Step by step
The question
Question: are (2, −1, 1), (1, 1, 2) and (0, 3, 3) linearly independent?
Independent means: the only way to make the zero vector from them is to take 0 of each.
Otherwise they are dependent.
1 Write the vector equation
We look for numbers c₁, c₂ and c₃ that make the zero vector.
Careful!
Taking 0 of each always gives 0.
That proves nothing.
The question is: 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.
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.
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.
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.
Row 2 says c₂ + 2c₃ = 0.
So c₂ = −2.
So v₁ − 2v₂ + v₃ = 0, and not all the numbers are 0.
So the vectors are linearly dependent.
6 Check
Let's check: we work out v₁ − 2v₂ + v₃.
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.