Step by step
The question
Question: find the rank of A.
The rank of A is how many leading 1s it has, after row reducing.
It tells how many columns are really new.
1 Row reduce
We row reduce A.
We call the rows R 1 and R 2.
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.
2 Ring the leading 1s
The first number that is not 0 in each row is a 1.
We call it a leading 1, and ring it.
The leading 1s are in columns 1 and 2.
Column 3 has none.
3 Count them
2 leading 1s, so the rank of A is 2.
Every column either has a leading 1 or is free: 2 + 1 = 3.
The rank is also the dimension of the column space: 2.
Careful!
Count the leading 1s, not the rows.
A has 2 rows, and here every row has one.
4 Check
Let's check that column 3 adds nothing new.
In the reduced matrix it holds 2 and 3.
So column 3 should be 2 times column 1 and 3 times column 2.
We get (2, 3): that is column 3 of A.
It is made from the leading columns, so it adds nothing new.
The answer is right!
Let's look back.
First: row reduce.
Then: ring the leading 1s.
Then: count them.
Last: check.