Step by step
The question
Question: find all the solutions of x + 3y = 5 and z = 2.
We want every x, y and z that make both true.
There are 3 letters, but only 2 equations.
1 Write the augmented matrix
Each row is one equation: the numbers in front of x, y and z, then the number on the right, after the bar.
We call the rows R 1 and R 2.
2 Row reduce
We use the three row moves: swap two rows, times a row by a number that is not 0, add a multiple of one row to another.
They never change the answers.
Our goal: a leading 1 in each row, with 0s above and below it.
The matrix already has its leading 1s, with 0s above and below.
No moves are needed.
3 Find the free letters
The first number that is not 0 in each row is a 1.
We call it a leading 1, and ring it.
The letters x and z have a leading 1 in their columns.
The letter y has none: it is free.
A free letter can be any number.
4 Write the leading letters with the free ones
Row 1 as an equation: x + 3y = 5.
We want x alone.
To remove + 3y, we take away 3y on both sides.
They cancel on the left.
So x = 5 − 3y.
Careful!
The + 3y moves to the other side, so its sign changes.
So x is not 5 + 3y.
Row 2 as an equation: z = 2.
So z = 2.
5 Write it as vectors
We stack the letters in a column.
The letter y is free, so it stays y.
We split it up.
The numbers alone make one vector: (5, 0, 2).
What goes with y makes another: (−3, 1, 0).
So every solution is one point, plus any amount of one direction.
One free letter: the solutions make a line.
6 Check
Let's check.
First the point: all free letters 0.
We put x = 5, y = 0 and z = 2 into every equation.
The point works in every equation.
Now the direction (−3, 1, 0).
Its left sides must give 0, so adding it changes nothing.
Every left side gives 0.
So adding any amount of it keeps every equation true.
The answer is right!
Let's look back.
First: write the augmented matrix.
Then: row reduce.
Then: find the free letters.
Then: write the leading letters with the free ones.
Then: write it as vectors.
Last: check.