Augmented matrix is a coefficient matrix having an extra column containing the constants of system of linear equations which is separated by a vertical line. Augmented matrix is useful for solving system of linear equations.

Augmented Matrix Definition

Augmented matrix is a matrix whose elements are the coefficients of a set of simultaneous linear equations with the constant terms of the equations entered in an added column. To solve the system of equations using augmented matrix, simple matrix row operations are used. Solutions to the system of linear equations using augmented matrix can have one solution (consistent), no solution (inconsistent) and Infinitely many solutions.

Augmented Matrix Form

Consider a system of linear equations having three equations and three variables.

$l_{1}$x + $m_{1}$y = $c_{1}$
$l_{2}$x + $m_{2}$y = $c_{2}$
$l_{3}$x + $m_{3}$y = $c_{3}$

The augmented matrix for the given system of equations is:

\begin{bmatrix}
l_{1} &m_{1} :c_{1}\\
l_{2} &m_{2} :c_{2}\\
l_{3} &m_{3} :c_{3}
\end{bmatrix}
Constant terms are represented using ":" symbol or a vertical bar.

How to Solve an Augmented Matrix?

Here are the few simple steps to be used while solving augmented matrix:
• For a given linear system of equations represent the given equations in an augmented matrix form.
• Row reduce the augmented matrix using row operations.
• The final matrix after row operations should be in the form of an Identity matrix.

Solving Augmented Matrix

Solved Examples

Question 1: Solve : 2x + 2y = 5 and 12x + 5y = 2
Solution:

The augmented matrix for the given system of equations is:
$\begin{bmatrix} 2 &2 :5\\ 12 &5 :2 \end{bmatrix}$

Divide the first row by 2.
$\begin{bmatrix} 1 &1 :\frac{5}{2}\\ 12 &5 :2 \end{bmatrix}$

Multiply the first row by 12.
$\begin{bmatrix} 12 &12 :30\\ 12 &5 :2 \end{bmatrix}$

Subtract the 1st row from the 2nd row and restore the first row.
$\begin{bmatrix} 1 &1 :\frac{5}{2}\\ 0 &-7 :-28 \end{bmatrix}$

Divide the second row by -7.
$\begin{bmatrix} 1 &1 :\frac{5}{2}\\ 0 &1 :4 \end{bmatrix}$

Consider subtracting second row from the first row.
$\begin{bmatrix} 1 &0 :\frac{-3}{2}\\ 0 &1 :4 \end{bmatrix}$

Therefore the given matrix has been reduced to an identity matrix. The solution is $x_{1}$ = $\frac{-3}{2}$ and $x_{2}$ = 4.

Question 2: Solve  x + 5y - z = -11
3z =  12
2x + 4y - 2z =  8
Solution:

The augmented matrix for the given system of equations is:
$\begin{bmatrix} 1 &5&-1 :-11\\ 0 &0&3 :12\\ 2&4&-2 :8 \end{bmatrix}$

Interchange rows 2 and 3
$\begin{bmatrix} 1 &5 &-1:-11\\ 2 &4&-2 :8\\ 0 &0& 3: 12 \end{bmatrix}$

Multiply row 3 by $\frac{1}{3}$

$\begin{bmatrix} 1 &5 &-1:-11\\ 2&4&-2 :8\\ 0 &0& 1: 4 \end{bmatrix}$

Multiply row 2 by $\frac{-1}{2}$

$\begin{bmatrix} 1 &5 &-1:-11\\ -1 &-2&1 :-4\\ 0 &0& 1: 4 \end{bmatrix}$

Add row 1 and row 2, place the result in second row.
$\begin{bmatrix} 1 &5 &-1:-11\\ 0 &3&0 :-15\\ 0 &0& 1: 4 \end{bmatrix}$

Multiply second row by $\frac{1}{3}$
$\begin{bmatrix} 1 &5 &-1:-11\\ 0 &1&0 :-5\\ 0 &0& 1: 4 \end{bmatrix}$

Multiply row 2 by -5 and add it to row 1. Replace row 1 with the result
$\begin{bmatrix} 1 &0 &-1:14\\ 0 &1&0 :-5\\ 0 &0& 1: 4 \end{bmatrix}$

Add row 3 to row 1 and row 1 with the result
$\begin{bmatrix} 1 &0 &0:18\\ 0 &1&0 :-5\\ 0 &0& 0: 4 \end{bmatrix}$

Therefore the given matrix has been reduced to an identity matrix. The solution is x =18, y = -5 and z = 4.