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

Step by step

The question

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

v₁ = [2; 1], v₂ = [1; 3]

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[2; 1] + c₂[1; 3] = [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.

= [2, 1, 0; 1, 3, 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, 0; 2, 1, 0]

→ [1, 3, 0; 0, −5, 0]

→ [1, 3, 0; 0, 1, 0]

→ [1, 0, 0; 0, 1, 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.

The vectors make a square matrix.

Its determinant must not be 0.

D = |2, 1; 1, 3|

So D = 2 × 3 − 1 × 1.

D = 2 × 3 − 1 × 1

2 times 3 is 6.

1 times 1 is 1.

D = 6 − 1

6 minus 1 is 5.

So D = 5.

D = 5

D is 5, not 0.

So the 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