Inverse transformations

25 min

Shoes off before socks

In the morning you put on your socks and then your shoes. In the evening you undo this in the reverse order: shoes off first, then socks. Walking home works the same way: if you went 3 km east and then 2 km north, you return by going 2 km south and then 3 km west. Undoing a transformation, and undoing a chain of them, follows exactly these rules.

Definition

The identity $I$ is the transformation that leaves every point where it is. The inverse of a transformation $f$ is the transformation $f^{- 1}$ with

f^{-1} \circ f = f \circ f^{-1} = I :

if $f$ maps $P$ to $P^{'}$ , then $f^{- 1}$ maps $P^{'}$ back to $P$ . A transformation that is its own inverse is called self-inverse.

Example: the inverse of "translate 3 units right" is "translate 3 units left".

Important

Transformation	Inverse	Reason
Translation $T_{\vec{v}}$	$T_{-\vec{v}}$	$\vec{v} + (-\vec{v}) = \vec{0}$
Central symmetry $S_{X}$	$S_{X}$ (self-inverse)	two half-turns make a full turn
Reflection $M_{m}$	$M_{m}$ (self-inverse)	reflecting twice in one mirror restores every point
Rotation $R_{X, \theta}$	$R_{X, -\theta}$	$\theta + (-\theta) = 0$
Enlargement $E (X, k)$	$E\!\left(X, \frac{1}{k}\right)$	$k \times \frac{1}{k} = 1$

The inverse of a composite undoes the steps in reverse order:

(g \circ f)^{-1} = f^{-1} \circ g^{-1} .

Check: $(f^{-1} \circ g^{-1}) \circ (g \circ f) = f^{-1} \circ (g^{-1} \circ g) \circ f = f^{-1} \circ f = I$ .

Worked example 1: working back to the object

Find the point $P$ in each case. (a) $T_{(3, -2)}$ maps $P$ to $(5, 1)$ . (b) $R_{O, 90^\circ}$ maps $P$ to $(- 4, 1)$ . (c) The reflection in $y = x$ maps $P$ to $(2, 7)$ .

Solution Apply the inverse to the image.

(a) $T_{(-3, 2)}$ : $P = (5 - 3,\; 1 + 2) = (2, 3)$ .

(b) $R_{O, -90^\circ} : (x, y) \mapsto (y, -x)$ : $P = (1, 4)$ . Check: $R_{O, 90^\circ}$ sends $(1, 4)$ to $(- 4, 1)$ . ✓

Worked example 2: the inverse of a composite

Let $f = R_{O, 90^\circ}$ and $g = T_{(2, 0)}$ , and $h = g \circ f$ . Find a rule for $h^{- 1}$ and check it.

Solution $h : (x, y) \mapsto (-y, x) \mapsto (-y + 2,\; x)$ .

$h^{-1} = f^{-1} \circ g^{-1}$ : first $T_{(-2, 0)}$ , then $R_{O, -90^\circ}$ : $(x, y) \mapsto (x - 2,\; y) \mapsto \big(y,\; -(x - 2)\big) = (y,\; 2 - x)$ .

Check: $h(y,\; 2 - x) = \big(-(2 - x) + 2,\; y\big) = (x, y)$ . ✓ Undoing in the wrong order, $R_{O, -90^\circ}$ first, gives $(y - 2,\; -x)$ , and $h(y - 2,\; -x) = (x + 2,\; y - 2)$ , which is not $(x, y)$ .

Worked example 3: finding the original triangle

A triangle was enlarged with centre $X (1, 1)$ and scale factor 2, and then translated by $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$ . The final image is $A^{''} (6, 5)$ , $B^{''} (10, 5)$ , $C^{''} (6, 7)$ . Find the original triangle.

Solution Undo in reverse order: first $T_{(-3, 0)}$ , then $E\!\left(X, \frac{1}{2}\right)$ .

Image	after $T_{(-3, 0)}$	$\frac{1}{2}\overrightarrow{XP}$	Original
$A^{''} (6, 5)$	$(3, 5)$	$\frac{1}{2}\begin{pmatrix} 2 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$	$A (2, 3)$
$B^{''} (10, 5)$	$(7, 5)$	$\frac{1}{2}\begin{pmatrix} 6 \\ 4 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$	$B (4, 3)$
$C^{''} (6, 7)$	$(3, 7)$	$\frac{1}{2}\begin{pmatrix} 2 \\ 6 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$	$C (2, 4)$

Check forwards with $A$ : $\overrightarrow{XA} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ , doubled $\begin{pmatrix} 2 \\ 4 \end{pmatrix}$ , gives $(3, 5)$ , then $(6, 5)$ . ✓

Worked example 4: an inverse from the rule

A transformation is $f : (x, y) \mapsto (y + 1,\; 3 - x)$ . Find $f^{- 1}$ , and describe both.

