 # Different Methods for Solving the System of Linear Equations ## Different Methods for Solving the System of Linear Equations

System of linear equations are solved with the help of various methods to determine the values of unknown variables of linear equations. There exist different methods for solving systems of linear equations, among which the most widely used are Gaussian Elimination, Gauss-Jordan Elimination, Cramer’s Rule, Back-substitution, and LU Decomposition.

In this blog post, we will explore the elimination method, Gaussian Elimination method, and Cramer’s Rule in more detail and provide a step-by-step guide on how to use these methods to solve linear equations.

### Elimination method

The elimination method is a way to solve problems in algebra that involve multiple equations with different variables. It helps us find values for the variables that satisfy all of the equations at the same time.

This technique involves changing the equations so that we can add or subtract them to eliminate one of the variables. By doing this, we can then solve for the remaining variables.

### Explanation Elimination Method

The elimination method involves manipulating equations to eliminate one variable and obtain a single equation that can be solved for the remaining variable.

To do this, we need to multiply one or both equations by a constant that will result in the coefficients of one of the variables being equal in both equations. Once we have this equality, we can subtract or add the two equations to eliminate one of the variables.

### Example of Elimination Method

Here’s an example to illustrate the elimination method:

Example:

Solve the following system of linear equations using the elimination method:

2x + 3y = 8

3x + 4y = 11

Solution:

To use the elimination method, we need to choose a variable to eliminate. In this case, we will eliminate the variable x. To do this, we need to multiply the first equation by 3 and the second equation by -2, which will result in the coefficients of x being equal but with opposite signs:

(3)(2x + 3y = 8)

6x + 9y = 24

(-2)(3x + 4y = 11)

-6x – 8y = -22

Now we can add the two equations together to eliminate the x variable:

6x + 9y = 24

-6x – 8y = -22

y = 2

Now that we have solved for y, we can substitute this value back into either of the original equations to solve for x:

2x + 3y = 8

2x + 3(2) = 8

2x + 6 = 8

2x = 2

x = 1 and y = 2

Note: Tips for Using the Elimination Method is to choose the variable you want to eliminate carefully. In some cases, it may be easier to eliminate a variable then other.

The problems of solving system of linear equations through elimination method can be solved easily with the help of an elimination calculator.

### Gaussian Elimination

Gaussian Elimination is the most common method, consisting of applying elementary row operations to a matrix to transform it into row echelon or reduced row echelon form.

### Example Gaussian Elimination

Consider the system of equations:

x + 2y + z = 4

2x – y + 3z = 1

3x + y – z = 3

Solution

To solve this system of equations using an augmented matrix:

Step 1: Write the augmented matrix:

[1 2 1 | 4]

[2 -1 3 | 1]

[3 1 -1 | 3]

Step 2: Use the first row to eliminate the x-coefficient in the second and third rows. We can multiply the first row by 2 and subtract it from the second row, and multiply the first row by 3 and subtract it from the third row:

[1 2 1 | 4]

[0 -5 1 | -7]

[0 -5 -4 | -9]

Step 3: Use the second row to eliminate the y-coefficient in the third row. We can multiply the second row by -1 and add it to the third row:

[1 2 1 | 4]

[0 -5 1 | -7]

[0 0 -15 |-16]

Step 4: Solve for the variables starting with the bottom row. We can solve for z:

-15z = -16

z = 16/15

Substitute z back into the second row to solve for y:

-5y + (16/15) = -7

-5y = -109/15

y = 109/75

Substitute y and z back into the first row to solve for x:

x + 2(109/75) + (16/15) = 4

x = 1/15

x = 1/15, y = 109/75, and z = 16/15.

### Cramer’s Rule

Cramer’s Rule is another method that employs determinants to solve systems of linear equations. Although less efficient than Gaussian elimination, it can be suitable for smaller systems. Back-substitution is a method that comes in handy when the system is already in upper triangular form, and it involves solving variables one by one starting from the last equation and working backward.

### Example Cramer’s Rule

To solve the system of equations:

2x – 3y = 1

4x + y = 7

Using Cramer’s Rule, we first need to find the determinant of the coefficient matrix and the determinants obtained by replacing each column with the constants in the equations.

Solution

The coefficient matrix is:

|2 -3| |4   1|

The determinant of the coefficient matrix is:

det(A) = (2 * 1) – (4 * (-3)) = 14

Next, we replace the x-column with the constants in the equations to obtain

Dx:  | 1   -3 | | 7   1 |

The determinant of Dx is:

det(Dx) = (1 * 1) – (7 * (-3)) = 22

Similarly, we replace the y-column with the constants in the equations to obtain

Dy:  | 2    1 | | 4   7 |

The determinant of Dy is:

det(Dy) = (2 * 7) – (4 * 1) = 10

Finally, we can find the values of x and y:

x = det(Dx)/det(A) = 22/14 = 11/7

x = 11/7 and y = 5/7.

## Conclusion

To sum up, the elimination method is an effective approach for solving systems of linear equations that involve multiple variables. This method requires adjusting the equations to eliminate one of the variables and obtain a single equation that can be solved for the remaining variable.

With dedication and attention to detail, anyone can learn how to use this method to solve algebraic problems with accuracy and efficiency. By mastering the elimination method, you can enhance your problem-solving skills and gain a better understanding of fundamental algebraic concepts.

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