Gaussian Elimination Method In 3 × 3 Matrices

Suppose I have a system of equations like

$\begin{eqnarray*} a_{11}x_1+a_{12}x_2+a_{13}x_3 &=& b_1\\ a_{21}x_1+a_{22}x_2+a_{23}x_3 &=& b_2\\\ a_{31}x_1+a_{32}x_2+a_{33}x_3 &=& b_3. \end{eqnarray*}$

How it would be if I want to write it in a matrix form?

Well, in the matrix form, it will be

$\begin{pmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{pmatrix}\begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix} = \begin{pmatrix} b_1\\ b_2\\ b_3 \end{pmatrix}.$

Here the coefficient matrix is

$\begin{pmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\end{pmatrix},$

the variable matrix is

$\begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}$

and the constant matrix is

$\begin{pmatrix} b_1\\ b_2\\ b_3 \end{pmatrix}.$

Now there are several methods to solve a system of equations using matrix analysis.

One of these methods is the Gaussian elimination method.

Since here I have three equations with three variables, I will use the Gaussian elimination method in 3 × 3 matrices.

If interested, you can also check out the Gaussian elimination method in 4 × 4 matrices.

METHOD

In this method, first of all, I have to pick up the augmented matrix.

The augmented matrix is the combined matrix of both coefficient and constant matrices.

In this case the augmented matrix $\textbf{A}_b$ is

$\textbf{A}_b = \begin{pmatrix} a_{11} & a_{12} & a_{13} & b_1\\ a_{21} & a_{22} & a_{23} & b_2\\ a_{31} & a_{32} & a_{33} & b_3 \end{pmatrix}.$

Now the job is to get an equivalent upper triangular matrix.

That will be similar to

$\begin{pmatrix} a_{11} & a_{12} & a_{13} & b_1\\ 0 & a & b & c\\ 0 & 0 & d & f \end{pmatrix}\text{or}~\begin{pmatrix} a_{21} & a_{22} & a_{23} & b_2\\ 0 & a & b & c\\ 0 & 0 & d & f \end{pmatrix}$

$\text{or}~\begin{pmatrix} a_{31} & a_{32} & a_{33} & b_3\\ 0 & a & b & c\\ 0 & 0 & d & f \end{pmatrix}.$

After that, I’ll use the backward substitution method to get the values of x_1, x_2, x_3 .

Now I’ll give you an example.

EXAMPLE

According to Stroud and Booth (2011)* “By the method of Gaussian elimination, solve the equations $\textbf{AX} = \textbf{b}$ where

$\textbf{A} = \begin{pmatrix} 1 & -2 & -4\\ 2 & 1 & -3\\ 1 & 3 & 2 \end{pmatrix}~\text{and}~\textbf{b} = \begin{pmatrix} -3\\ 4\\ 5 \end{pmatrix}.$

”

SOLUTION

In this example, the set of equations is $\textbf{AX} = \textbf{b}$ .

Also, I know that the coefficient matrix $\textbf{A}$ is

$\textbf{A} = \begin{pmatrix} 1 & -2 & -4\\ 2 & 1 & -3\\ 1 & 3 & 2 \end{pmatrix}.$

In the same way, I also know that the constant matrix $\textbf{b}$ is

$\textbf{b} = \begin{pmatrix} -3\\ 4\\ 5 \end{pmatrix}.$

Now I have to solve this set of equations. This means I have to get the value of the matrix $\textbf{X}$ .

Let me choose $\textbf{X}$ as

$\textbf{X} = \begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}.$

As I have mentioned earlier, the first step is to convert the augmented matrix $\textbf{A}_b$ to an upper triangular matrix.

STEP 1

Now the augmented matrix $\textbf{A}_b$ is

$\textbf{A}_b = \begin{pmatrix} 1 & -2 & -4&-3\\ 2 & 1 & -3&4\\ 1 & 3 & 2&5 \end{pmatrix}.$

First of all, I’ll subtract twice row 1 from row 2.

Simultaneously, I’ll also subtract row 1 from row 3.

In mathematical term, I’ll write it like this:

Row 2 – 2(Row 1), Row 3 – Row 1.

Thus the equivalent matrix will be

$\textbf{A}_b \thicksim \begin{pmatrix} 1 & -2 & -4&-3\\ 0 & 5 & 5&10\\ 0& 5 & 6&8 \end{pmatrix}.$

Next, I’ll divide row 2 by 5.

$\cfrac{1}{5}$ Row 2 gives

$\textbf{A}_b \thicksim \begin{pmatrix} 1 & -2 & -4&-3\\ 0 & 1 & 1&2\\ 0& 5 & 6&8 \end{pmatrix}.$

Then I’ll subtract 5 times row 2 from row 3.

Thus Row 3 – 5 (Row 2) gives

$\textbf{A}_b \thicksim \begin{pmatrix} 1 & -2 & -4&-3\\ 0 & 1 & 1&2\\ 0& 0 & 1&-2 \end{pmatrix}.$

This is the upper triangular form of the matrix.

Therefore the system of equations in the matrix form is

$\begin{pmatrix} 1 & -2 & -4\\ 0 & 1 & 1\\ 0& 0 & 1 \end{pmatrix}\begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix} = \begin{pmatrix} -3\\ 2\\ -2 \end{pmatrix}.$