The equations $x-y+2z=4$,$3x+y+4z=6$,and $x+y+z=1$ have

Use the product $\left[\begin{array}{lll}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\left[\begin{array}{lll}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]$ to solve the system of equations:
$x-y+2z=1$
$2y-3z=1$
$3x-2y+4z=2$

For a system of simultaneous linear equations,if $A X=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]$,$\operatorname{Adj} A=\left[\begin{array}{ccc}1 & -1 & -1 \\ 1 & 1 & -1 \\ 1 & 1 & 1\end{array}\right]$ and $\operatorname{det} A>0$,then $X=$

The linear system of equations $\begin{cases} 8x - 3y - 5z = 0 \\ 5x - 8y + 3z = 0 \\ 3x + 5y - 8z = 0 \end{cases}$ has

If the system of equations $kx + 2y - z = 2, (k - 1)x + ky + z = 1, x + (k - 1)y + kz = 3$ has only one solution,then the number of possible real value$(s)$ of $k$ is -

Let $a, b, c, d, e$ be five numbers satisfying the system of equations:
$2a + b + c + d + e = 6$
$a + 2b + c + d + e = 12$
$a + b + 2c + d + e = 24$
$a + b + c + 2d + e = 48$
$a + b + c + d + 2e = 96$
Then $|c|$ is equal to: