Composite and inverse enlargements

25 min

Copy, then copy the copy

A photocopier's "A4 → A5" setting reduces a page by a scale factor of about 0.71, and its "A5 → A4" setting enlarges by about 1.41. Reduce a page and then enlarge the copy, and you are back to (almost exactly) the original size, because $0.71 \times 1.41 \approx 1.00$ : the second enlargement undoes the first. Enlarge twice with 1.41 instead and every length is multiplied by $1.41 \times 1.41 \approx 2$ , which turns A5 into A3. Doing one transformation after another, and undoing one, are the two ideas of this lesson.

Definition

A composite transformation is one transformation followed by another. If $f$ is done first and then $g$ , the composite is written $g \circ f$ (read " $g$ after $f$ "): $(g \circ f)(P) = g\big(f(P)\big)$ .

The inverse $f^{- 1}$ of a transformation $f$ undoes it: if $f$ maps $P$ to $P^{'}$ , then $f^{- 1}$ maps $P^{'}$ back to $P$ .

Example: if $f$ is "enlarge with centre $O$ , scale factor 2", then $f \circ f$ enlarges with scale factor 4, and $f^{- 1}$ enlarges with scale factor $\frac{1}{2}$ .

Important

Write $E (X, k)$ for the enlargement with centre $X$ and scale factor $k$ .

1. The inverse: $E(X, k)^{-1} = E\!\left(X, \frac{1}{k}\right)$ . Reason: $\overrightarrow{XP'} = k\,\overrightarrow{XP}$ is the same as $\overrightarrow{XP} = \frac{1}{k}\,\overrightarrow{XP'}$ .

2. Same centre: $E(X, k_2) \circ E(X, k_1) = E(X, k_1k_2)$ . Reason: $\overrightarrow{XP''} = k_2\,\overrightarrow{XP'} = k_2k_1\,\overrightarrow{XP}$ .

3. Different centres: every length is multiplied by $k_{1}$ and then by $k_{2}$ . If $k_1k_2 \neq 1$ , the composite is a single enlargement with scale factor $k_{1} k_{2}$ , and its centre lies on the line through the two centres. If $k_{1} k_{2} = 1$ , it is a translation.

Worked example 1: two enlargements from one centre

$P (1, 2)$ is enlarged with $E (O, 2)$ and the image is then enlarged with $E (O, - 3)$ . Find the final image and the single transformation.

Solution $(1, 2) \to (2, 4) \to (-6, -12)$ . The single transformation is $E(O, 2 \times (-3)) = E(O, -6)$ . Check: $-6 \times (1, 2) = (-6, -12)$ . ✓

Worked example 2: undoing an enlargement

The enlargement $E (X, 3)$ with $X (2, 1)$ maps a point $P$ to $P^{'} (11, 7)$ . Write down the inverse and use it to find $P$ .

Solution The inverse is $E\!\left(X, \frac{1}{3}\right)$ . $\overrightarrow{XP'} = \begin{pmatrix} 9 \\ 6 \end{pmatrix}$ , so $\overrightarrow{XP} = \frac{1}{3}\begin{pmatrix} 9 \\ 6 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$ and $P (5, 3)$ .

Check forwards: $\overrightarrow{XP} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$ , times 3 is $\begin{pmatrix} 9 \\ 6 \end{pmatrix}$ , which lands on $(11, 7)$ . ✓

Worked example 3: two different centres

Triangle $T$ with vertices $(1, 1)$ , $(2, 1)$ , $(1, 2)$ is mapped by $E (O, 2)$ to $T^{'}$ , and $T^{'}$ is mapped by $E (Y, 3)$ with $Y (3, 0)$ to $T^{''}$ . Find $T^{''}$ and describe the single transformation from $T$ to $T^{''}$ .

Solution $T^{'}$ : $(2, 2)$ , $(4, 2)$ , $(2, 4)$ . Then $\overrightarrow{YP''} = 3\,\overrightarrow{YP'}$ : for $(2, 2)$ , $\overrightarrow{YP'} = \begin{pmatrix} -1 \\ 2 \end{pmatrix}$ , so $\overrightarrow{YP''} = \begin{pmatrix} -3 \\ 6 \end{pmatrix}$ and $P^{''} (0, 6)$ . Likewise $(4, 2) \mapsto (6, 6)$ and $(2, 4) \mapsto (0, 12)$ .

