Are (1, 2), (3, 4), (5, 6) independent Step by Step

Step by step

The question

Question: are (1, 2), (3, 4) and (5, 6) linearly independent?

v₁ = [1; 2], v₂ = [3; 4], v₃ = [5; 6]

Independent means: the only way to make the zero vector from them is to take 0 of each.

Otherwise they are dependent.

independent: the only way to make 0

is c₁ = c₂ = c₃ = 0

1 Write the vector equation

We look for numbers c₁, c₂ and c₃ that make the zero vector.

c₁[1; 2] + c₂[3; 4] + c₃[5; 6] = [0; 0]

Careful!

Taking 0 of each always gives 0.

That proves nothing.

The question is: is it the only way?

Careful: c₁ = c₂ = c₃ = 0 always works: is it the ONLY way?

A shortcut: 3 vectors with only 2 entries each are always dependent.

Let's see why.

3 vectors in R²: more vectors than entries

2 Write the augmented matrix

Each vector goes in as a column: the first entries make row 1, the second entries row 2.

After the bar: 0s.

Every move keeps them 0.

= [1, 3, 5, 0; 2, 4, 6, 0]

We call the rows R 1 and R 2.

3 Row reduce

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, 3, 5, 0; 0, −2, −4, 0]

→ [1, 3, 5, 0; 0, 1, 2, 0]

→ [1, 0, −1, 0; 0, 1, 2, 0]

4 Count 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.

Column 3 has no leading 1.

So c₃ is free: it can be any number, not only 0.

column 3: no leading 1, so c₃ is free

5 Find a way to make 0

We pick c₃ = 1.

Each row then gives a leading c.

Row 1 says c₁ − c₃ = 0.

So c₁ = 1.

c₁ = 1

Row 2 says c₂ + 2c₃ = 0.

So c₂ = −2.

c₂ = −2

So v₁ − 2v₂ + v₃ = 0, and not all the numbers are 0.

v₁ − 2v₂ + v₃ = 0

So the vectors are linearly dependent.

= dependent

6 Check

Let's check: we work out v₁ − 2v₂ + v₃.

= [1; 2] − 2 × [3; 4] + [5; 6]

= [1; 2] − [2 × 3; 2 × 4] + [5; 6]

= [1; 2] − [6; 8] + [5; 6]

= [1 − 6 + 5; 2 − 8 + 6]

= [0; 0]

We get the zero vector.

So the vectors are dependent: the answer is right!

Let's look back.

First: write the vector equation.

Then: write the augmented matrix.

Then: row reduce.

Then: count the leading 1s.

Then: find a way to make 0.

Last: check.

Answer