Systems of Three Equations

Solving using Matrices and Row Reduction

Solving using Matrices and Row Reduction

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

a 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

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:

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) .