In general: $(x, y) \mapsto (2x, 2y) \mapsto \big(3 + 3(2x - 3),\; 3 \times 2y\big) = (6x - 6,\; 6y)$ . The scale factor is $2 \times 3 = 6$ . The centre $Z$ is the point that does not move: $6 z - 6 = z$ gives $z = 1.2$ , and $6 y = y$ gives $y = 0$ . So $T \to T''$ is the enlargement $E\big((1.2, 0), 6\big)$ , and its centre is on the $x$ -axis, the line through $O$ and $Y$ .

Worked example 4: when the scale factors cancel

Describe $E\!\left((6, 0), \frac{1}{2}\right) \circ E(O, 2)$ .

Solution $(x, y) \mapsto (2x, 2y) \mapsto \left(6 + \frac{1}{2}(2x - 6),\; \frac{1}{2} \times 2y\right) = (x + 3,\; y)$ .

Every point moves 3 units to the right: the composite is the translation by $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$ . This fits the rule: $2 \times \frac{1}{2} = 1$ , so no length changes, yet the centres differ, so points still move.

Warning

Order matters. $g \circ f$ means $f$ first. With different centres, $E(Y, 3) \circ E(O, 2)$ and $E(O, 2) \circ E(Y, 3)$ have the same scale factor but different centres.

The inverse of $E (X, k)$ is not $E (X, - k)$ . Undoing a multiplication by $k$ needs a multiplication by $\frac{1}{k}$ .

Tip

For any composite of enlargements, find the scale factor first by multiplying. Then find the centre as the one point that stays where it is: set the composite rule equal to $(x, y)$ and solve.

Maths Genie: Negative enlargement and combined transformations. Watch the second half, where two transformations are replaced by a single one, as in Worked examples 3 and 4.

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Practice

The last two are harder.

Find the single transformation equal to $E (O, 4)$ followed by $E\!\left(O, \frac{1}{2}\right)$ , and the image of $(3, - 1)$ .
Write down the inverse of $E (O, - 2)$ , and find the point whose image under $E (O, - 2)$ is $(4, - 6)$ .
Write down the inverse of $E\big((1, 3), 4\big)$ . Which point is mapped to $(9, 7)$ by $E\big((1, 3), 4\big)$ ?
What single transformation is $E (O, 3)$ followed by $E\!\left(O, -\frac{1}{3}\right)$ ?
Show that $E\!\left((4, 0), \frac{1}{2}\right) \circ E(O, 2)$ is a translation, and give its vector.
Find the scale factor and the centre of $E\big((2, 0), 2\big) \circ E(O, 3)$ .
Explain why $E (X, k)$ followed by $E\!\left(X, \frac{1}{k}\right)$ , in either order, leaves every point where it was.

Show answersHide answers

$E (O, 2)$ ; $(6, - 2)$ .
$E\!\left(O, -\frac{1}{2}\right)$ ; the point is $-\frac{1}{2}(4, -6) = (-2, 3)$ .
$E\!\left((1, 3), \frac{1}{4}\right)$ ; $\overrightarrow{XP'} = \begin{pmatrix} 8 \\ 4 \end{pmatrix}$ , so $\overrightarrow{XP} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ and the point is $(3, 4)$ .
$E (O, - 1)$ , the half-turn about $O$ .
$(x, y) \mapsto (2x, 2y) \mapsto \left(4 + \frac{1}{2}(2x - 4),\; y\right) = (x + 2,\; y)$ : the translation by $\begin{pmatrix} 2 \\ 0 \end{pmatrix}$ .
$(x, y) \mapsto (3x, 3y) \mapsto (2 + 2(3x - 2),\; 6y) = (6x - 2,\; 6y)$ . Scale factor 6; centre: $6 x - 2 = x$ gives $x = 0.4$ , so $(0.4, 0)$ .
Same centre, so the composite is $E\!\left(X, k \times \frac{1}{k}\right) = E(X, 1)$ , and $\overrightarrow{XP'} = 1 \times \overrightarrow{XP}$ means $P^{'} = P$ for every point.

Sources and credits

Written by Ganzaa.
Video: Negative Enlargement and Combined Transformations, Maths Genie, embedded from YouTube.

The diagrams, worked examples and practice items were written by Ganzaa. The copier factors 0.71 and 1.41 are the usual A4 → A5 and A5 → A4 settings (71% and 141%).

1Copy, then copy the copy

Copy, then copy the copy