Is (3, 1), (6, 2) a basis for R^2 Step by Step

Step by step

The question

Question: are (3, 1) and (6, 2) a basis for R²?

v₁ = [3; 1], v₂ = [6; 2]

A basis for R² is a set of vectors that are independent, and that make every vector of R² as a linear combination.

basis for R²: independent, and every vector

is a linear combination of them

1 Count the vectors

There are 2 vectors, and R² needs exactly 2.

Good.

Now: are they independent?

2 Write the matrix

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

We test independence, so the right side is 0s.

Every move keeps them 0, so we leave them out.

= [3, 6; 1, 2]

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, 2; 3, 6]

→ [1, 2; 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.

Careful!

For 2 vectors in R², independent is enough.

Then they also reach every vector.

Careful: 2 vectors in R²: independent is enough

Column 2 has no leading 1.

So the vectors are dependent.

column 2: no leading 1: dependent

So they are not a basis.

= not a basis

5 Check

Let's check with the determinant of the same matrix.

A basis: it is not 0.

Not a basis: it is 0.

D = |3, 6; 1, 2|

So D = 3 × 2 − 1 × 6.

D = 3 × 2 − 1 × 6

3 times 2 is 6.

1 times 6 is 6.

D = 6 − 6

6 minus 6 is 0.

So D = 0.

D = 0

D is 0.

So they are not a basis: the answer is right!

Let's look back.

First: count the vectors.

Then: write the matrix.

Then: row reduce.

Then: count the leading 1s.

Last: check.

Answer