Is (5, 3) in the span of (1, 1), (2, 1), (1, 0) Step by Step

Step by step

The question

Question: is (5, 3) in the span of (1, 1), (2, 1) and (1, 0)?

v₁ = [1; 1], v₂ = [2; 1], v₃ = [1; 0]

We call them v₁, v₂ and v₃, and the vector we test b.

b = [5; 3]

The span of some vectors is every linear combination of them: any numbers times them, added.

span: every c₁v₁ + c₂v₂ + c₃v₃

1 Write the vector equation

b is in the span if some numbers c₁, c₂ and c₃ make this true.

We look for them.

c₁[1; 1] + c₂[2; 1] + c₃[1; 0] = [5; 3]

2 Write the augmented matrix

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

b goes after the bar.

= [1, 2, 1, 5; 1, 1, 0, 3]

We call the rows R 1 and R 2.

Careful!

Each vector is a column of the matrix, not a row.

Row 1 holds the first entries of v₁, v₂ and v₃, and of b.

Careful: the vectors are the COLUMNS

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, 1, 5; 0, −1, −1, −2]

→ [1, 2, 1, 5; 0, 1, 1, 2]

→ [1, 0, −1, 1; 0, 1, 1, 2]

4 Read the answer

Each row is an equation again.

Row 1 says c₁ = 1.

c₁ = 1

Row 2 says c₂ = 2.

c₂ = 2

c₃ is free: there are many ways.

We pick c₃ = 0.

So yes: b = v₁ + 2v₂.

b is in the span.

b = v₁ + 2v₂

5 Check

Let's check: we work out v₁ + 2v₂.

= [1; 1] + 2 × [2; 1]

= [1; 1] + [2 × 2; 2 × 1]

= [1; 1] + [4; 2]

= [1 + 4; 1 + 2]

= [5; 3]

We get (5, 3): that is b.

So the answer is right!

Let's look back.

First: write the vector equation.

Then: write the augmented matrix.

Then: row reduce.

Then: read the answer.

Last: check.

Answer