Worked examples
The question
Question: is (5, 5) in the span of (1, 2) and (3, 1)?
v1 = [1; 2], v2 = [3; 1]
We call them v₁ and v₂, and the vector we test b.
b = [5; 5]
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] + c2[3; 1] = [5; 5]
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, 3, 5; 2, 1, 5]
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 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, 3, 5; 0, −5, −5]
→ [1, 3, 5; 0, 1, 1]
→ [1, 0, 2; 0, 1, 1]
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] + [3; 1]
= [2 × 1; 2 × 2] + [3; 1]
= [2; 4] + [3; 1]
= [2 + 3; 4 + 1]
= [5; 5]
We get (5, 5): 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.
▶ Watch it solved
The question
Question: is (2, 1, 3) a linear combination of (1, 0, 1) and (0, 1, 1)?
v1 = [1; 0; 1], v2 = [0; 1; 1]
We call them v₁ and v₂, and the vector we test b.
b = [2; 1; 3]
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; 0; 1] + c2[0; 1; 1] = [2; 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, 2; 0, 1, 1; 1, 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 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, 0, 2; 0, 1, 1; 0, 1, 1]
→ [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; 0; 1] + [0; 1; 1]
= [2 × 1; 2 × 0; 2 × 1] + [0; 1; 1]
= [2; 0; 2] + [0; 1; 1]
= [2 + 0; 0 + 1; 2 + 1]
= [2; 1; 3]
We get (2, 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.
▶ Watch it solved
The question
Question: is (1, 4, 3) in the span of (1, 1, 1) and (0, 1, 1)?
v1 = [1; 1; 1], v2 = [0; 1; 1]
We call them v₁ and v₂, and the vector we test b.
b = [1; 4; 3]
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; 1; 1] + c2[0; 1; 1] = [1; 4; 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; 1, 1, 4; 1, 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 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, 0, 1; 0, 1, 3; 0, 1, 2]
→ [1, 0, 1; 0, 1, 3; 0, 0, −1]
4 Read the answer
Look at row 3: 0s, then −1 after the bar.
As an equation it says 0c₁ + 0c₂ = −1.
The left side is always 0, so this is false for every choice of numbers.
0c1 + 0c2 = −1
So no numbers work: (1, 4, 3) is not in the span.
b = is not in the span
5 Check
Let's check.
The other rows give c₁ = 1 and c₂ = 3.
They are the only numbers that fit those rows.
We try them.
= [1; 1; 1] + 3 × [0; 1; 1]
= [1; 1; 1] + [3 × 0; 3 × 1; 3 × 1]
= [1; 1; 1] + [0; 3; 3]
= [1 + 0; 1 + 3; 1 + 3]
= [1; 4; 4]
We get (1, 4, 4), not (1, 4, 3).
So b is not in the span: 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 b = is not in the span
▶ Watch it solved
