Worked examples
The question
Question: find what A does to the unit square.
A = [2, 0; 0, 1]
Times A turns every point into a new point.
So A moves the whole plane.
The unit square has corners (0, 0), (1, 0), (1, 1) and (0, 1).
A × (point) = new point:
A moves every point of the plane
1 Where (1, 0) goes
First, where does (1, 0) go?
A times (1, 0).
= [2, 0; 0, 1] × [1; 0]
= [2 × 1 + 0 × 0; 0 × 1 + 1 × 0]
= [2; 0]
(2, 0): that is the first column of A.
2 Where (0, 1) goes
Now (0, 1).
A times (0, 1).
= [2, 0; 0, 1] × [0; 1]
= [2 × 0 + 0 × 1; 0 × 0 + 1 × 1]
= [0; 1]
(0, 1): the second column of A.
Careful!
The columns of A show where (1, 0) and (0, 1) go.
Not the rows.
Careful: columns, not rows
3 Draw the new shape
The unit square, dashed.
(1, 0) goes to (2, 0).
(0, 1) goes to (0, 1).
The square goes to the shape with these two sides.
Its far corner is (2, 1).
4 Name it
(0, 1) stays, and (1, 0) gets 2 times longer.
Everything is stretched across by 2.
So A is a stretch across by 2.
= a stretch across by 2
The determinant tells the area: det A = 2.
The square's area 1 becomes 2.
det A = 2 × 1 − 0 × 0 = 2
area: 1 → 2
5 Check
Let's check the far corner: A times (1, 1).
= [2, 0; 0, 1] × [1; 1]
= [2 × 1 + 0 × 1; 0 × 1 + 1 × 1]
= [2; 1]
We get (2, 1): the far corner of the new shape, column 1 plus column 2.
So the picture is right!
Let's look back.
First: where (1, 0) goes.
Then: where (0, 1) goes.
Then: draw the new shape.
Then: name it.
Last: check.
Answer = a stretch across by 2
▶ Watch it solved
The question
Question: find what A does to the unit square.
A = [1, 2; 0, 1]
Times A turns every point into a new point.
So A moves the whole plane.
The unit square has corners (0, 0), (1, 0), (1, 1) and (0, 1).
A × (point) = new point:
A moves every point of the plane
1 Where (1, 0) goes
First, where does (1, 0) go?
A times (1, 0).
= [1, 2; 0, 1] × [1; 0]
= [1 × 1 + 2 × 0; 0 × 1 + 1 × 0]
= [1; 0]
(1, 0): that is the first column of A.
2 Where (0, 1) goes
Now (0, 1).
A times (0, 1).
= [1, 2; 0, 1] × [0; 1]
= [1 × 0 + 2 × 1; 0 × 0 + 1 × 1]
= [2; 1]
(2, 1): the second column of A.
Careful!
The columns of A show where (1, 0) and (0, 1) go.
Not the rows.
Careful: columns, not rows
3 Draw the new shape
The unit square, dashed.
(1, 0) goes to (1, 0).
(0, 1) goes to (2, 1).
The square goes to the shape with these two sides.
Its far corner is (3, 1).
4 Name it
(1, 0) stays.
(0, 1) slides across to (2, 1).
The square leans over: a shear.
Its height stays 1.
So A is a shear across by 2.
= a shear across by 2
The determinant tells the area: det A = 1.
The square's area 1 becomes 1.
det A = 1 × 1 − 0 × 2 = 1
area: 1 → 1
5 Check
Let's check the far corner: A times (1, 1).
= [1, 2; 0, 1] × [1; 1]
= [1 × 1 + 2 × 1; 0 × 1 + 1 × 1]
= [3; 1]
We get (3, 1): the far corner of the new shape, column 1 plus column 2.
So the picture is right!
Let's look back.
First: where (1, 0) goes.
Then: where (0, 1) goes.
Then: draw the new shape.
Then: name it.
Last: check.
Answer = a shear across by 2
▶ Watch it solved
The question
Question: find where A sends (3, 1).
A = [0, −1; 1, 0], v = [3; 1]
The image of v is where A sends it: A times v.
image of v: the new point A × v
1 Multiply
We work out A times v.
= [0, −1; 1, 0] × [3; 1]
= [0 × 3 + (−1) × 1; 1 × 3 + 0 × 1]
We work out every times.
= [0 − 1; 3 + 0]
= [−1; 3]
So the image is (−1, 3).
A v = [−1; 3]
2 Draw it
v is the arrow to (3, 1).
A v is the arrow to (−1, 3).
A is a turn of 90° counterclockwise, and the arrow moved that way.
Careful!
Each entry of A v is a row of A times v: first times first, second times second.
Careful: row times column, in order
3 Check
Let's check another way: A v is 3 times column 1, plus 1 times column 2.
= 3 × [0; 1] + [−1; 0]
= [3 × 0; 3 × 1] + [−1; 0]
= [0; 3] + [−1; 0]
= [0 + (−1); 3 + 0]
= [−1; 3]
We get (−1, 3) again.
So the answer is right!
Let's look back.
First: multiply.
Then: draw it.
Last: check.
▶ Watch it solved
