Today it’s all about Gaussian elimination method. More specifically, I’ll talk about Gaussian elimination method in 3 × 3 matrices.
Gaussian elimination method in 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 eequations using matrix analysis.
One of these methods in Gaussian elimination method.
Since here I have three equations with three variables, I will use Gaussian elimination method in matrices.
In this method, first of all, I have to pick up the augmented matrix.
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
Now I’ll give you an example.
Example of Gaussian elimination method in 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.
According to Stroud and Booth (2011)* “By the method of Gaussian elimination, solve the equations where
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.
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.
Here I’ll use the backward substitution to solve these equations.
This means I’ll start from the bottom.
Now the equations are
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 my 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!!