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

Step by step

The question

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

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

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

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

The matrix already has its leading 1s, with 0s above and below.

No moves are needed.

4 Read the answer

Each row is an equation again.

Row 1 says c₁ = 3.

c₁ = 3

Row 2 says c₂ = 3.

c₂ = 3

Row 3 says c₃ = 3.

c₃ = 3

So yes: b = 3v₁ + 3v₂ + 3v₃.

b is in the span.

b = 3v₁ + 3v₂ + 3v₃

5 Check

Let's check: we work out 3v₁ + 3v₂ + 3v₃.

= 3 × [1; 0; 0] + 3 × [0; 1; 0] + 3 × [0; 0; 1]

= [3 × 1; 3 × 0; 3 × 0] + [3 × 0; 3 × 1; 3 × 0] + [3 × 0; 3 × 0; 3 × 1]

= [3; 0; 0] + [0; 3; 0] + [0; 0; 3]

= [3 + 0 + 0; 0 + 3 + 0; 0 + 0 + 3]

= [3; 3; 3]

We get (3, 3, 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