# Solving a System of Linear Equations Using Gaussian Elimination ## Problem 24

Solve the following system of linear equations using Gaussian elimination.
\begin{align*}
x+2y+3z &=4 \\
5x+6y+7z &=8\\
9x+10y+11z &=12
\end{align*} Add to solve later

### 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.

First, the augmented matrix of the system is
$A=\left[ \begin{array}{rrr|r} 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 9 & 10 & 11 & 12 \end{array} \right].$ We apply elementary row operations as follows to reduce the system to row echelon form.

$A \xrightarrow{R_3 -9R_1} \left[\begin{array}{rrr|r} 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 0 & -8 & -16 & -24 \end{array}\right] \xrightarrow{-\frac{1}{8}R_3} \left[\begin{array}{rrr|r} 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 0 & 1 & 2 & 3 \end{array}\right]$ $\xrightarrow{R_2-5R_1} \left[\begin{array}{rrr|r} 1 & 2 & 3 & 4 \\ 0 & -4 & -8 & -12 \\ 0 & 1 & 2 & 3 \end{array}\right] \xrightarrow{-\frac{1}{4} R_2} \left[\begin{array}{rrr|r} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 1 & 2 & 3 \end{array}\right]$ $\xrightarrow{R_3-R_2} \left[\begin{array}{rrr|r} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right]$ The last matrix is in row echelon form.

The corresponding system of linear equations of it is
\begin{align*}
x+2y+3z &=4\\
y+2z&=3 \\
0z&=0
\end{align*}
The last equation $0z=0$ means that $z$ can be any number.
(More systematically, the variables corresponding to leading $1$’s in the echelon form matrix are dependent variables, and the rests are independent (free) variables.)

So let us say that $t$ is a value for $z$, namely $z=t$.
Then from the second equation, we have $y=-2t+3$.
From the first equation, we have
$x=-2y-3z+4=-2(-2t+3)-3t+4=t-2.$ Thus the solution set is
$(x,y,z)=(t-2, -2t+3, t)$ for any $t$.

## Comment.

You may want to check whether the answer is correct by substituting this solution to the original equations.

Also, if you further reduce the matrix into reduced row echelon form, the last system becomes simpler (and simplest in a sense).
This procedure, to reduce a matrix until reduced row echelon form, is called the Gauss-Jordan elimination.

## Related Question.

For a similar problem, you may want to check out Solve a system of linear equations by Gauss-Jordan elimination. Add to solve later

### 1 Response

1. 07/30/2016

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