Gaussian elimination method in 4 × 4 matrices. Hello friends, today it’s all about the Gaussian elimination method in 4 × 4 matrices. Have a look!!

Gaussian elimination method in 4 × 4 matrices_compressed

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

First of all, I’ll give a brief description of this method.

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

So 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 four equations with four variables, I will use the Gaussian elimination method in 4 × 4 matrices.

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

**Example of the Gaussian elimination method in 4 × 4 matrices**

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

So here is the example.

**Example **

According to Stroud and Booth (2011)* “Solve the following set of equations by Gaussian elimination:

**Solution**

Here the given set of equations is

Now I can rewrite it in a matrix form as

Thus the coefficient matrix is

Also, the constant matrix is

Therefore, the augmented matrix will be

As my earlier post on Gaussian elimination method in 3 × 3 matrices, here also I’ll start with the augmented matrix.

First of all, I’ll reduce the augmented matrix to an upper triangular matrix.

Now for that, I’ll get three zero in the first column of the augmented matrix.

**Step 1 **

So the augmented matrix is

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

Simultaneously, I’ll also subtract 4 times row 1 from row 3.

Also, I’ll subtract row 1 from row 4.

Thus in the mathematical term, I’ll write it like this:

Row 2 – 3(Row 1), Row 3 – 4(Row 1), Row 4 – Row 1.

Therefore the equivalent matrix will be

Now I have already got three zeros in the first column.

Next, I’ll try to get two zeros in the second column. As I can see, I already have zeros in the first column.

**Step 2**

Therefore, as a next step, I’ll subtract twice row 2 from row 3.

Also, I’ll subtract th times row 3 from row 4.

Thus in the mathematical term, it will be Row 2 – 2(Row 3), Row 4 – (Row 3).

Therefore the equivalent matrix will be

Now I’ll divide row 3 with .

Therefore the equivalent matrix will be

Next I’ll interchange between row 2 and 3.

Therefore the equivalent matrix will be

Now I have got two zeros in the second column where I already had zeros in the first column.

My next task is to get one zero in the third column where the elements at both the first and second columns are zero.

**Step 3**

Therefore, as a next step, I’ll subtract th times row 4 from row 3.

So in the mathematical term, it will be Row 3 – Row 4.

Therefore the equivalent matrix will be

Next I’ll interchange between row 3 and 4.

Therefore the equivalent matrix will be

Thus finally I have got an upper triangular matrix.

Therefore the system of equations in the matrix form is

Now my next job is to solve this system.

**Step 4**

So here I’ll use the backward substitution to solve the system of equations.

Now the equations are

(1)

(2)

(3)

(4)

Since I’ll use the backward substitution method, I’ll start with the equation (4).

So I’ll simplify equation (4) to get

Next, I’ll substitute in equation (3) to get

Now I’ll simplify it to get the value of . Thus it will be

So this means

Then I’ll put and in equation (2) to get

Now I’ll simplify it to get the value of

Thus it will be

So this gives

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

Thus it will be

If I simplify it, I get

So this means

Hence I can conclude that and is the solution of the given set of equations.

And this is the answer to this given example.

Dear friends, this is the end of today’s post on the Gaussian elimination method in 4 × 4 matrices. 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!!

Aqsa Reza says

The post is very helpful .The explanation is simple and to the point.

Dr. Aspriha Peters says

Dear Aqsa,

Thank you very much. This means a lot to me.

Faizan says

i want to solve these 4 equation simultaneously( Elimination Method) can u do a solution for me?