How to solve this linear equations using gauss jordan method? How to solve the linear equations in Gauss elimination method
$$2x+3y+z=1\\
x+y+z=3\\
3x+4y+2z=4$$
 A: The system has no unique solution, because it's linearly dependent ($III = I+II$), this allows you to drop one equation (say $III$) and find a basis for the solution space, by putting the system into the form
$$\begin{align*}
x + a z & = b \\
y + c z & = d
\end{align*}$$
The solutions will then be of the form $(b - at, d-ct, t)$ where $t\in\mathbb R$ can be chosen.
Hint
Look at the equations $I - 2\cdot II$ and $I - 3 \cdot II$.

Pursuing the hint you will arrive at
$$\begin{align*}
x + 2z &= 8\\
y - z &= -5
\end{align*}$$
Thus for any $t\in\mathbb R$ the vector $(8-2t, -5+t,t)$ will solve your system. Chosing $t=4$ you get one possible of the infinitely many solutions, namely the one you provided,
$$(x,y,z) = (8-2\cdot 4, -5+4,4) = (0,-1,4)$$
A: \begin{align*}
\left(\begin{array}{ccc|c}
1&1&1&3\\
2&3&1&1\\
3&4&2&4
\end{array}\right)\\\\
\left(\begin{array}{ccc|c}
1&1&1&3\\
0&1&-1&-5\\
0&1&-1&-5
\end{array}\right)\\\\
\begin{array}{ccc}
x&y&z
\end{array}\qquad\qquad\\
\left(\begin{array}{ccc|c}
1&1&1&3\\
0&1&-1&-5\\
0&0&0&0
\end{array}\right)
\end{align*}
Z is your free variable .continue from here .
