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176 LearnersLast updated on October 24, 2025

An augmented matrix is formed by combining the columns of two matrices. This matrix is used to solve simple linear equations. The number of variables present in a system of linear equations equals the number of columns in the coefficient matrix. Let’s explore the topic in more detail.
One of the two matrices combined to form an augmented matrix is a coefficient matrix of the variables, and the other contains constants from the equations. If one matrix has m columns and the other has n columns, then the augmented matrix formed has m + n columns. The number of rows in the matrix will be the same as the number of equations.
For example, consider a system of equations:
2x+3y=8
4x-y=2
The coefficient matrix is:
The constant matrix is:
So, the augmented matrix is:
Augmented matrices provide an efficient way to solve systems of linear equations using row operations. Augmented matrices represent an entire system as one matrix to which row operations can be applied for efficient solutions. They are also foundational for row operations, matrix forms, and understanding the identity matrix and inverse matrices.
The properties of augmented matrices help us understand how they are structured, what operations can be used to solve them, and how they show if a system has no solutions, one solution or infinitely many solutions. These properties are:
Simplifying an augmented matrix helps find the solution to a system of linear equations. This is done using row operations to transform the matrix into reduced row echelon form (RREF) using the Gauss-Jordan elimination method.
Row transformations change the first part of the augmented matrix into the identity matrix. The values of the last column obtained are the solution for the system of linear equations.
Suppose we are given a system of linear equations:
a1x+b1y+c1z=d1
a2x+b2y+c2z=d2
a3x+b3y+c3z=d3
We can represent this system using an augmented matrix as follows:
By applying elementary row operations, we simplify the matrix into row-reduced echelon form (RREF), where the left side becomes the identity matrix:
This final matrix shows that the system has a unique solution, given by:
x=p, y=1, z=r
The values in the last column of the augmented matrix are the values that solve the original system of equations.
For finding augmented matrix, the steps are as follows -
Step 1: At first, the coefficient matrix is formed by the coefficients from each equation.
Step 2: The constant matrix has to be placed on the right-hand side of each equation.
Step 3: For Augmented matrix, combine both coefficient matrix and constant matrix side-by-side.
An augmented matrix is formed by combining two matrices from a system of linear equations: the coefficient matrix and the constant matrix (column matrix). They are placed side by side and separated by a dotted line.
For instance, for a system of linear equations:
2x+3y-z=5
4x-y+2z=6
-3x+2y+z=-4
The coefficient matrix A is:
The constant matrix B is:
The augmented matrix is:
For a general system of linear equations with coefficients aij and variables x1, x2,. . ., xn
The system can be written as:
a11x1+a12x2+ . . . +a1nxn=b1
a21x1+a22x2+ . . . +a2nxn=b2
. .
. .
. .
am1x1+am2x2+ . . . +amnxn=bm
The augmented matrix for this system is:
Augmented matrices are often the first application of matrices students come across. To build an error free foundation for matrix calculations, avoid the following mistakes:
Augmented matrices are key mathematical tools used in algebraic calculations. Listed below are some real life applications of this tool:
Solve the system using an augmented matrix: x+y=4 2x-y=1
x=5/3, y=7/3
Write the augmented matrix:
Use row operations: R2 → R2 - 2R1
R2 = [2,-1,1] - [2× [1,1,4] = [0,-3,-7]
The new matrix is:
From row 2, -3y = -7
y=7/3
Substituting this into row 1, we get
x+7/3=4
x=4-7/3=5/3
Solve x-2y+z=3 2x+y-3z=-1 -3x+y+2z=4
0=6The system has no solution and is inconsistent.
Augmented matrix:
Row operations:
R2→R2-2R1
R3→R3+3R1
R2= [2,1,-3,-1-21], [-2,1,3] = 0,5,-5,-7
R3= [-3,1,2,4] + [31,-2,1,3] = [-3,1,2,4] +[3,-6,3,9] = [0,-5,5,13]
Now the matrix is:
R3→R3+R2
R3= [0,-5,5,13] + [0,5,-5,-7] = 0,0,0,6
So, the matrix is:
The last row is: 0x+0y+0z=6
⇒ 0=6
Write the augmented matrix for: 3x-y+2z=7 2y-z=4 x+y+z=6
3x-y+2z=7
0x+2y-z=4
x+y+z=6
Rewrite missing variables with zero coefficients:
3x-y+2z=7
0x+2y-z=4
x+y+z=6
Write the augmented matrix for the given system of equations. x+2z=5 2x-y+z=4 y-z=3
x+0y+2z=5
2x-y+z=4
0x+y-z=3
Rewrite with all variables (x, y, z)
x+0y+2z=5
2x-y+z=4
0x+y-z=3
Solve: x+y+z=6 2x+3y+z=14 x+2y+3z=14
x=1, y=4, z=2
The augmented matrix for this system is:
Remove row below 1st column:
R2→2R1
R3→R3-R1
R2= [2,3,1,14] - [21,1,1,6] = 0,1,-1,2
R3= [1,2,3,14] - [1,1,1,6] = 0,1,2,8
The matrix now is:
R3→R3-R2
R3 = [0,1,2,8] - [0,1,-1,2] = 0,0,3,6
R3→1/3R3 = [0,0,1,2]
Using back substitution:
R2→R2+R3 = [0,1,0,4]
R1→R1-R3 = [1,1,0,4]
R1→R1-R2 = [1,0,0,1]
Final matrix:
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.






