Rank of [2, 4; 1, 2; 3, 6] Step by Step

Step by step

The question

Question: find the rank of A.

A = [2, 4; 1, 2; 3, 6]

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.

= [2, 4; 1, 2; 3, 6]

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, 2; 2, 4; 3, 6]

→ [1, 2; 0, 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 column 1.

Column 2 has none.

leading 1s in column 1

3 Count them

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

rank A = 1

Every column either has a leading 1 or is free: 1 + 1 = 2.

The rank is also the dimension of the column space: 1.

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

Careful!

Count the leading 1s, not the rows.

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

Careful: count the leading 1s, not the rows

4 Check

Let's check that column 2 adds nothing new.

In the reduced matrix it holds 2.

So column 2 should be 2 times column 1.

= 2 × [2; 1; 3]

= [2 × 2; 2 × 1; 2 × 3]

= [4; 2; 6]

We get (4, 2, 6): that is column 2 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