Solution Write $x^{'} = y + 1$ and $y^{'} = 3 - x$ , and solve for $x$ and $y$ : $y = x^{'} - 1$ and $x = 3 - y^{'}$ . So $f^{-1} : (x, y) \mapsto (3 - y,\; x - 1)$ .

The fixed point of $f$ : $x = y + 1$ and $y = 3 - x$ give $(2, 1)$ . Relative to it, $\begin{pmatrix} u \\ v \end{pmatrix} \mapsto \begin{pmatrix} v \\ -u \end{pmatrix}$ , so $f = R_{(2, 1), -90^\circ}$ , and $f^{-1} = R_{(2, 1), +90^\circ}$ , as the table predicts.

Warning

Reverse the order, not only the steps. $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$ , never $g^{-1} \circ f^{-1}$ (unless, as for two translations, the order does not matter).

The inverse of a rotation keeps its centre. Only the angle changes sign.

Tip

Always check an inverse by going forwards again: apply the original transformation to the point you found. If you do not get back the given image, something is wrong.

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Practice

The last two are harder.

Write down the inverse of: (a) $T_{(-4, 7)}$ ; (b) $R_{O, 30^\circ}$ ; (c) the reflection in $x = 2$ ; (d) $E (O, - 3)$ .
$R_{O, -90^\circ}$ maps $P$ to $(3, - 5)$ . Find $P$ .
The central symmetry about $(2, - 1)$ maps $Q$ to $(7, 4)$ . Find $Q$ .
Find the inverse of $(x, y) \mapsto (x - 3,\; -y)$ , and describe the transformation.
$h = T_{(1, 1)} \circ R_{O, 180^\circ}$ . Find rules for $h$ and $h^{- 1}$ . What do you notice?
Find the inverse of $(x, y) \mapsto (2x - 3,\; 2y + 1)$ , and describe both transformations.
Show that $(f^{-1} \circ g^{-1}) \circ (g \circ f) = I$ , and explain in words why the order of the inverses is reversed.

Show answersHide answers

(a) $T_{(4, -7)}$ ; (b) $R_{O, -30^\circ}$ ; (c) the same reflection; (d) $E\!\left(O, -\frac{1}{3}\right)$ .
Apply $R_{O, 90^\circ} : (x, y) \mapsto (-y, x)$ : $P = (5, 3)$ . Check: $(5, 3) \mapsto (3, -5)$ . ✓
$S$ is self-inverse: $Q = (4 - 7,\; -2 - 4) = (-3, -6)$ .
$x = x^{'} + 3$ , $y = - y^{'}$ : the inverse is $(x, y) \mapsto (x + 3,\; -y)$ . The transformation is a glide reflection along the $x$ -axis, 3 units left; its inverse glides 3 units right.
$h : (x, y) \mapsto (-x, -y) \mapsto (1 - x,\; 1 - y)$ . $h^{-1} = R_{O, 180^\circ} \circ T_{(-1, -1)} : (x, y) \mapsto (x - 1,\; y - 1) \mapsto (1 - x,\; 1 - y)$ . The rules are the same: $h$ is the half-turn about $\left(\frac{1}{2}, \frac{1}{2}\right)$ , which is self-inverse.
$x = \frac{x' + 3}{2}$ , $y = \frac{y' - 1}{2}$ : the inverse is $(x, y) \mapsto \left(\frac{x + 3}{2},\; \frac{y - 1}{2}\right)$ . The fixed point is $(3, - 1)$ , so the map is $E\big((3, -1), 2\big)$ and its inverse is $E\!\left((3, -1), \frac{1}{2}\right)$ .
$(f^{-1} \circ g^{-1}) \circ (g \circ f) = f^{-1} \circ (g^{-1} \circ g) \circ f = f^{-1} \circ I \circ f = f^{-1} \circ f = I$ . The last step done ( $g$ ) must be the first one undone, as with shoes and socks.

1Shoes off before socks

Shoes off before socks