Coordinates of (5, 2) in the basis (1, 0), (1, 1) Step by Step

Step by step

The question

Question: find the coordinates of (5, 2) in the basis (1, 0) and (1, 1).

b₁ = [1; 0], b₂ = [1; 1]

We call the basis vectors b₁ and b₂, in this order, and the vector v.

v = [5; 2]

The coordinates of v are the numbers that make v from the basis vectors, in the basis order.

coordinates: the numbers c₁, ..., c₂ in

v = c₁b₁ + c₂b₂

1 Write the vector equation

We look for c₁ and c₂.

c₁[1; 0] + c₂[1; 1] = [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.

v goes after the bar.

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

We call the rows R 1 and R 2.

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, 3; 0, 1, 2]

4 Read the coordinates

Each row is an equation again.

Row 1: c₁ = 3.

c₁ = 3

Row 2: c₂ = 2.

c₂ = 2

We write them as a column, in the basis order: (3, 2).

[v]_B = [3; 2]

Careful!

The order follows the basis.

c₁ goes with b₁, so it comes first.

Careful: not (2, 3) ✗

5 Check

Let's check: 3b₁ + 2b₂ must give v.

= 3 × [1; 0] + 2 × [1; 1]

= [3 × 1; 3 × 0] + [2 × 1; 2 × 1]

= [3; 0] + [2; 2]

= [3 + 2; 0 + 2]

= [5; 2]

We get (5, 2): that is v.

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 coordinates.

Last: check.

Answer