91Ó°ÊÓ

First of all we will make the augmented matrix to the corresponding system of equations.

Solving system of equations using gauss Jordan’s elimination method.

$\left|\begin{array}{c}{\mathrm{x}}_1+2{\mathrm{x}}_3+4{\mathrm{x}}_4=-8\\ {\mathrm{x}}_2-3\mathrm{x}-{\mathrm{x}}_4=6\\ 3{\mathrm{x}}_1+4{\mathrm{x}}_2-6{\mathrm{x}}_3+8{\mathrm{x}}_4=0\\ -{\mathrm{x}}_2+3{\mathrm{x}}_3+4{\mathrm{x}}_4=-12\end{array}\right|$

Corresponding Augmented matrix

$\left[\begin{array}{cccc}1& 0& 2& 4\\ 0& 1& -3& -1\\ 3& 4& -6& 8\\ 0& -1& 3& 4\end{array}\overline{)\begin{array}{c}-8\\ 6\\ 0\\ -12\end{array}}\right]$

$$\begin{gathered} \left[\begin{array}{cccc}1& 0& 2& 4\\ 0& 1& -3& -1\\ 3& 4& -6& 8\\ 0& -1& 3& 4\end{array}\overline{)\begin{array}{c}-8\\ 6\\ 0\\ -12\end{array}}\right] \\ {R}_3→R_3-3R_1 \end{gathered}$$

Proceeding for solving the augmented matrix

$$\begin{gathered} \left[\begin{array}{cccc}1& 0& 2& 4\\ 0& 1& -3& -1\\ 0& 4& -12& -4\\ 0& -1& 3& 4\end{array}\overline{)\begin{array}{c}-8\\ 6\\ 24\\ -12\end{array}}\right] \\ {R}_3→\frac{1}{4}{R}_3 \\ {R}_4→R_4+R_2 \\ {R}_3→R_3-R_2 \\ \left[\begin{array}{cccc}1& 0& 2& 4\\ 0& 1& -3& -1\\ 0& 0& 0& 0\\ 0& 0& 0& 3\end{array}\overline{)\begin{array}{c}-8\\ 6\\ 0\\ -6\end{array}}\right] \end{gathered}$$

$$\begin{gathered} R_4→\frac{1}{3}{R}_4 \\ {R}_1→R_1-4R_4 \\ {R}_2→R_2+R_4 \\ \left[\begin{array}{cccc}1& 0& 2& 0\\ 0& 1& -3& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 1\end{array}\overline{)\begin{array}{c}0\\ 4\\ 0\\ -2\end{array}}\right] \end{gathered}$$

Corresponding system of equations

$\left|\begin{array}{c}{x}_1+2x_3=0\\ x_2-3x_3=4\\ x_4=-2\end{array}\right|=\left|\begin{array}{c}{x}_1=2x_3\\ x_2=4+3x_3\\ x_4=-2\end{array}\right|$

Here, we have $x_3$is free variable.

Since for free variables we can let any value for that.

We have free variable$x_3$.

Let the value of$x_3=1$

$⇒x_1=-2,x_2=7,x_4=-2$

For the given system of equation is free variable $x_3$and$x_1=-2x_3,x_2=4+3x_3,x_4=-2$.