## Deploy Apps Close to Your Users with Section

This Engineering Education (EngEd) Program is supported by Section.

In just one minute, you can deploy apps as close as possible to your users. Section is affordable, simple and powerful.

# System of Linear Equations using Gaussian Elimination Algorithm

##### July 15, 2022

A system of linear equations is a collection of one or more linear equations involving the same set of variables. Linear systems are encountered when building regression models in machine learning.

There are various techniques for solving this problem. Some methods are complex, while others are easy to understand and implement. The Gaussian Elimination method is one of the best solutions to these systems.

This article will look at the intuition behind the Gaussian Elimination method, carry out a handy computation, and finally illustrate how we can implement this method in R.

### Prerequisites

The reader is required to have:

1. A basic understanding of Elementary Linear Algebra.
2. R programming skills.
3. R installed on their computer.

### Understanding the Gaussian Elimination Algorithm

We need to these steps to solve a system of linear equations using the Gaussian Elimination algorithm.

Suppose we are given a system of linear equations as shown below. ### Step 1:

Represent the above system of linear equations in a matrix form, i.e., Assign A, X and b to the coefficient matrix, variables vector, and a vector of solutions, respectively.

That is:   ### Step 2:

Using matrices A and b, we create an augmented matrix, i.e., attach b to matrix A as the last column. Now, to reduce the above matrix \${C}\$ to a form that is simple to solve for unknowns, we need to perform some operations. These operations should not change the solution of the linear system.

Some of the allowed operations are:

1. Change the order of the rows.
2. Scale up a row, i.e., multiplying with a constant.
3. To eliminate particular values, you can multiply one row with a constant and add the output to another row.

So using these operations, we will learn how to reduce a system of linear equations using the Gaussian Elimination Algorithm.

### Step 3:

In this step, we keep the first element of the first row. This value is called a pivot. Now, using our \$3^{rd}\$ operation outlined above, we make all the values below the pivot value zeros. This is as shown in the matrix below. ### Step 4:

We keep the first value after the zero in the second row in the second iteration. Then, as we did in the first iteration, make all values below this value zeros.

This is as shown below: ### Step 5:

Repeat the operations above until we obtain an upper triangular matrix. The matrix we got in the previous step is already in upper triangular form.

The next step is to find the solution to our original system using this reduced matrix. From our reduced matrix, we can write down the following system of linear equations. This new system of equations is much easier to solve than the original system. To find the solution to our original system, we will solve these equations we just derived from the upper triangular matrix. This is very easy and fast compared to the original system’s computing solution.

Now, on the system above, all we need to do is to perform a back substitution. The back substitution is performed in the order outlined below: Note, we first solved for the last variable \$(x_3)\$ and then incorporated its solution in solving for the preceding variable until we got to \$x_1\$. As we all know, systems of linear equations in real-life data can consist of millions of equations. It is impractical to solve those systems manually. This requires us to make use of computation software. In the last section of this article, we shall see how we can implement this method.

### R implementation of Gaussian Elimination Algorithm

Here we will need to create a matrix, the one we used to explain this concept with, which we will then write a code to reduce it into an upper triangular matrix. Below is an implementation process of this method.

``````# create a matrix
A <- matrix(c(-3,2,-1,6,-6,7,3,-4,4),byrow = T,nrow=3,ncol=3)
A # print a matrix
b <- matrix(c(-1,-7,-6),nrow=3,ncol=1)
b # print matrix b
# dimension of matrix A
nrow <- nrow(A)
nrow
# concatinante matrix A and vector b
Ugmt.mtx <- cbind(A,b)
Ugmt.mtx

Ugmt.mtx[1,] <- Ugmt.mtx[1,]/Ugmt.mtx[1,1]

for (i in 2:nrow){ # loop over rows
for (j in i:nrow) { # loop over columns
Ugmt.mtx[j, ] <- Ugmt.mtx[j, ] - Ugmt.mtx[i-1, ] * Ugmt.mtx[j, i-1] # replace the row values at jth position with left hand computions
}
Ugmt.mtx[i,] <- Ugmt.mtx[i,]/Ugmt.mtx[i,i]
}
# print ouput
Ugmt.mtx
``````

Executing the code we obtain:

``````     [,1]       [,2]       [,3]       [,4]
[1,]    1 -0.6666667  0.3333333  0.3333333
[2,]    0  1.0000000 -2.5000000  4.5000000
[3,]    0  0.0000000  1.0000000 -1.0000000
``````

Note, to find the values of our variables; we need to perform a back substitution using this output. However, to simplify things further, we can take the above matrix and make elements on the upper triangular as zeros. This will ensure we do not need to perform the back substitution on the final output, which might be computationally expensive. Rather than creating an identity matrix with respect to the output variables.

This method of reducing a matrix is called the Gauss-Jordan Elimination Method. To further understand how this method works, I recommend visiting this blog.

This method is implemented in R as follows:

``````A <- matrix(c(-3,2,-1,6,-6,7,3,-4,4),byrow = T,nrow=3,ncol=3)
A
b <- matrix(c(-1,-7,-6),nrow=3,ncol=1)
b
# dimension of matrix A
nrow <- nrow(A)
nrow
# concatinante matrix A and vector b
Ugmt.mtx <- cbind(A,b)
Ugmt.mtx

Ugmt.mtx[1,] <- Ugmt.mtx[1,]/Ugmt.mtx[1,1]

for (i in 2:nrow){
for (j in i:nrow) {
Ugmt.mtx[j, ] <- Ugmt.mtx[j, ] - Ugmt.mtx[i-1, ] * Ugmt.mtx[j, i-1]
}
Ugmt.mtx[i,] <- Ugmt.mtx[i,]/Ugmt.mtx[i,i]
}
for (i in p:2){
for (j in i:2-1) {
Ugmt.mtx[j, ] <- Ugmt.mtx[j, ] - Ugmt.mtx[i, ] * Ugmt.mtx[j, i]
}
}
Ugmt.mtx
``````

This code returns.

``````     [,1] [,2] [,3] [,4]
[1,]    1    0    0    2
[2,]    0    1    0    2
[3,]    0    0    1   -1
``````

As we can see, the returned output contains the exact values for the variables we are solving.

### Conclusion

This article introduced the concept of solving systems of linear equations using the Gaussian elimination method. Using the reduced matrix, we determined the solution for our variables using the concept of back substitution. Finally, we implemented this process in R.

Since the final output still required us to solve for unknowns, we went a step further using the back substitution and demonstrated an upgraded version of our previous approach, the Jordan-Gaussian Elimination method. We demonstrated how to implement this method where it returned the exact values of the unknown variables.

##### Written by: Stacy Jelagat

Peer review by: Paul Odhiambo