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

Step by step

The question

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

v₁ = [1; 2; 0], v₂ = [1; 1; 2]

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

b = [3; 5; 2]

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

span: every c₁v₁ + c₂v₂

1 Write the vector equation

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

We look for them.

c₁[1; 2; 0] + c₂[1; 1; 2] = [3; 5; 2]

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

We call the rows R 1, R 2 and R 3.

Careful!

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

Row 1 holds the first entries of v₁ and v₂, 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, 1, 3; 0, −1, −1; 0, 2, 2]

→ [1, 1, 3; 0, 1, 1; 0, 2, 2]

→ [1, 0, 2; 0, 1, 1; 0, 0, 0]

4 Read the answer

Each row is an equation again.

Row 1 says c₁ = 2.

c₁ = 2

Row 2 says c₂ = 1.

c₂ = 1

So yes: b = 2v₁ + v₂.

b is in the span.

b = 2v₁ + v₂

5 Check

Let's check: we work out 2v₁ + v₂.

= 2 × [1; 2; 0] + [1; 1; 2]

= [2 × 1; 2 × 2; 2 × 0] + [1; 1; 2]

= [2; 4; 0] + [1; 1; 2]

= [2 + 1; 4 + 1; 0 + 2]

= [3; 5; 2]

We get (3, 5, 2): 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