Determinant and area scale | Year 11

01 / LEARN

Understand the idea.

Area from the image of a unit square

The columns of a linear transformation matrix are the images of the unit square’s two edge vectors. Their determinant gives signed area; its absolute value is the ordinary area multiplier for every region, not just the unit square.

Use $image area = ∣ det A ∣ \times original area$ . If the original area is already a triangle’s area, its factor of $\frac{1}{2}$ has already been included. Do not halve again when applying the transformation multiplier.

Orientation and loss of dimension

For an invertible map, a positive determinant preserves orientation and a negative determinant reverses it. The magnitude controls area, not individual lengths: stretching one direction and compressing another can preserve area while changing the shape.

A zero determinant collapses the plane to a line or point, so every image region has zero area. For an invertible map, the inverse has area multiplier $\frac{1}{∣ det A ∣}$ because it must undo the original area change.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A reflected triangle

A triangle has area $7 {cm}^{2}$ . Find its image area under $(\begin{array}{cccccccccc} 2 & 1 \\ 0 & - 3 \end{array})$ .

Worked solution

1
Calculate the signed determinant.
$det A = 2 (- 3) - 1 (0) = - 6$
Subtract the off-diagonal product from the main-diagonal product.
2
Convert it to the physical area multiplier.
$∣ det A ∣ = ∣ - 6 ∣ = 6$
Area is non-negative; the negative determinant instead records orientation reversal.
3
Scale the given triangle area.
$image area = 6 (7) = 42 {cm}^{2}$
The original value is already an area, so no additional factor of $\frac{1}{2}$ is needed.

Answer

$42 {cm}^{2}$ .

EXAMPLE 02

Area without length preservation

Describe the effects of $(\begin{array}{cccccccccc} 2 & 0 \\ 0 & \frac{1}{2} \end{array})$ .

Worked solution

1
Read the coordinate rule.
$(x, y) \mapsto (2 x, \frac{y}{2})$
The diagonal entries scale the horizontal and vertical directions independently.
2
Calculate the area and orientation effects.
$det A = 2 (\frac{1}{2}) - 0 = 1$
Positive determinant 1 means unchanged area and preserved orientation.
3
Compare unit-vector lengths.
$∥ A e_{1} ∥ = 2, ∥ A e_{2} ∥ = \frac{1}{2}$
The original unit lengths become unequal, so the map does not preserve lengths in general.
4
Check an angle using perpendicular inputs.
$\begin{aligned} (1, 1) \cdot (1, - 1) & = 0 \\ (2, \frac{1}{2}) \cdot (2, - \frac{1}{2}) & = 4 - \frac{1}{4} = \frac{15}{4} \neq 0 \end{aligned}$
These initially perpendicular directions no longer meet at a right angle, so angles are not generally preserved.

Answer

Areas are preserved, but general lengths and angles are not.

03 / YOUR TURN

Try it yourself.

Check your understanding

A region of area 12 maps to area 3. Find the absolute determinant.

Show answer & reasoning

$\frac{1}{4}$ .

The absolute determinant is the ratio of image area to original area: $∣ det A ∣ = \frac{3}{12} = \frac{1}{4}$ . The areas alone do not determine the sign, so they do not establish whether orientation is preserved.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Find the area factor of $(\begin{array}{cccccccccc} 3 & 0 \\ 0 & 2 \end{array})$ .
2 marks
An area-5 region is mapped by $(\begin{array}{cccccccccc} 1 & 2 \\ 3 & 4 \end{array})$ . Find its image area and orientation change.
2 marks
Triangle $(0, 0), (2, 0), (0, 3)$ is dilated by factor 4. Find its image area.
2 marks
If $det A = - 5$ , state the determinant and area factor of $A^{- 1}$ .
2 marks
Find the area effect of $(\begin{array}{cccccccccc} 1 & 2 \\ 2 & 4 \end{array})$ .
2 marks
The diagonal matrix $(\begin{array}{cccccccccc} k & 0 \\ 0 & 3 \end{array})$ sends an area-4 region to area 24. Find all real $k$ .
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the area factor of $(\begin{array}{cccccccccc} 3 & 0 \\ 0 & 2 \end{array})$ .
2 marks
The determinant is 6.
$6$ .
An area-5 region is mapped by $(\begin{array}{cccccccccc} 1 & 2 \\ 3 & 4 \end{array})$ . Find its image area and orientation change.
2 marks
The determinant is $- 2$ .
The area factor is 2.
Area 10; orientation reverses.
Triangle $(0, 0), (2, 0), (0, 3)$ is dilated by factor 4. Find its image area.
2 marks
Its original area is $\frac{1}{2} (2) (3) = 3$ .
The determinant of $4 I$ is 16, so the image area is 48.
48 square units.
If $det A = - 5$ , state the determinant and area factor of $A^{- 1}$ .
2 marks
The inverse determinant is the reciprocal.
Take the absolute value for area.
Determinant $- \frac{1}{5}$ ; area factor $\frac{1}{5}$ .
Find the area effect of $(\begin{array}{cccccccccc} 1 & 2 \\ 2 & 4 \end{array})$ .
2 marks
The determinant is zero, so the map is singular.
Every image region has area zero.
The diagonal matrix $(\begin{array}{cccccccccc} k & 0 \\ 0 & 3 \end{array})$ sends an area-4 region to area 24. Find all real $k$ .
2 marks
The area factor is 6, so $∣ 3 k ∣ = 6$ .
Thus the magnitude of $k$ is 2.
$k = 2$ or $k = - 2$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: matrices and transformations; original lessons follow the approved school-sized chapter structure.

Official curriculum source ↗

In this lesson

Understand the idea.

Area from the image of a unit square

Orientation and loss of dimension

See the method in action.

A reflected triangle

Calculate the signed determinant.

Convert it to the physical area multiplier.

Scale the given triangle area.

Area without length preservation

Read the coordinate rule.

Calculate the area and orientation effects.

Compare unit-vector lengths.

Check an angle using perpendicular inputs.

Try it yourself.

Check your understanding

Keep practising