The principles involved in row reduction of matrices are equivalent to those we used in the elimination method of solving systems of equations. That is, we are allowed to

1. Multiply a row by a non-zero constant.

2. Add one row to another.

3. Interchange between rows

4. Add a multiple of one row to another.

How do we use this to solve systems of equations? We follow the steps:

Step 1. Write the augmented matrix of the system.

Step 2. Row reduce the augmented matrix.

Step 3. Write the new, equivalent, system that is defined by the new, row reduced, matrix.

Step 4. Solution is found by going from the bottom equation

Example: solve the system of equations using the row reduction method

3x + 2y - z = 1

x - 2y + z = 0

2x + y - 3z = -1

Solution:

Step 1: Write the augmented matrix of the system:

Step 2: Row reduce the augmented matrix:

The symbols we used above the arrows are short for:

R₁ <--> R₂ Interchange Rows 1 and 2.

R₂ = R₂ - 3R₂ New Row2 = old Row2 minus 3 times Row1

R₂ = R₂/-4 Divide Row 2 by -4.

Step 3:

Rewrite the system using the row reduced matrix:

x + 2y + -2z = 0

y + z = -1/4

z = 7/8

Step 4:

And the solution is found by going from the bottom equation up:

Try yourself