Let $X = \begin{bmatrix} a \\ b \\ c \end{bmatrix}$,$A = \begin{bmatrix} 1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1 \end{bmatrix}$,and $B = \begin{bmatrix} 3 \\ 1 \\ 4 \end{bmatrix}$. If $AX = B$,then the value of $2a - 3b + 4c$ is:

The sum of three numbers is $6$. If we multiply the third number by $3$ and add the second number to it,we get $11$. By adding the first and third numbers,we get double the second number. Represent this algebraically and find the numbers using the matrix method.

If ${a_1}x + {b_1}y + {c_1}z = 0, {a_2}x + {b_2}y + {c_2}z = 0, {a_3}x + {b_3}y + {c_3}z = 0$ and $\left| \begin{matrix} {a_1} & {b_1} & {c_1} \\ {a_2} & {b_2} & {c_2} \\ {a_3} & {b_3} & {c_3} \end{matrix} \right| = 0$,then the given system has

The number of non-trivial solutions of the system $x-y+z=0, x+2y-z=0, 2x+y+3z=0$ is

If $AX=B$,where $A=\begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 4 \\ 1 & 3 & 4 \end{bmatrix}$,$X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ and $B=\begin{bmatrix} 12 \\ 15 \\ 13 \end{bmatrix}$,then $x^{2}+y^{2}+z^{2}=$

If the system of linear equations $2x + 3y - z = -2$; $x + y + z = 4$; $x - y + |\lambda|z = 4\lambda - 4$ (where $\lambda \in R$) has no solution,then: