91影视

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

$\left|\begin{array}{c}{\mathrm{x}}_4+2{\mathrm{x}}_5-{\mathrm{x}}_6=2\\ {\mathrm{x}}_1+2{\mathrm{x}}_2+{\mathrm{x}}_5-{\mathrm{x}}_6=0\\ {\mathrm{x}}_1+2{\mathrm{x}}_2+2{\mathrm{x}}_{3-}{\mathrm{x}}_5-{\mathrm{x}}_6=2\end{array}\right|$

Corresponding Augmented matrix

$\left[\begin{array}{cccccc}0& 0& 0& 1& 2& 1\\ 1& 2& 0& 0& 1& -1\\ 1& 2& 2& 0& -1& 1\end{array}\overline{)\begin{array}{c}2\\ 0\\ 2\end{array}}\right]$

$\left[\begin{array}{cccccc}0& 0& 0& 1& 2& 1\\ 1& 2& 0& 0& 1& -1\\ 1& 2& 2& 0& -1& 1\end{array}\overline{)\begin{array}{c}2\\ 0\\ 2\end{array}}\right]$

We have three linear equation which means there are three free variables.

$\left[\begin{array}{cccccc}0& 0& 0& 1& 2& 1\\ 1& 2& 0& 0& 1& -1\\ 1& 2& 2& 0& -1& 1\end{array}\overline{)\begin{array}{c}2\\ 0\\ 2\end{array}}\right]$

$R_1鈫�R_3$

Proceeding for solving the augmented matrix

$$\begin{gathered} \left[\begin{array}{cccccc}0& 0& 0& 1& 2& 1\\ 1& 2& 0& 0& 1& -1\\ 1& 2& 2& 0& -1& 1\end{array}\overline{)\begin{array}{c}2\\ 0\\ 2\end{array}}\right] \\ {R}_2鈫�R_2-R_1 \\ {R}_2鈫�\frac{-1}{2}{R}_2 \\ \left[\begin{array}{cccccc}0& 2& 2& 0& -1& 1\\ 0& 0& 1& 0& -1& 1\\ 0& 0& 0& 1& 2& -1\end{array}\overline{)\begin{array}{c}2\\ 1\\ 2\end{array}}\right] \end{gathered}$$

System of equation to the corresponding above matrix will be

$\left|\begin{array}{c}{x}_1+2x_2+2x_3-x_5+x_6=2\\ x_3-x_5+x_6=1\\ x_4-x_5+x_6=2\end{array}\right|=\left|\begin{array}{c}{x}_1=2-2x_2-2x_3+x_5-x_6\\ x_3=1+x_5-x_6\\ x_4=2--x_5+x_6\end{array}\right|$

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

We have free variables ${\mathrm{x}}_2,{\mathrm{x}}_5,{\mathrm{x}}_6.$.

Let the value of $鈬�x_5=1,x_4=1,x_1=-4$.

For the given system of equation ${\mathrm{x}}_2,{\mathrm{x}}_{6,}{\mathrm{x}}_5$are free variables and ${\mathrm{x}}_3=1-{\mathrm{x}}_6+{\mathrm{x}}_5,{\mathrm{x}}_4=2-2{\mathrm{x}}_5+{\mathrm{x}}_6,{\mathrm{x}}_1=-2{\mathrm{x}}_2+{\mathrm{x}}_{6,}-{\mathrm{x}}_5$.