### 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:

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).