Gaussian elimination method in 3 × 3 matrices. Today it’s all about the Gaussian elimination method in 3 × 3 matrices. Have a look!!

Want to read more on the other methods of solving a system of equations using matrices?? Please check these out…

**Gaussian elimination method in 4×4 matrices**

**Solve a system of equations using the inverse matrix method****The row transformation method to solve a system of equations****The triangular decomposition method in 3×3 matrices**

**Gaussian elimination method in 3 × 3 matrices**

Suppose I have a system of equations like

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

Well, in the matrix form, it will be

Here the coefficient matrix is

the variable matrix is

and the constant matrix is

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.

**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 is

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

That will be similar to

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

Now I’ll give you an example.

**Example of the Gaussian elimination method in 3 × 3 matrices**

Disclaimer: This is not my own example. I have chosen it from some book. I have also given the due reference at the end of the post.

So here is the example.

**Example 1**

According to Stroud and Booth (2011)* “By the method of Gaussian elimination, solve the equations where

”

**Solution**

In this example, the set of equations is .

Also, I know that the coefficient matrix is

In the same way, I also know that the constant matrix is

Now I have to solve this set of equations. This means I have to get the value of the matrix .

Let me choose as

As I have mentioned earlier, the first step is to convert the augmented matrix to an upper triangular matrix.

**Step 1**

Now the augmented matrix is

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

Next, I’ll divide row 2 by 5.

Row 2 gives

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

Thus Row 3 – 5 (Row 2) gives

This is the upper triangular form of the matrix.

Therefore the system of equations in the matrix form is

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)

(2)

(3)

Therefore from equation (3), I can say that

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

Thus it will be

Now I’ll simplify it to get the value of

At the end, I’ll substitute and in equation (1) to get

Hence I can conclude that the variable matrix is

This is the solution to this example.

Dear friends, this is the end of today’s post. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!

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Good new method METHOD IN 3 × 3 MATRICES