Is (4, 1, 3) in the span of (1, 0, 1), (0, 1, 0), (1, 1, 2) Step by Step

Step by step

The question

Question: is (4, 1, 3) in the span of (1, 0, 1), (0, 1, 0) and (1, 1, 2)?

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

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

b = [4; 1; 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; 0; 1] + c₂[0; 1; 0] + c₃[1; 1; 2] = [4; 1; 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, 0, 1, 4; 0, 1, 1, 1; 1, 0, 2, 3]

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, 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, 0, 1, 4; 0, 1, 1, 1; 0, 0, 1, −1]

→ [1, 0, 0, 5; 0, 1, 0, 2; 0, 0, 1, −1]

4 Read the answer

Each row is an equation again.

Row 1 says c₁ = 5.

c₁ = 5

Row 2 says c₂ = 2.

c₂ = 2

Row 3 says c₃ = −1.

c₃ = −1

So yes: b = 5v₁ + 2v₂ − v₃.

b is in the span.

b = 5v₁ + 2v₂ − v₃

5 Check

Let's check: we work out 5v₁ + 2v₂ − v₃.

= 5 × [1; 0; 1] + 2 × [0; 1; 0] − [1; 1; 2]

= [5 × 1; 5 × 0; 5 × 1] + [2 × 0; 2 × 1; 2 × 0] − [1; 1; 2]

= [5; 0; 5] + [0; 2; 0] − [1; 1; 2]

= [5 + 0 − 1; 0 + 2 − 1; 5 + 0 − 2]

= [4; 1; 3]

We get (4, 1, 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