In mathematics, the system of linear equations is the set of two or more linear equations involving the same variables. Here, linear equations can be defined as the equations of the first order, i.e., the highest power of the variable is 1. Linear equations can have one variable, two variables, or three variables. Thus, we can write linear equations with n number of variables. In this article, you will get the definition of the system of linear equations, different methods of solving these systems of linear equations and solved examples.
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System of Linear Equations Definition
Let
a_{11}x_{1} + a_{12}x_{2} + a_{13}x_{3} + …. + a_{1n}x_{n} = b_{1}
a_{21}x_{1} + a_{22}x_{2} + a_{23}x_{3} + …. + a_{2n}x_{n} = b_{2}
…………………………………..
…………………………………..
a_{n1}x_{1} + a_{n2}x_{2} + a_{n3}x_{3} + …. + a_{nn}x_{n} = b_{n}
be a system of “n” linear equations in “n” variables x_{1}, x_{2}, x_{3},…., x_{n}.
Where,
a_{11}, a_{12}, …, a_{21}, a_{22},…, a_{n1}, a_{n2},…, a_{nn} are the coefficients of variables, x_{1}, x_{2},…., x_{n}.
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System of Linear Equations Matrix
As we know, the system of linear equations can be written in matrix form. Thus, the system of linear equations in n variables can be written as:
\(\begin{array}{l}\begin{bmatrix}a_{11} & a_{12} & a_{13} &…. & a_{1n} \\a_{21} &a_{22} & a_{23} &…. & a_{2n} \\… &… & … & …. &… \\… &… &… &…. &… \\a_{n1} & a_{n2} &a_{n3} &…. &a_{nn} \\\end{bmatrix}\ \begin{bmatrix}x_1 \\ x_2\\ …\\ …\\x_n\end{bmatrix}=\begin{bmatrix}b_1 \\ b_2\\ …\\ …\\b_n\end{bmatrix}\end{array} \)
(or)
AX = B
Here,
\(\begin{array}{l}A=\begin{bmatrix}a_{11} & a_{12} & a_{13} &…. & a_{1n} \\a_{21} &a_{22} & a_{23} &…. & a_{2n} \\… &… & … & …. &… \\… &… &… &…. &… \\a_{n1} & a_{n2} &a_{n3} &…. &a_{nn} \\\end{bmatrix},\ X= \begin{bmatrix}x_1 \\ x_2\\ …\\ …\\x_n\end{bmatrix},\ B=\begin{bmatrix}b_1 \\ b_2\\ …\\ …\\b_n\end{bmatrix}\end{array} \)
Thus, A is called the coefficient matrix.
Solutions to System of Linear Equations
Any set of values of x_{1}, x_{2}, x_{2},…x_{n} which simultaneously satisfies the system of linear equations given above is called a solution of the system. If the system of equations has one or more solutions, the equations are called consistent. Also, if the system of equations does not admit any solution, then the equations are called inconsistent.
Consider the system of equations AX = B and these equations are said to be hom*ogeneous if B = 0 and Nonhom*ogeneous if B ≠ 0.
How to Solve System of Linear Equations?
The following methods of solving system of linear equations AX = B, are applicable only when the coefficient matrix A is nonsingular, i.e., A ≠ 0.
 Cramers method
 Inverse method
 GaussJordan method
 Gauss Elimination method
 LU Decomposition method of factorisation (or) Method of Triangularisation
System of Linear Equations in Two Variables
The system of linear equations in two variables is the set of equations that contain only two variables. For example, 2x + 3y = 4; 3x + 5y = 12 are the system of equations in two variables. There are several methods of solving linear equations in two variables, such as:
 Graphical method
 Substitution Method
 Elimination Method
 CrossMultiplication Method
 Matrix method
Click here to learn more about linear equations in two variables.
Solving System of Linear Equations by Elimination
Let’s learn how to solve the system of linear equations by the elimination method here.
Consider the following system of linear equations in three variables.
2x – y + 3z = 9
x – 3y – 2z = 0
3x + 2y – z = 1
Step 1: Let us write the given equations in the form of AX = B.
\(\begin{array}{l}\begin{bmatrix}2 & 1 & 3 \\1 & 3 & 2 \\ 3& 2 & 1 \\\end{bmatrix}\ \begin{bmatrix}x \\ y\\z\end{bmatrix}=\begin{bmatrix}9 \\0 \\1\end{bmatrix}\end{array} \)
Step 2: Now, write the augmented matrix [A : B].
\(\begin{array}{l}\begin{bmatrix}2 & 1 & 3 & 9 \\1 & 3 & 2 & 0\\ 3& 2 & 1 & 1\\\end{bmatrix}\end{array} \)
Step 3: Using elementary row operations, eliminate the unknown x from all the equations, except the first. Eliminate y from the third equation. This can be done as follows.
\(\begin{array}{l}\begin{bmatrix}2 & 1 & 3 & 9 \\1 & 3 & 2 & 0\\ 3& 2 & 1 & 1\\\end{bmatrix}\end{array} \)
Interchange R_{1} and R_{2}, i.e., R_{1} ↔ R_{2}.
\(\begin{array}{l}\begin{bmatrix}1 & 3 & 2 & 0\\2 & 1 & 3 & 9 \\ 3& 2 & 1 & 1\\\end{bmatrix}\end{array} \)
Now, perform R_{2} = R_{2} – 2R_{1} and R_{3} = R_{3} – 3R_{1}.
\(\begin{array}{l}\begin{bmatrix}1 & 3 & 2 & 0\\0 & 5 & 7 & 9 \\0 & 11 & 5 & 1\\\end{bmatrix}\end{array} \)
By performing R_{3} = R_{3} – (11/5)R_{2}, we get;
\(\begin{array}{l}\begin{bmatrix}1 & 3 & 2 & 0\\0 & 5 & 7 & 9 \\0 & 0 & \frac{52}{5} & \frac{104}{5}\\\end{bmatrix}\end{array} \)
From the above matrix, we get;
x – 3y – 2z = 0….(1)
5y + 7z = 9….(2)
(52/5)z = 104/5….(3)
Solving equation (3), we get z = 2.
Substituting z = 2 in equation (2), we get;
5y + 7(2) = 9
5y + 14 = 9
5y = 9 – 14
y = 5/5
y = 1
Substituting y = 1 and z = 2 in equation (1), we get;
x – 3(1) – 2(2) = 0
x + 3 – 4 = 0
x = 1
Therefore, the solution set of given system of linear equations is x = 1, y = 1 and z = 2.
System of Linear Equations Problems
 Solve the system of linear equations:
x + y + z = 6
3x – 2y – z = 4
2x + 3y – 2z = 2  Solve the following system of linear equations by elimination method.
3x + 5y + 6z = 7
x + 3y – 2z = 5
2x + 4y + 3z = 8  Solve: 3x – y + 14z = 7; 2x + 2y + 3z = 0; x – 12y – 18z = 33
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