Rank of [1, 0, 1; 0, 1, 1; 1, 1, 3] Step by Step

Step by step

The question

Question: find the rank of A.

A = [1, 0, 1; 0, 1, 1; 1, 1, 3]

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, 0, 1; 0, 1, 1; 1, 1, 3]

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, 0, 1; 0, 1, 1; 0, 1, 2]

→ [1, 0, 1; 0, 1, 1; 0, 0, 1]

→ [1, 0, 0; 0, 1, 0; 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, 2 and 3.

leading 1s in columns 1, 2 and 3

3 Count them

3 leading 1s, so the rank of A is 3.

rank A = 3

Every column either has a leading 1 or is free: 3 + 0 = 3.

The rank is also the dimension of the column space: 3.

rank + free columns = columns: 3 + 0 = 3

Careful!

Count the leading 1s, not the rows.

A has 3 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, 0, 1; 0, 1, 1; 1, 1, 3|1, 0; 0, 1; 1, 1

Down: 3.

down: 1 × 1 × 3 + 0 × 1 × 1 + 1 × 0 × 1 = 3

Up: 2.

up: 1 × 1 × 1 + 1 × 1 × 1 + 3 × 0 × 0 = 2

Down minus up: D = 1.

D = 3 − 2 = 1

D is 1, 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