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Are (1, 0, 2), (2, 1, 0), (0, 1, 1) linearly independent

Step by step, with the reason for every step · Linear independence · Linear Algebra

Step by step

The question

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

v1 = [1; 0; 2], v2 = [2; 1; 0], v3 = [0; 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 = c3 = 0

1 Write the vector equation

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

c1[1; 0; 2] + c2[2; 1; 0] + c3[0; 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 = c3 = 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, 2, 0, 0; 0, 1, 1, 0; 2, 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, 2, 0, 0; 0, 1, 1, 0; 0, −4, 1, 0]
→ [1, 0, −2, 0; 0, 1, 1, 0; 0, 0, 5, 0]
→ [1, 0, −2, 0; 0, 1, 1, 0; 0, 0, 1, 0]
→ [1, 0, 0, 0; 0, 1, 0, 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 = c₃ = 0.

That is the only way.

a leading 1 in every column: c1 = c2 = c3 = 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 = |1, 2, 0; 0, 1, 1; 2, 0, 1|1, 2; 0, 1; 2, 0

Down: 5.

down: 1 × 1 × 1 + 2 × 1 × 2 + 0 × 0 × 0 = 5

Up: 0.

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

Down minus up: D = 5.

D = 5 − 0 = 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
= independent

Your turn

Try this one yourself, then check:

Are [1; 0; 2], [4; 2; 1], [0; 2; 3] independent?
Show the answer
independent

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