Gaussian Elimination Method: Step-by-Step Guide with Examples

The Gaussian Elimination Method is a systematic procedure used to solve systems of linear equations by transforming the system's matrix into a row echelon form through elementary row operations. This method simplifies finding solutions, identifying inconsistencies, and determining if a system has a unique, infinite, or no solution.

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1. Introduction to the Gaussian Elimination Method:

The Gaussian Elimination Method is one of the most important techniques in linear algebra used to solve systems of linear equations. Developed by the German mathematician Carl Friedrich Gauss, this method systematically transforms a matrix into a simpler form to find the solutions of a system of equations. This method is widely used in various fields like engineering, physics, and computer science, making it an essential tool for anyone involved in mathematical computations.

2. What is Gaussian Elimination Method:

The Gaussian Elimination Method is a process for solving systems of linear equations by converting a system into a row-echelon form using elementary row operations. The key idea is to simplify the system of equations step by step, reducing it to a form where the solution can be easily obtained through back-substitution.

In this method, the matrix of coefficients of the system is transformed into an upper triangular matrix (or row echelon form), where all elements below the main diagonal are zeroes. Once this is achieved, you can easily solve for the variables by starting with the last equation and substituting back into the previous equations.

3. How to use the Gaussian Elimination Method:

The process of solving a system of equations using Gaussian Elimination involves the following steps:

1. Form the Augmented Matrix: Convert the system of linear equations into an augmented matrix, where the coefficients and constants are represented in matrix form.

2. Apply Elementary Row Operations: Use the following operations to transform the augmented matrix into row echelon form:

• Swap rows: Interchanging two rows of the matrix.

• Multiply a row by a non-zero scalar: This operation maintains the equality of the equation.

• Add or subtract a multiple of one row to/from another row: Used to create zeros below the pivot elements.

3. Transform to Row Echelon Form: Continue performing row operations until the matrix is in upper triangular form (or row echelon form).

4. Back Substitution: Once in row echelon form, solve for the variables starting with the last row (which should have one variable) and substitute upwards.

4. Rules for Gaussian Elimination Method:

Certain rules need to be followed to ensure that the Gaussian Elimination method works correctly:

1. No Dividing by Zero: Avoid dividing by zero during row operations. If a pivot element is zero, swap rows to place a non-zero element as the pivot.

2. Row Operations Should Maintain the System’s Integrity: The three elementary row operations (row swapping, multiplying by a scalar, adding/subtracting multiples of rows) must be used to maintain the system's equality.

3. Simplify to Row Echelon Form: The ultimate goal is to simplify the augmented matrix into row echelon form, where the pivot elements (leading non-zero in each row) are 1, and all elements below the pivot are zeroes.

5. Properties of Gaussian Elimination Method:

The Gaussian Elimination method has several important properties that make it useful:

1. Direct Solution Method: It provides an exact solution to systems of linear equations, provided the system has a unique solution.

2. Works for Consistent and Inconsistent Systems: Gaussian Elimination can identify whether a system of equations is consistent (has solutions) or inconsistent (no solution).

3. Can Handle Infinite Solutions: If a system has infinitely many solutions, the method can help express the solution parametrically.

4. Transforms to Upper Triangular Matrix: The method always transforms the matrix into an upper triangular matrix or row echelon form, making it easy to apply back substitution.

5. Not Limited to Square Matrices: It can be applied to both square and non-square matrices.

6. Gaussian Elimination Method Solved Examples:

Question: 1
Solving a System of ${2}$ Equations with ${2}$ Variables
Solve the system: $\space \begin{aligned} x 2y = 3 \\ 3x 4y = 7 \end{aligned}$

Solution:

Step 1: Set up the augmented matrix $\space \begin{bmatrix} 1 & 2 & 3 \\ 3 & 4 & 7 \end{bmatrix}$

Step 2: Eliminate the ${x}$-term from the second row

• Subtract ${3}$ times the first row from the second row to eliminate the ${x}$-term: $\quad{R_2 = R_2 - 3R_1 \space \rArr} \begin{bmatrix} 1 & 2 & 3 \\ 0 & -2 & -2 \end{bmatrix}$

