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### 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.$|\begin{array}{c}3x_1+6x_2+9x_3+5x_4+25x_5&=53\\ 7x_1+14x_2+21x_3+9x_4+53x_5&=105\\ -4x_1-8x_2-12x_3+5x_4-10x_5&=11\end{array}|$

For the given system of equation we have ${x}_{2},{x}_{3},{x}_{5}$ are free variables and ${x}_{1}=134-2{x}_{2}-3{x}_{3}-5{x}_{5},{x}_{4}=7-{x}_{5}$ .

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.

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

Corresponding augmented matrix

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

## Step 2: Solving the Augmented matrix

$\left[\begin{array}{ccccc}3& 6& 9& 5& 25\\ 7& 14& 21& 9& 53\\ -4& -8& -12& 5& -10\end{array}\overline{)\begin{array}{c}53\\ 105\\ 11\end{array}}\right]\phantom{\rule{0ex}{0ex}}{R}_{1}↔{R}_{3}\phantom{\rule{0ex}{0ex}}{R}_{1}\to \left(-1\right){R}_{1}-{R}_{3}\phantom{\rule{0ex}{0ex}}~\left[\begin{array}{cccccc}1& 2& 3& -10& -15& -64\\ 7& 14& 21& 9& 53& 105\\ 3& 6& 9& 5& 25& 53\end{array}\right]$

${R}_{2}\to {R}_{2}-7{R}_{1}\phantom{\rule{0ex}{0ex}}{R}_{3}\to {R}_{3}-3{R}_{1}\phantom{\rule{0ex}{0ex}}\left[\begin{array}{ccccc}1& 2& 3& -10& -15\\ 0& 0& 0& 79& 158\\ 0& 0& 0& 35& 70\end{array}\overline{)\begin{array}{c}-64\\ 553\\ 245\end{array}}\right]\phantom{\rule{0ex}{0ex}}{R}_{2}\to \frac{1}{79}{R}_{2}\phantom{\rule{0ex}{0ex}}{R}_{1}\to \frac{1}{35}{R}_{3}\phantom{\rule{0ex}{0ex}}\left[\begin{array}{ccccc}1& 2& 3& -10& -15\\ 0& 0& 0& 1& 2\\ 0& 0& 0& 1& 2\end{array}\overline{)\begin{array}{c}-64\\ 7\\ 7\end{array}}\right]\phantom{\rule{0ex}{0ex}}{R}_{1}\to {R}_{1}+10{R}_{2}\phantom{\rule{0ex}{0ex}}{R}_{3}\to {R}_{3}-{R}_{2}\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cccccc}1& 2& 3& 0& 5& 134\\ 0& 0& 0& 1& 2& 7\\ 0& 0& 0& 0& 0& 0\end{array}\right]$

## Step 3: Value of variables

Corresponding system of equations will be

$\left|\begin{array}{c}{x}_{1}+2{x}_{2}+3{x}_{3}+5{x}_{5}=134\\ {x}_{4}+{x}_{5}=7\end{array}\right|=\left|\begin{array}{c}{x}_{1}=134-2{x}_{2}-3{x}_{3}-5{x}_{5}\\ {x}_{4}=7-{x}_{5}\end{array}\right|$

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 ${x}_{1}=134-2{x}_{2}-3{x}_{3}-5{x}_{5},{x}_{4}=7-{x}_{5}$.