$,$A \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$,and $A \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$. If $X = (x_1, x_2, x_3)^T$ and $I$ is an identity matrix of order $3$,then the system $(A - 2I)X = \begin{bmatrix} 4 \\ 1 \\ 1 \end{bmatrix}$ has:

If the values $x=\alpha, y=\beta, z=\gamma$ satisfy all the $3$ equations $x+2y+3z=4$,$3x+y+z=3$ and $x+3y+3z=2$,then $3\alpha+\gamma=$

Let $M$ be a $3 \times 3$ matrix such that $M \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$, $M \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}$ and $M \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$. If $M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 11 \end{pmatrix}$, then $x + y + z$ equals :

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