# 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) .

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