Systems with three equations and three variables can also be solved
using matrices and row
reduction. First, arrange the system in
the following form:
a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3
where
a1, 2, 3, b1, 2, 3, and
c1, 2, 3 are the
x,
y,
and
z coefficients, respectively, and
d1, 2, 3 are constants.
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:
Note: If the system row reduces to
then the system is
inconsistent.
If the system row reduced to
then the system has multiple solutions.
Example: Solve the following system:
5x + 3y = 2z - 4
2x + 2z + 2y = 0
3x + 2y + z = 1
First, arrange the equations:
5x + 3y - 2z = - 4
2x + 2y + 2z = 0
3x + 2y + 1z = 1
Next, form the
3×4 matrix:
Finally, row reduce the matrix:
Thus,
(x, y, z) = (3, - 5, 2).
= 
