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Found in: Page 19

### Linear Algebra With Applications

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

# Solve the linear system of equations. You may use technology.$|\begin{array}{c}3X+6y+14z=22\\ 7X+14y+30z=46\\ 4X+8y+7z=6\end{array}|$

For the given system of equation there is unique solution. $Y=-6,y=2,z=2$.

See the step by step solution

## Step 1: Augmented matrix

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.

$\left|\begin{array}{c}3X+6y+14Z=22\\ 7X+14y+30z=46\\ 4X+8y+7Z=6\end{array}\right|$

Corresponding augmented matrix

$\left[\begin{array}{ccc}3& 6& 14\\ 7& 14& 30\\ 4& 8& 7\end{array}\overline{)\begin{array}{c}22\\ 46\\ 6\end{array}}\right]$

## Step 2: Solving the Augmented matrix

$\left[\begin{array}{ccc}3& 6& 14\\ 7& 14& 30\\ 4& 8& 7\end{array}\overline{)\begin{array}{c}22\\ 46\\ 6\end{array}}\right]\phantom{\rule{0ex}{0ex}}{R}_{1}↔{R}_{3}\phantom{\rule{0ex}{0ex}}{R}_{1}\to {R}_{1}-{R}_{3}\phantom{\rule{0ex}{0ex}}\left[\begin{array}{ccc}1& 2& -7\\ 7& 14& 30\\ 3& 6& 14\end{array}\overline{)\begin{array}{c}-16\\ 46\\ 22\end{array}}\right]$

${R}_{2}\to {R}_{2}-7{R}_{1}\phantom{\rule{0ex}{0ex}}{R}_{3}\to {R}_{3}-3{R}_{1}\phantom{\rule{0ex}{0ex}}\left[\begin{array}{ccc}1& 2& -7\\ 0& 0& 79\\ 0& 0& 35\end{array}\overline{)\begin{array}{c}-16\\ 2\\ 2\end{array}}\right]\phantom{\rule{0ex}{0ex}}{R}_{2}\to \frac{1}{79}{R}_{2}\phantom{\rule{0ex}{0ex}}{R}_{3}\to \frac{1}{35}-{R}_{3}\phantom{\rule{0ex}{0ex}}\left[\begin{array}{ccc}1& 2& -7\\ 0& 0& 1\\ 0& 0& 1\end{array}\overline{)\begin{array}{c}-16\\ 2\\ 2\end{array}}\right]$

## Step 3:Value of variables

Corresponding system of equations will be

$\left|\begin{array}{c}X=-2y+7Z-16\\ y=2\\ Z=2\end{array}\right|$

Hence, the solution of equation will be$x=-6,y=2z=2$ .

For the given system of equation there is unique solution $x=-6,y=2z=2$.