Rank of [1, 2; 3, 4] Step by Step | Rank and column space

Step by step

The question

Question: find the rank of A.

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

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.

swap two rows

× a row by a number (not 0)

add a multiple of one row to another

→ [1, 2; 0, −2]

→ [1, 2; 0, 1]

→ [1, 0; 0, 1]

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.

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 + 0 = 2.

The rank is also the dimension of the column space: 2.

rank + free columns = columns: 2 + 0 = 2

Careful!

Count the leading 1s, not the rows.

A has 2 rows, and here every row has one.

Careful: count the leading 1s, not the rows

4 Check

Let's check another way.

Every column has a leading 1, so the columns are independent.

Then the determinant is not 0.

D = |1, 2; 3, 4|

So D = 1 × 4 − 3 × 2.

D = 1 × 4 − 3 × 2

1 times 4 is 4.

3 times 2 is 6.

D = 4 − 6

4 minus 6 is −2.

So D = −2.

D = −2

D is −2, not 0.

So the answer is right!

Let's look back.

First: row reduce.

Then: ring the leading 1s.

Then: count them.

Last: check.

Answer