Rank of [1, 1, 1; 1, 2, 3; 2, 3, 4] Step by Step

Step by step

The question

Question: find the rank of A.

A = [1, 1, 1; 1, 2, 3; 2, 3, 4]

The rank of A is how many leading 1s it has, after row reducing.

It tells how many columns are really new.

rank: how many leading 1s after row reducing

1 Row reduce

We row reduce A.

= [1, 1, 1; 1, 2, 3; 2, 3, 4]

We call the rows R 1, R 2 and R 3.

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.

swap two rows

× a row by a number (not 0)

add a multiple of one row to another

→ [1, 1, 1; 0, 1, 2; 0, 1, 2]

→ [1, 0, −1; 0, 1, 2; 0, 0, 0]

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.

leading 1s in columns 1 and 2

3 Count them

2 leading 1s, so the rank of A is 2.

rank A = 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.

rank + free columns = columns: 2 + 1 = 3

Careful!

Count the leading 1s, not the rows.

A has 3 rows, but only 2 of them have a leading 1.

Careful: count the leading 1s, not the rows

4 Check

Let's check that column 3 adds nothing new.

In the reduced matrix it holds −1 and 2.

So column 3 should be −1 times column 1 and 2 times column 2.

= −[1; 1; 2] + 2 × [1; 2; 3]

= −[1; 1; 2] + [2 × 1; 2 × 2; 2 × 3]

= −[1; 1; 2] + [2; 4; 6]

= [−1 + 2; −1 + 4; −2 + 6]

= [1; 3; 4]

We get (1, 3, 4): 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.

Answer