System Of Linear Equations Solver

System Of Linear Equations Solver is a free online calculator which help you to find out the solutions of linear equations if they exist. This calculator is very useful in finding the multi variable linear equations which can take multiple hours to solve by hand. Go ahead and try this calculator to solve your linear equations problems.

System Of Linear Equations are the set of two or more than two linear equations which contains the same variables i.e. x,y,z,.....,k. Linear Equations are those equations in which the variables are raised to the power of one.The path of the linear equations are straight. Equations in which variables are raised to any other powers rather then one are non-linear equations. The path of the non-linear functions are curve. The general form of 2-variable system of linear equations is given by:

Similarly, the general form of 3-variable system of linear equations looks is :

In the above examples x,y,z are the variables and are the coefficients.These coefficients are used to compute the value of the variables. The value of these variables is called the solutions of the linear equations.

We can solve system of linear equations by using various methods.In this calculator we offer only three methods of calculations.
1. Row Echelon Method or Gauss–Jordan elimination
2. Cramers Method
3. Matrix Inversion Method

Row Echelon Method→Cramers Method→Matrix Inversion Method→

Row Echelon Method or Gauss Elimination method?

Row Echelon Method or Gauss Elimination method is a technique in which row of system of linear equation are converted into a Row Echelon Form or Upper Triangular Matrix. After that the solution is calculated from the echelon form Matrix using back substitution method.
Let us suppose, we have a system of linear equations represented below in Augmented Matrix form:

1	2	3	8
7	6	3	1
8	9	7	4

Swapping rows for better numerical stability.

∼

8	9	7	4
7	6	3	1
1	2	3	8

Row1 ↔ Row3

Normalizing pivot to 1.

∼

1	9/8	7/8	1/2
7	6	3	1
1	2	3	8

Row1 = Row1 / 8

Eliminating entry in row 2 using row 1.

∼

1	9/8	7/8	1/2
0	-15/8	-25/8	-5/2
1	2	3	8

Row2 = Row2 - (7 × Row1)

Eliminating entry in row 3 using row 1.

∼

1	9/8	7/8	1/2
0	-15/8	-25/8	-5/2
0	7/8	17/8	15/2

Row3 = Row3 - (1 × Row1)

Normalizing pivot to 1.

∼

1	9/8	7/8	1/2
0	1	5/3	4/3
0	7/8	17/8	15/2

Row2 = Row2 / -15/8

Eliminating entry in row 3 using row 2.

∼

1	9/8	7/8	1/2
0	1	5/3	4/3
0	0	2/3	19/3

Row3 = Row3 - (7/8 × Row2)

Normalizing pivot to 1.

∼

1	9/8	7/8	1/2
0	1	5/3	4/3
0	0	1	19/2

Row3 = Row3 / 2/3

After several row operations we get the Row Echelon Form.From the last row of Row Echelon Matrix we can find the value of one variable. Then we substitute the value of it in the second last row and find the value of another variable. We continue this process until all the values of the variables are found.

From the above row echelon Form we find that:

Now, Put that value in second row:

Substituting the value of we get:

Similarly, putting the value of in the first row we get:

Types of Solutions in System of Linear Equations.

1. Unique Solution

If the determinant of the coefficient matrix is not equal to zero then we will have unique solution. We can also say that thenumber of variables is equal to the number of equations and no rows in the augmented matrix is dependent to each other or redundant or filled with zero The rank of the coefficient matrix is equal to the rank of the augmented matrix.
For Example:

1	0	3	4
6	3	1	5
3	4	5	6

2. No Solution

If the determinant of the coefficient matrix is equal to zero then we have no solution. We can also determine this by comparing the rank of augmented matrix and coefficient matrix. If the rank of the augmented matrix is greater than the rank of the coefficient matrix then we have no solution.
For Example:

1	0	3	4
6	3	1	5
0	0	0	6

3. Infinite Solutions

If the determinant of the coefficient matrix is equal to zero and and at the same time if the rank of the augmented matrix is equal to the rank of the coefficient matrix then we have infinite many solutions. We can also determine this if all the elements in a row of augmented matrix are zero or if the rows are duplicates or dependent to each other.
For Example:

Augmented matrix with a row with all zeros