Step 3: Solve for ${y}$

• Divide the second row by ${-2}$ to simplify: ${\space R_2 = \dfrac{R_2}{-2} \space \rArr \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \end{bmatrix}}$

• Now, the second row tells us: $\space{y=1}$

Step 4: Substitute ${y=1}$ into the first row to solve for ${x}$ $\quad{x 2(1)=3 \space \rArr \space x 2=3 \space \rArr \space x=1}$

The solution is: $\space{x=1, \space y=1}$

Question: 2
Solving a System of ${2}$ Equations with ${2}$ Variables
Solve the system: $\space \begin{aligned} 2x 3y = 8 \\ x - y = 1 \end{aligned}$

Solution: Step 1: Set up the augmented matrix $\space \begin{bmatrix} 2 & 3 & 8 \\ 1 & -1 & 1 \end{bmatrix}$

Step 2: Eliminate the ${x}$-term from the second row

• Subtract ${\frac{1}{2}}$ times the first row from the second row: $\space{R_2 = R_2 -\dfrac{1}{2} R_1 \space \rArr} \begin{bmatrix} 2 & 3 & 8 \\ 0 & -\frac{5}{2} & -3 \end{bmatrix}$

Step 3: Solve for ${y}$

• Multiply the second row by ${-\frac{2}{5}}$ to simplify: ${\space R_2 = \dfrac{R_2}{-\frac{5}{2}} \space \rArr \begin{bmatrix} 2 & 3 & 8 \\ 0 & 1 & \frac{6}{5} \end{bmatrix}}$

• Now, the second row tells us: $\space{y=\dfrac{6}{5}}$

Step 4: Substitute ${y=\frac{6}{5}}$ into the first row to solve for ${x}$ ${\space 2x 3 \left( \dfrac{6}{5} \right) = 8 \space \rArr \space 2x \dfrac{18}{5} = 8}$

Multiply by ${5}$: $\space{10x 18=40 \space \rArr \space 10x=22 \space \rArr \space x=\dfrac{11}{5}}$

The solution is: $\space{x=\dfrac{11}{5}, \space y=\dfrac{6}{5}}$

Question: 3
Solving a Consistent System of ${3}$ Variables
Solve the system: $\space\begin{aligned} x y z = 9 \\ 2x 3y z = 16 \\ x 2y 3z = 14 \end{aligned}$

Solution:

Step 1: Set up the augmented matrix $\space \begin{bmatrix} 1 & 1 & 1 & 9 \\ 2 & 3 & 1 & 16 \\ 1 & 2 & 3 & 14 \end{bmatrix}$

Step 2: Eliminate the ${x}$ -terms from the second and third rows

• Subtract ${2}$ times the first row from the second row: ${\space R_2 = R_2 - 2R_1 \space \rArr \begin{bmatrix} 1 & 1 & 1 & 9 \\ 0 & 1 & -1 & -2 \\ 1 & 2 & 3 & 14 \end{bmatrix}}$

• Subtract the first row from the third row: ${\space R_3 = R_3 - R_1 \space \rArr \begin{bmatrix} 1 & 1 & 1 & 9 \\ 0 & 1 & -1 & -2 \\ 0 & 1 & 2 & 5 \end{bmatrix}}$

Step 3: Eliminate the ${y}$-term from the third row

• Subtract the second row from the third row: ${\space R_3 = R_3 - R_2 \space \rArr \begin{bmatrix} 1 & 1 & 1 & 9 \\ 0 & 1 & -1 & -2 \\ 0 & 0 & 3 & 7 \end{bmatrix}}$

Step 4: Solve for ${z,y,}$ and ${x}$

• From the third row: ${3z=7 \space \rArr \space z=\frac{7}{3}}$

• Substitute ${z=\frac{7}{3}}$ into the second row: ${\space y-\dfrac{7}{3}=-2 \space \rArr \space y=\dfrac{-6 7}{3}=\dfrac{1}{3}}$

