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

$\left|\begin{array}{c}4x_1+3x_2-2x_3-x_4=4\\ 5x_1+4x_2-3x_3-x_4=4\\ -2x_1-2x_2-x_3-2x_4=-3\\ 11x_1+6x_2+4x_3-x_4=11\end{array}\right|$

Corresponding Augmented matrix

$\left[\begin{array}{cccc}4& 3& 2& -1\\ 5& 4& 3& -1\\ -2& -2& -1& 2\\ 11& 6& 4& 1\end{array}\overline{)\begin{array}{c}4\\ 4\\ -3\\ 11\end{array}}\right]$

$$\begin{gathered} \left[\begin{array}{cccc}4& 3& 2& -1\\ 5& 4& 3& -1\\ -2& -2& -1& 2\\ 11& 6& 4& 1\end{array}\overline{)\begin{array}{c}4\\ 4\\ -3\\ 11\end{array}}\right] \\ \left[\begin{array}{cccc}4& 3& 2& -1\\ 5& 4& 3& -1\\ -2& -2& -1& 2\\ 11& 6& 4& 1\end{array}\overline{)\begin{array}{c}4\\ 4\\ -3\\ 11\end{array}}\right] \\ {R}_1↔R_2 \end{gathered}$$

Proceeding for solving the augmented matrix

role="math" localid="1659355005979" $$\begin{gathered} \left[\begin{array}{cccc}4& 3& 2& -1\\ 5& 4& 3& -1\\ -2& -2& -1& 2\\ 11& 6& 4& 1\end{array}\overline{)\begin{array}{c}4\\ 4\\ -3\\ 11\end{array}}\right] \\ {R}_2→R_2-4R_1 \\ {R}_3→R_3+2R_1 \\ {R}_4→R_4+11R_1 \\ \left[\begin{array}{cccc}4& 3& 2& -1\\ 5& 4& 3& -1\\ -2& -2& -1& 2\\ 11& 6& 4& 1\end{array}\overline{)\begin{array}{c}4\\ 4\\ -3\\ 11\end{array}}\right] \\ {R}_2→\left(-1\right)R_1 \\ {R}_1→R_1+R_2 \\ {R}_4→R_4+5R_2 \\ \left[\begin{array}{cccc}1& 0& -1& -1\\ 0& 1& 2& 1\\ 0& 0& 1& 0\\ 0& 0& 3& 6\end{array}\overline{)\begin{array}{c}4\\ -4\\ -3\\ 9\end{array}}\right] \\ {R}_1→R_1-R_3 \\ {R}_2→R_2+2R_3 \\ {R}_4→R_4+3R_3 \\ \left[\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& 0& -3\\ 0& 0& 1& 2\\ 0& 0& 0& 0\end{array}\overline{)\begin{array}{c}1\\ 2\\ -3\\ 0\end{array}}\right] \\ \mathrm{Corresponding}\mathrm{system}\mathrm{of}\mathrm{equations} \\ \left|\begin{array}{c}{\mathrm{x}}_1+{\mathrm{x}}_4=1\\ {\mathrm{x}}_2-3{\mathrm{x}}_4=2\\ {\mathrm{x}}_3+2{\mathrm{x}}_4=-3\end{array}\right|=\left|\begin{array}{c}{\mathrm{x}}_1=1-{\mathrm{x}}_4\\ {\mathrm{x}}_2=2+3{\mathrm{x}}_4\\ {\mathrm{x}}_3=-3-2{\mathrm{x}}_4\end{array}\right| \\ \mathrm{Here},\mathrm{we}\mathrm{have}{\mathrm{x}}_4\mathrm{is}\mathrm{free}\mathrm{variable}. \end{gathered}$$

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

We have free variable $x_4$.

Let the value of$x_4=1$

$⇒x_3=-5,x_2=5,x_1=0$

For the given system of equation$x_4$ is free variable and ${\mathrm{x}}_1=1-{\mathrm{x}}_{4,}{\mathrm{x}}_2=2+3{\mathrm{x}}_4,{\mathrm{x}}_3=-3-2{\mathrm{x}}_4$.