Systems with three equations and three variables can also be solved using matrices and row reduction. First, arrange the system in the following form:
where a_{1, 2, 3}, b_{1, 2, 3}, and c_{1, 2, 3} are the x, y, and z coefficients, respectively, and d_{1, 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:
First, arrange the equations:
Next, form the 3×4 matrix:

Finally, row reduce the matrix:

Thus, (x, y, z) = (3, - 5, 2).