• Substitute ${y=\frac{1}{3}}$ and ${z=\frac{7}{3}}$ into the first row: $\space x \dfrac{1}{3} \dfrac{7}{3} = 9 \space \rArr \space x \dfrac{8}{3} = 9 \space \rArr \space x = \dfrac{27 - 8}{3} = \dfrac{19}{3}$

The solution is: $\space{x=\dfrac{19}{3}, \space y=\dfrac{1}{3}, \space z=\dfrac{7}{3}}$

Question: 4
Solving a System with a Unique Solution
Solve the system: $\space\begin{aligned} x y 2z = 7 \\ 2x 3y 5z = 18 \\ 3x 2y 4z = 16 \end{aligned}$

Step 1: Set up the augmented matrix $\space \begin{bmatrix} 1 & 1 & 2 & 7 \\ 2 & 3 & 5 & 18 \\ 3 & 2 & 4 & 16 \end{bmatrix}$

Step 2: Eliminate the ${x}$ -terms from the second and third rows

• Subtract ${2}$ times the first row from the second row: ${\space R_2 = R_2 - 2R_1 \space \rArr \begin{bmatrix} 1 & 1 & 2 & 7 \\ 0 & 1 & 1 & 4 \\ 3 & 2 & 4 & 16 \end{bmatrix}}$

• Subtract ${3}$ times the first row from the third row: ${\space R_3 = R_3 - 3R_1 \space \rArr \begin{bmatrix} 1 & 1 & 2 & 7 \\ 0 & 1 & 1 & 4 \\ 0 & -1 & -2 & -5 \end{bmatrix}}$

Step 3: Eliminate the ${y}$-term from the third row

• Add the second row to the third row: ${\space R_3 = R_3 R_2 \space \rArr \begin{bmatrix} 1 & 1 & 2 & 7 \\ 0 & 1 & 1 & 4 \\ 0 & 0 & -1 & -1 \end{bmatrix}}$

Step 4: Solve for ${z,y,}$ and ${x}$

• From the third row: ${-z = -1 \space \rArr \space z=1}$

• Substitute ${z=1}$ into the second row: ${\space y 1=4 \space \rArr \space y=3}$

• Substitute ${y=3}$ and ${z=1}$ into the first row: ${\space x 3 2(1)=7 \space \rArr \space x 5=7 \space \rArr x=2}$

The solution is: $\space{x=2, \space y=3, \space z=1}$

Question: 5
Solving a Consistent System
Solve the system: $\space\begin{aligned} x y 2z = 7 \\ 2x 3y z = 18 \\ 3x 2y 4z = 16 \end{aligned}$

Step 1: Set up the augmented matrix $\space \begin{bmatrix} 1 & 1 & 2 & 7 \\ 2 & 3 & 1 & 18 \\ 3 & 2 & 4 & 16 \end{bmatrix}$

Step 2: Eliminate the ${x}$ -terms from the second and third rows

• Subtract ${2}$ times the first row from the second row: ${\space R_2 = R_2 - 2R_1 \space \rArr \begin{bmatrix} 1 & 1 & 2 & 7 \\ 0 & 1 & -3 & 4 \\ 3 & 2 & 4 & 16 \end{bmatrix}}$

• Subtract ${3}$ times the first row from the third row: ${\space R_3 = R_3 - 3R_1 \space \rArr \begin{bmatrix} 1 & 1 & 2 & 7 \\ 0 & 1 & -3 & 4 \\ 0 & -1 & -2 & -5 \end{bmatrix}}$

Step 3: Eliminate the ${y}$-term from the third row

• Add the second row to the third row: ${\space R_3 = R_3 R_2 \space \rArr \begin{bmatrix} 1 & 1 & 2 & 7 \\ 0 & 1 & -3 & 4 \\ 0 & 0 & -5 & -1 \end{bmatrix}}$

