# Solve a System of Linear Equations by Gauss-Jordan Elimination

## Problem 27

Solve the following system of linear equations using Gauss-Jordan elimination.

\begin{align*}

6x+8y+6z+3w &=-3 \\

6x-8y+6z-3w &=3\\

8y \,\,\,\,\,\,\,\,\,\,\,- 6w &=6

\end{align*}

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### We use the following notation.

**Elementary row operations.**

The three elementary row operations on a matrix are defined as follows.

(1) Interchanging two rows:

$R_i \leftrightarrow R_j$ interchanges rows $i$ and $j$.

(2) Multiplying a row by a non-zero scalar (a number):

$tR_i$ multiplies row $i$ by the non-zero scalar (number) $t$.

(3) Adding a multiple of one row to another row:

$R_j+tR_i$ adds $t$ times row $i$ to row $j$.

## Solution.

The augmented matrix of the system is

\[A=\left[ \begin{array}{rrrr|r}

6 & 8 & 6 & 3 & -3 \\

6 & -8 & 6 & -3 & 3\\

0 & 8 & 0 & -6 & 6

\end{array} \right].\]
We apply elementary row operations as follows to reduce the system to a matrix in reduced row echelon form.

\[A \xrightarrow{R_2 – R_1}

\left[\begin{array}{rrrr|r}

6 & 8 & 6 & 3 & -3 \\

0 & -16 & 0 & -6 & 6\\

0 & 8 & 0 & -6 & 6

\end{array}\right]
\xrightarrow[R_2+2R_3]{R_1-R_3}

\left[\begin{array}{rrrr|r}

6 & 0 & 6 & 9 & -9 \\

0 & 0 & 0 & -18 & 18\\

0 & 8 & 0 & -6 & 6

\end{array}\right]
\]
\[\xrightarrow[\frac{1}{2}R_3]{\frac{1}{3}R_1, \frac{-1}{18}R_2}

\left[\begin{array}{rrrr|r}

2 & 0 & 2 & 3 & -3 \\

0 & 0 & 0 & 1 & -1\\

0 & 4 & 0 & -3 & 3

\end{array}\right]
\xrightarrow[R_3+3R_2]{R_1-3R_2}

\left[\begin{array}{rrrr|r}

2 & 0 & 2 & 0 & 0 \\

0 & 0 & 0 & 1 & -1\\

0 & 4 & 0 & 0 & 0

\end{array}\right]
\]

\[

\xrightarrow[\frac{1}{4}R_3]{\frac{1}{2}R_1}

\left[\begin{array}{rrrr|r}

1 & 0 & 1 & 0 & 0 \\

0 & 0 & 0 & 1 & -1\\

0 & 1 & 0 & 0 & 0\end{array}\right]
\xrightarrow{R_2 \leftrightarrow R_3}

\left[\begin{array}{rrrr|r}

1 & 0 & 1 & 0 & 0 \\

0 & 1 & 0 & 0 & 0 \\

0 & 0 & 0 & 1 & -1

\end{array}\right]
\]
The last matrix is in reduced row echelon form.

The corresponding system of linear equations is

\begin{align*}

x+z &=0\\

y&=0 \\

w&=-1

\end{align*}

Let $z=t$ be a free variable. Then the solution is $(x,y,z,w)=(-t,0,t,-1)$ for any number $t$.

### Similar Problem.

Another similar problem is Solving a system of linear equations using Gaussian elimination.

The instruction of the problem says to use Gaussian elimination, but try to solve it using Gauss-Jordan elimination as well.

Add to solve later

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