Question

In Exercises 1 through 12, find all solutions of the equations

with paper and pencil using Gauss–Jordan elimination.

Show all your work.

$\left|\begin{array}{c}2x_1-3x_3+7x_5+7x_6=0\\ -2x_1+x_2+6x_5-12x_6=0\\ x_2-3x_3+x_5+5x_6=0\\ -2x_2+x_4+x_5+x_6=0\\ 2x_1+x_2-3x_3+8x_5+7x_6=0\end{array}\right|$

Short answer

For the given system of equation is free variable $x_5$and$x_1=-7x_5,x_2=-x_3=0,x_4=-3x_5,x_6=0$.

Step-by-step solution

Augmented matrix

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

$\left|\begin{array}{c}2{\mathrm{x}}_1-3{\mathrm{x}}_3+7{\mathrm{x}}_5+7{\mathrm{x}}_6=0\\ -2{\mathrm{x}}_1+{\mathrm{x}}_2+6{\mathrm{x}}_5-12{\mathrm{x}}_6=0\\ {\mathrm{x}}_2-3{\mathrm{x}}_3+{\mathrm{x}}_5+5{\mathrm{x}}_6=0\\ -2{\mathrm{x}}_2+{\mathrm{x}}_4+{\mathrm{x}}_5+{\mathrm{x}}_6=0\\ 2{\mathrm{x}}_1+{\mathrm{x}}_2-3{\mathrm{x}}_3+8{\mathrm{x}}_5+7{\mathrm{x}}_6=0\end{array}\right|$

Corresponding Augmented matrix

$\left[\begin{array}{cccccc}2& 0& -3& 0& 7& 7\\ -2& 1& 6& 0& -6& -12\\ 0& 1& -3& 0& 1& 5\\ 0& -2& 0& 1& 1& 1\\ 2& 1& -3& 0& 8& 7\end{array}\overline{)\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\end{array}}\right]$

Solving the Augmented matrix

$$\begin{gathered} \left[\begin{array}{cccccc}2& 0& -3& 0& 7& 7\\ -2& 1& 6& 0& -6& -12\\ 0& 1& -3& 0& 1& 5\\ 0& -2& 0& 1& 1& 1\\ 2& 1& -3& 0& 8& 7\end{array}\overline{)\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\end{array}}\right] \\ {R}_2\to R_2+R_1 \\ {R}_5\to R_5-R_1 \end{gathered}$$

Proceeding for solving the augmented matrix

$$\begin{gathered} \left[\begin{array}{cccccc}2& 0& -3& 0& 7& 7\\ 0& 1& 3& 0& 1& -5\\ 0& 1& -3& 0& 1& 5\\ 0& -2& 0& 1& 1& 1\\ 0& 1& 0& 0& 1& 0\end{array}\overline{)\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\end{array}}\right] \\ {R}_3\to R_3-R_2 \\ {R}_4\to R4+R2 \\ {R}_5\to R_5-R_2 \\ \left[\begin{array}{cccccc}2& 0& -3& 0& 7& 7\\ 0& 1& 3& 0& 1& -5\\ 0& 0& -6& 0& 0& 10\\ 0& 0& 6& 1& 3& -9\\ 2& 0& -3& 0& 0& 5\end{array}\overline{)\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\end{array}}\right] \end{gathered}$$

$$\begin{gathered} R_4\to R_4+R_3 \\ {R}_5\to -2R_5+R3 \\ \left[\begin{array}{cccccc}2& 0& -3& 0& 7& 7\\ 0& 1& 3& 0& 1& -5\\ 0& 0& -6& 0& 0& 10\\ 0& 0& 0& 1& 3& 1\\ 0& 0& 0& 1& 3& -19\end{array}\overline{)\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\end{array}}\right] \end{gathered}$$

$$\begin{gathered} R_5\to R_5-R_4 \\ \left[\begin{array}{cccccc}2& 0& -3& 0& 7& 7\\ 0& 1& 3& 0& 1& -5\\ 0& 0& -6& 0& 0& 10\\ 0& 0& 0& 1& 3& 1\\ 0& 0& 0& 0& 0& -20\end{array}\overline{)\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\end{array}}\right] \end{gathered}$$

Corresponding system of equations

$\left|\begin{array}{c}{x}_1=-7x_5\\ x_2=-x_5\\ x_3=0\\ x_4=-3x_5\\ x_6=0\end{array}\right|$

Here, we have $x_5$is free variable.

Values of variables 

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

We have free variable$x_5$.

Let the value of $x_5=1$

$\Rightarrow x_1=-7,x_2=-1,x_3=0,x_4=-3,x_6=0$

Final answer

For the given system of equation is free variable $x_5$and$x_1=-7x_5,x_2=-1,x_3=0x_4=-3,x_6=0$.