Systems with three equations and three variables can also be solved
using matrices and row
reduction. First, arrange the system in
the following form:

wherea_{1}x+b_{1}y+c_{1}z=d_{1}

a_{2}x+b_{2}y+c_{2}z=d_{2}

a_{3}x+b_{3}y+c_{3}z=d_{3}

Next, create a
3×4
matrix, placing the
*x*
coefficients in the
1st column, the
*y*
coefficients in the 2nd column, the
*z*
coefficients
in the 3rd column, and the constants in the 4th column, with a line
separating the 3rd column and the 4th column:

This is equivalent to writing

= |

which is equivalent to the original three equations (check the multiplication yourself).

Finally, row reduce the
3×4
matrix using the elementary row
operations. The result should be
the identity matrix on the left side of the line and a column of
constants on the right side of the line. These constants are the
solution to the system of equations:

then the system is inconsistent. If the system row reduced to

then the system has multiple solutions.

*Example*: Solve the following system:

5First, arrange the equations:x+ 3y= 2z- 4

2x+ 2z+ 2y= 0

3x+ 2y+z= 1

5Next, form the 3×4 matrix:x+ 3y- 2z= - 4

2x+ 2y+ 2z= 0

3x+ 2y+ 1z= 1

Finally, row reduce the matrix:

Thus, (