Are (1, 2, 0) and (3, 1, 1) independent Step by Step

Step by step

The question

Question: are (1, 2, 0) and (3, 1, 1) linearly independent?

v₁ = [1; 2; 0], v₂ = [3; 1; 1]

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 = 0

1 Write the vector equation

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

c₁[1; 2; 0] + c₂[3; 1; 1] = [0; 0; 0]

Careful!

Taking 0 of each always gives 0.

That proves nothing.

The question is: is it the only way?

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

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

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

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, 0; 0, −5, 0; 0, 1, 0]

→ [1, 3, 0; 0, 1, 0; 0, −5, 0]

→ [1, 0, 0; 0, 1, 0; 0, 0, 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.

Every column has a leading 1.

So each row says one c is 0: c₁ = c₂ = 0.

That is the only way.

a leading 1 in every column: c₁ = c₂ = 0

So the vectors are linearly independent.

= independent

5 Check

Let's check another way.

Two vectors are dependent only if one is a number times the other.

Entry 1 needs k = 3, but then entry 2 gives 6, not 1.

v₂ = k × v₁? entry 1: k = 3, entry 2: 3 × 2 = 6 ≠ 1

No number works.

So the two vectors are independent: 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.

Last: check.

Answer