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Q11E

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Found in: Page 18

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# In Exercises 1 through 12, find all solutions of the equationswith paper and pencil using Gauss–Jordan elimination.Show all your work. $|\begin{array}{c}{x}_{1}+2{x}_{3}+4{x}_{4}=-8\\ {x}_{2}-3x-{x}_{4}=6\\ 3{x}_{1}+4{x}_{2}-6{x}_{3}+8{x}_{4}=0\\ -{x}_{2}+3{x}_{3}+4{x}_{4}=-12\end{array}|$

For the given system of equation we have free variable is ${x}_{3}$and${x}_{1}=-2{x}_{3},{x}_{2}=4+3{x}_{3},{x}_{4}=-2$.

See the step by step solution

## Step 1: 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.

$|\begin{array}{c}{\mathrm{x}}_{1}+2{\mathrm{x}}_{3}+4{\mathrm{x}}_{4}=-8\\ {\mathrm{x}}_{2}-3\mathrm{x}-{\mathrm{x}}_{4}=6\\ 3{\mathrm{x}}_{1}+4{\mathrm{x}}_{2}-6{\mathrm{x}}_{3}+8{\mathrm{x}}_{4}=0\\ -{\mathrm{x}}_{2}+3{\mathrm{x}}_{3}+4{\mathrm{x}}_{4}=-12\end{array}|$

Corresponding Augmented matrix

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

## Step 2: Solving the Augmented matrix

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

Proceeding for solving the augmented matrix

$\left[\begin{array}{cccc}1& 0& 2& 4\\ 0& 1& -3& -1\\ 0& 4& -12& -4\\ 0& -1& 3& 4\end{array}\overline{)\begin{array}{c}-8\\ 6\\ 24\\ -12\end{array}}\right]\phantom{\rule{0ex}{0ex}}{R}_{3}\to \frac{1}{4}{R}_{3}\phantom{\rule{0ex}{0ex}}{R}_{4}\to {R}_{4}+{R}_{2}\phantom{\rule{0ex}{0ex}}{R}_{3}\to {R}_{3}-{R}_{2}\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cccc}1& 0& 2& 4\\ 0& 1& -3& -1\\ 0& 0& 0& 0\\ 0& 0& 0& 3\end{array}\overline{)\begin{array}{c}-8\\ 6\\ 0\\ -6\end{array}}\right]$

${R}_{4}\to \frac{1}{3}{R}_{4}\phantom{\rule{0ex}{0ex}}{R}_{1}\to {R}_{1}-4{R}_{4}\phantom{\rule{0ex}{0ex}}{R}_{2}\to {R}_{2}+{R}_{4}\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cccc}1& 0& 2& 0\\ 0& 1& -3& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 1\end{array}\overline{)\begin{array}{c}0\\ 4\\ 0\\ -2\end{array}}\right]$

Corresponding system of equations

$\left|\begin{array}{c}{x}_{1}+2{x}_{3}=0\\ {x}_{2}-3{x}_{3}=4\\ {x}_{4}=-2\end{array}\right|=\left|\begin{array}{c}{x}_{1}=2{x}_{3}\\ {x}_{2}=4+3{x}_{3}\\ {x}_{4}=-2\end{array}\right|$

Here, we have ${x}_{3}$is free variable.

## Step 3: Values of variables

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${x}_{1}=-2{x}_{3},{x}_{2}=4+3{x}_{3},{x}_{4}=-2$.