Direct Methods Gaussian Elimination

Direct Method of Gaussian Elimination is a numerical method of solving a system of linear equations AX = B. A represents the coefficient matrix of order m × n, X is the column matrix of order n × 1, which represents the unknowns of the linear equations. B is a column vector of order m × 1, obtained by multiplication of A and X. Hence, AX = B represents the m linear equation of n unknowns.

There are two numerical methods of solving these systems of linear equations, one direct method another is the iterative method.

Direct methods are more concise without the error of approximation obtained in a finite number of steps. However, iterative methods start with an approximate solution and then generate a sequence of solutions that modify the previous one to get an approximate answer.

Here we will discuss the direct method of Gaussian Elimination to find the solution of a system of linear equations with appropriate examples.

The Gaussian Elimination Method

The Gaussian elimination method is one of the efficient direct methods used to solve a given system of linear equations. It is attributed to the German mathematician Carl Fedrick Gauss. The idea of this method is based on the elimination of one unknown among the given simultaneous equations.

Let the given system of linear equations be

a₁₁x₁ + a₁₂x₂ + … + a_1nx_n = b₁

a₂₁x₁ + a₂₂x₂ + … + a_2nx_n = b₂

….

….

a_m1x₁ + a_m2x₂ + … + a_mnx_n = b_m

If we write all these equations in matrix form then

Then as per the Gaussian elimination method, the augmented matrix [A : B] is reduced by elementary row operations to a matrix U such that

where A’ is an upper triangular matrix.

Now, we get the system A’X = B’. By backward substitution of the unknowns in A’X = B’, we get the required solution of the given system of linear equations AX = B.

Steps of Gaussian Elimination Method

• By the given system of linear equations in n variables form the matrices A, X and B.
• Now reduce the augmented matrix [A : B] by elementary row operations to get [A’ : B’].
• We get A’ as an upper triangular matrix.
• By the backward substitution in A’X = B’, we get the solution of the given system of linear equations.

Also Read:

• Cramer’s Rule
• Consistent and Inconsistent System
• Elementary Transformations
• Inverse of Matrix

Solved Examples on Gaussian Elimination Method

Example 1:

Solve the system of linear equations:

2x + 3y – z = 5

4x + 4y – 3z = 3

–2x + 3y – z = 1

by Gaussian elimination method.

Solution:

The given system of equations in matrix form is AX = B where

Now, applying elementary row operations on

[A : B] =

Applying R₂ → R₂ + (-2) R₁ and R₃ → R₃ + R₁

Applying R₃ → R₃ + 3R₂

Thus, we get

Now, we get the system of linear equations as

2x + 3y – z = 5 ………(1)

–2y – z = –7 ………(2)

–5z = –15 ………(3)

By backward substitution, we have x = 1, y = 2, z = 3, which is the solution of the given system of equations.

Example 2:

Solve the system of linear equations:

x + y + z = 6

x – y + z = 2

2x – y + 3z = 9

by Gaussian elimination method.

Solution:

The given system of equations in matrix form is AX = B where

Now, applying elementary row operations on

[A : B] =

Applying R₂ → R₂ + (-1) R₁ and R₃ → R₃ + (-2) R₁

Applying R₂ → (–½)R₂ and R₃ → R₃ + 3 R₂

Thus, we get

Now, we get the system of linear equations as

x + y + z = 6 ………(1)

y = 2 ………(2)

z = 3 ………(3)

Thus, we have x = 1, y = 2, z = 3, which is the solution of the given system of equations.