Step 4: Solve for ${z,y,}$ and ${x}$

• From the third row: ${-5z = -1 \space \rArr \space z=\frac{1}{5}}$

• Substitute ${z=\frac{1}{5}}$ into the second row: $\space{y - 3 \left ( \dfrac{1}{5} \right ) = 4 \space \rArr \space y = 4 \dfrac{3}{5} = \dfrac{20}{5} \dfrac{3}{5} = \dfrac{23}{5}}$

• Substitute ${y=\frac{23}{5}}$ and ${z=\frac{1}{5}}$ into the first row: $\space{x \dfrac{23}{5} 2 \left ( \dfrac{1}{5} \right ) = 7 \space \rArr \space x \dfrac{25}{5} = 7 \space \rArr \space x=7-5=2}$

The solution is: $\space{x=2, \space y=\dfrac{23}{5}, \space z=\dfrac{1}{5}}$

7. Practice Questions on Gaussian Elimination Method:

Q.1 Solve the following system using Gaussian Elimination: $\space \begin{aligned} x y 2z & = 4 \\ 2x 3y z & = 5 \\ 3x y 3z & = 6 \end{aligned}$

Q.2 Apply Gaussian Elimination to solve: $\space \begin{aligned} x - 2y z & = 3 \\ 2x y 3z & = 7 \\ x 3y - z & = 2 \end{aligned}$

Q.3 Determine if the system has a unique solution, no solution, or infinite solutions: $\space \begin{aligned} x y = 2 \\ 2x 2y = 5 \end{aligned}$

8. FAQs on Gaussian Elimination Method:

What is the purpose of the Gaussian Elimination Method?

The Gaussian Elimination Method is used to solve systems of linear equations by transforming the matrix of coefficients into a row echelon form. This allows for easy back-substitution to find the values of the unknown variables.

Can Gaussian Elimination be used for systems with more equations than variables?

Yes, Gaussian Elimination can be applied to overdetermined systems (more equations than variables) and underdetermined systems (fewer equations than variables). However, the solutions in underdetermined systems may not be unique, and overdetermined systems might not have a solution.

What are the three elementary row operations used in Gaussian Elimination?

The three elementary row operations are:

1. Swapping two rows.
2. Multiplying a row by a non-zero constant.
3. Adding or subtracting a multiple from one row to/from another.

How is Gaussian Elimination different from Gauss-Jordan Elimination?

Gaussian Elimination transforms the matrix into row echelon form, while Gauss-Jordan Elimination continues to reduce the matrix further into reduced row echelon form (RREF), where all leading coefficients are 1, and all elements above and below the leading 1s are zeros.

What happens if a pivot element is zero during Gaussian Elimination?

If a pivot element is zero, you must swap rows to place a non-zero element in that position. The system might not have a unique solution if no such row exists.

Can Gaussian Elimination be used to find the inverse of a matrix?

Yes, Gaussian Elimination can be used to find the inverse of a matrix by augmenting the matrix with an identity matrix and then applying row operations until the original matrix becomes the identity matrix and the augmented part becomes the inverse.

What are the limitations of Gaussian Elimination?

Gaussian Elimination can become computationally expensive for very large matrices or systems of equations, and it is prone to numerical instability if the matrix contains very large or very small values.

What are the real-world applications of the Gaussian Elimination Method?

Gaussian Elimination is used in various fields, including engineering (for solving electrical circuit problems), computer graphics (for transformations), economics (in optimization models), and physics (for solving systems of equations in simulations).

9. Real-Life Application of Gaussian Elimination Method:

Gaussian Elimination is widely used in various fields:

• Engineering: Solving complex systems of linear equations in structural analysis, electrical circuits, and more.

• Computer Graphics: In graphics transformations and projections.

• Economics: Solving systems of equations that represent economic models.

• Physics: Used in simulations and to solve large systems in mechanics and dynamics.

10. Conclusion:

The Gaussian Elimination Method is a powerful tool for solving systems of linear equations. It is fundamental in mathematics and applicable to many real-world scenarios. This method simplifies complex problems into easily solvable systems by following systematic row operations, making it an indispensable method in linear algebra.

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