Last updated on August 26th, 2025
A system of linear equations is called homogeneous when all the equations have zero as their constant on the right-hand side (RHS). Homogeneous equations have at least one or infinitely many solutions. In this article, we will learn more about homogeneous systems of linear equations, formulas, and real-life applications.
A homogeneous system of linear equations is a set of linear equations where each equation is equal to zero on the RHS. For example, 3x - y = 0. If the right side of the equation has a non-zero number (x + 2y = 5) it would be a non-homogeneous system.
A homogeneous system of linear equations is a system where all equations are equal to zero on the right-hand side. It has special properties, which make it different from other types of systems.
A homogeneous system of linear equations can be written in matrix form to make it simpler to solve the equation. In this form, the system is written as Ax=0, where all equations are equal to zero.
For example:
x - y = 0 (1)
2x + y = 0 (2)
Step 1:
We want to write this system as a matrix. Each equation gives the coefficients of variables x, and y.
Ax = 0
Let's write the coefficient matrix:
Coefficient matrix A:
Each variable has a coefficient in the matrix A
A = 21 1-1
Variable vector:
We have 2 variables = x and y
X = yx
Zero vector:
0 = 00
Matrix form of the system:
211-1 yx = 00
Step 2:
Solve the system
From equation (1)
x - y = 0
x = y
Substitute into equation (2)
2x + y = 0
2y + y = 0
3y = 0
y = 0
Now since x = y, we get
x = 0
Let’s now confirm if det(A) 0.
det(A) = 1·1 - (-1)·2 = 1 + 2 = 3 0.
Since det(A) 0, we only have the trivial solution: x = 0, y=0
A homogeneous linear system has either a unique solution or an infinite number of solutions. To find out if it’s a trivial solution or infinitely many solutions, we take a look at the determinant of the coefficient matrix, which is denoted as A.
The system has a unique solution if det(A) ≠ 0.
The system has an infinite number of solutions if det(A) = 0.
For example:
x + 2y + z = 0
2x + 3y + 4z = 0
x + y + z = 0
Step 1:
Coefficient Matrix: Write the coefficient matrix from the system.
1 2 1
2 3 4
1 1 1
A =
Step 2: Calculate the determinant
Use the formula for a 3 × 3 determinant:
det(A) = 1 (3 1 - 4 1) - 2 (2 1 - 4 1) + 1 (2 1 - 3 1)
= 1 (3 - 4) - 2 (2 - 4) + 1 (2 - 3)
= 1 (-1) - 2 (-2) + 1 (-1)
= -1 + 4 -1 = 2
Step 3:
Interpret the result
So, det(A) = 2 0
This means the homogeneous system has only the trivial solution:
x = 0, y = 0, and z = 0.
A homogeneous system is a system of linear equations where all constant terms are equal to zero. It is usually written as Ax = 0, and is solved using methods like row reduction. Homogeneous systems always have a trivial solution. If the determinant ≠ 0, only the trivial solution exists. Here are the steps to solve a homogeneous system:
Make an augmented matrix using the coefficients from the system of linear equations.
Use row operations (like subtracting rows) to simplify the matrix.
Solve for variables, start from the bottom row, and go up.
Check if the system has only the trivial solution (where all variables are zero) or infinitely many solutions.
For example:
x + y = 0
2x + 2y = 0
Step 1: Use 1st equation:
x + y = 0
x = -y
Step 2: Substitute into 2nd equation:
2(-y) + 2y = 0
Let y = t and x = -t
(x, y) = (-t, t)
In solving a system of linear equations, first find the values of the variables that make all the equations true at the same time. For example, solve the system:
2x + y = 8
3x -2y = -1
Step 1: Write it as augmented matrix
2 1 | 8
3 -2 | -1
Step 2: Replace the first element of the matrix with 1
To do that, we divide the first row by 2:
R1 = 12 R1
1 0.5 | 4
3 -2 | -1
Step 3: Eliminate the 3 below the leading 1
Using row operation:
R2 = R2 - 3 × R1
R2 = 3,-2,-1 - 3 1, 0.5, 4 = 3-3,-2-1.5,-1-12 = 0,-3.5,-13
So now, the new matrix is:
1 0.5 | 4
0 -3.5 | -13
Step 4: Make the second row’s middle into 1
Divide row 2 by -3.5
R2 = 1-3.5 R2 = 0,1,-13/-3.5 = 0,1,3.714...
Or as fraction: 133.5 = 13035 = 267
1 0.5 | 4
0 1 | 267
Step 5: make 0.5 into 0
R1= R1 -0.5 × R2
1 0 | 157
0 1 | 267
x = 157 and y = 267
A homogeneous system of linear equations is a system where all the constant terms are zero in equations. It always has at least one solution — a trivial solution.
The formula for a homogeneous system of linear equations in a matrix is Ax = 0
Where A is the coefficient of the matrix
x is the column of variables
0 is the zero vector
Homogeneous Systems of linear equations, where each of the equations is set to zero, are used to model real-world phenomena.
Students often make mistakes while dealing with a homogeneous system of linear equations. Learning about a few common mistakes and ways to avoid them will help us avoid such mistakes.
Solve the homogeneous system: x + y = 0 (1) 2x + 3y = 0 (2)
x = 0 and y = 0
Step 1: From the first equation
x + y = 0
x = -y
Step 2: substitute into the second equation:
2(-y) + 3y = 0
-2y + 3y = 0
y = 0
x = 0
Solve the system x + y + z = 0 2x + 2y + 2z = 0
(-s -t, s, t)
Step 1: Divide the second equation by 2:
x + y + z = 0
Now we see that both equations are identical, so the system effectively gives us only one independent equation.
Step 2: Let y = s, z = t,
x = -y -z
x = -s -t
(x, y, z) = (-s -t, s, t)
Solve: x + y + z = 0 2x + 3y + z = 0 3x + 4y + 2z = 0
(-2t, t, t)
Step 1: Use row operation
1 1 1 | 0
2 3 1 | 0
3 4 2 | 0
R2 R2 - 2R1
1 1 1 | 0
0 1 -1 | 0
0 1 -1 | 0
R3 R3 -R2
1 1 1 | 0
0 1 -1 | 0
0 0 0 | 0
Step 2:
Back-substitute
From row 2:
y -z = 0
y = z
From Row 1:
x + y + z = 0
x + z + z = 0
x = -2z
Let z = t
y = t
x = -2t
Solve x + y = 0 x - y = 0
x = 0 and y = 0
Step 1: Solve the second equation
x -y = 0
x = y
Step 2: Plug into the first equation
x + y = 0
y + y = 0
y =0
x = 0
Solve x + 2y + 3z = 0 2x + 4y + 6z = 0
-2s - 3t, s, t
Step 1: Observe the system
The second equation is just 2 times the first:
2(x + 2y + 3z) = (2x + 4y + 6z)
So we have only one independent equation:
x + 2y + 3z = 0
Step 2: Solve using parameters
Let y = s
Let z = t
Now solve for x:
x + 2s + 3t = 0
x = -2s - 3t
So the final answer is (x, y, z) = (−