91Ó°ÊÓ

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

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

$\left|\begin{array}{c}3\mathrm{x\_}1+6\mathrm{x\_}2+9\mathrm{x\_}3+5\mathrm{x\_}4+25\mathrm{x\_}5\&=53\\ 7\mathrm{x\_}1+14\mathrm{x\_}2+21\mathrm{x\_}3+9\mathrm{x\_}4+53\mathrm{x\_}5\&=105\\ -4\mathrm{x\_}1-8\mathrm{x\_}2-12\mathrm{x\_}3+5\mathrm{x\_}4-10\mathrm{x\_}5\&=11\end{array}\right|$

Corresponding augmented matrix

$\left[\begin{array}{ccccc}3& 6& 9& 5& 25\\ 7& 14& 21& 9& 53\\ -4& -8& -12& 5& -10\end{array}\overline{)\begin{array}{c}53\\ 105\\ 11\end{array}}\right]$

$$\begin{gathered} \left[\begin{array}{ccccc}3& 6& 9& 5& 25\\ 7& 14& 21& 9& 53\\ -4& -8& -12& 5& -10\end{array}\overline{)\begin{array}{c}53\\ 105\\ 11\end{array}}\right] \\ {R}_1↔R_3 \\ {R}_1→\left(-1\right)R_1-R_3 \\ ~\left[\begin{array}{cccccc}1& 2& 3& -10& -15& -64\\ 7& 14& 21& 9& 53& 105\\ 3& 6& 9& 5& 25& 53\end{array}\right] \end{gathered}$$

$$\begin{gathered} R_2→R_2-7R_1 \\ {R}_3→R_3-3R_1 \\ \left[\begin{array}{ccccc}1& 2& 3& -10& -15\\ 0& 0& 0& 79& 158\\ 0& 0& 0& 35& 70\end{array}\overline{)\begin{array}{c}-64\\ 553\\ 245\end{array}}\right] \\ {R}_2→\frac{1}{79}{R}_2 \\ {R}_1→\frac{1}{35}{R}_3 \\ \left[\begin{array}{ccccc}1& 2& 3& -10& -15\\ 0& 0& 0& 1& 2\\ 0& 0& 0& 1& 2\end{array}\overline{)\begin{array}{c}-64\\ 7\\ 7\end{array}}\right] \\ {R}_1→R_1+10R_2 \\ {R}_3→R_3-R_2 \\ \left[\begin{array}{cccccc}1& 2& 3& 0& 5& 134\\ 0& 0& 0& 1& 2& 7\\ 0& 0& 0& 0& 0& 0\end{array}\right] \end{gathered}$$

Corresponding system of equations will be

$\left|\begin{array}{c}{x}_1+2x_2+3x_3+5x_5=134\\ x_4+x_5=7\end{array}\right|=\left|\begin{array}{c}{x}_1=134-2x_2-3x_3-5x_5\\ x_4=7-x_5\end{array}\right|$

Here,$x_2,x_3,x_5$are free variables.

Let the value of $x_2,x_3,x_5=1$,$⇒x_1=124,x_4=6$

For the given system of equation we have $x_2,x_3,x_5$are free variables and $x_1=134-2x_2-3x_3-5x_5,x_4=7-x_5$.