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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}_bto 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 

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

Now my next job is to solve this system.

STEP 2

Here I’ll use the backward substitution to solve these equations.

This means I’ll start from the bottom.

Now the equations are

(1) \begin{equation*} (1). x_1 + (-2). x_2 + (-4). x_3 = (-3), \end{equation*}

(2) \begin{equation*} (1). x_2 + (1). x_3 = 2, \end{equation*}

(3) \begin{equation*} (1). x_3 = -2. \end{equation*}

Therefore from equation (3), I can say that

\[x_3 = -2.\]

Next, I’ll substitute x_3 = -2 in equation (2).

Thus it will be

\begin{equation*} (1). x_2 + (1). (-2) = 2. \end{equation*}

Now I’ll simplify it to get the value of x_2.

\begin{eqnarray*} (1). x_2 + (1). (-2) &=& 2\\ x_2 -2 &=& 2\\ x_2 &=& 4. \end{eqnarray*}

At the end, I’ll substitute x_2 = 4 and x_3 = -2 in equation (1) to get

\begin{eqnarray*} (1). x_1 + (-2). x_2 + (-4). x_3 &=& (-3)\\ x_1 + (-2). (4) + (-4). (-2) &=& -3\\ x_1 - 8 +8 &=& -3\\ x_1 &=& -3. \end{eqnarray*}

Hence I can conclude that the variable matrix \textbf{X} is [-3, 4, -2]^T.

This is the solution to this example.