SME
Grade 12 Math · Problem type
The image of a point, and undoing it
Use this for “Where does [[2, 1], [1, 1]] send (3, −1)? Which point does it send to (5, 2)?”
Diagram
A is 2 × 2 and B is 2 × 1, so AB is 2 × 1.
Row 1 of A times column 1 of B gives entry (1, 1): c₁₁ = 2 × 3 + 1 × (−1) = 5.
Tap an entry of AB to see its row and column.
This is an example. Type your own values to start a new problem.
=
D
Determinant
worked out (not typed)
Formulas
x
′
=
a
×
x
+
b
×
y
5 = 2 × 3 + 1 × (−1)
y
′
=
c
×
x
+
d
×
y
2 = 1 × 3 + 1 × (−1)
D
=
a
×
d
−
b
×
c
1 = 2 × 1 − 1 × 1
x
=
d
×
x
′ −
b
×
y
′
D
3
=
1 × 5 − 1 × 2
1
y
=
a
×
y
′ −
c
×
x
′
D
−1
=
2 × 2 − 1 × 5
1
Assumptions
•
The matrix [[a, b], [c, d]] sends (x, y) to (ax + by, cx + dy).
•
Type x′ and y′ to undo it: the inverse matrix sends the image back, when D = ad − bc is not 0.
•
D = 0 flattens the whole plane onto a line or a point, so the move can’t be undone.
Step-by-step
WE KNOW
x = 3, y = −1, a = 2, b = 1, c = 1, d = 1
FIND
x of the image (x′), y of the image (y′), determinant (D)
Step 1 · Find x of the image (x′)
Use x′ = a × x + b × y
Row
1
times
the
column
(
x
,
y
).
x
′
=
2
×
3
+
1
×
(−1)
x
′
=
6
−
1
x
′
=
5
Step 2 · Find y of the image (y′)
Use y′ = c × x + d × y
Row
2
times
the
column
(
x
,
y
).
y
′
=
1
×
3
+
1
×
(−1)
y
′
=
2
Step 3 · Find determinant (D)
Use D = a × d − b × c
The determinant says whether the move can be undone: not when it is 0.
D
=
2
×
1
−
1
×
1
D
=
1
Check
Put the numbers back in. Both sides should match.
5 = 2 × 3 + 1 × (−1) ✓
2 = 1 × 3 + 1 × (−1) ✓
1 = 2 × 1 − 1 × 1 ✓
3
=
1 × 5 − 1 × 2
1
✓
−1
=
2 × 2 − 1 × 5
1
✓
Topic check
See what you know: 5 questions on this topic
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Related lessons
Matrices as transformations of the plane
Main lesson
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