Linear Transformations in 3D

Everything from the 2D lesson lifts cleanly into three dimensions: a linear map of space is multiplication by a $3\times3$3×3 matrix, the columns are still the images of the basis vectors, composition is still multiplication (with order reversed), and the determinant still measures the scaling — now of volume rather than area. What is richer in 3D is the geometry of rotations: in the plane there was only one axis to spin about (the line through the origin perpendicular to the plane), but in space a rotation needs a stated axis, and the three coordinate-axis rotations have distinct matrices you must know cold. Reflections, too, are now reflections in planes rather than lines. This lesson assembles the full 3D toolkit — the standard reflection and rotation matrices derived from the master key, the volume-and-orientation reading of $\det M$detM, composition of 3D transformations, and the diagnostic skill of recognising an unknown $3\times3$3×3 matrix as a named transformation, which is exactly what examiners ask.

---

1. Where this sits in AQA 7367

3D linear transformations are compulsory pure content for Papers 1 and 2. Writing down or applying a standard matrix is AO1; interpreting $\det M$detM as a volume scale factor with orientation, or identifying an unknown matrix as a particular rotation/reflection, is AO2; and multi-step questions — compose two transformations then describe the single result, or find an axis of rotation — are AO3. The topic leans hard on $3\times3$3×3 determinants (Lesson 3), on matrix multiplication (Lesson 2), and it connects to vectors, lines and planes in A-Level Maths: a reflection plane or rotation axis is a plane/line through the origin.

---

2. The master key in 3D

A linear transformation $T:\mathbb{R}^3\to\mathbb{R}^3$T:R3→R3 is multiplication by a $3\times3$3×3 matrix $M$M whose columns are the images of the standard basis vectors:

$$M = \bigl(\,T(\mathbf{e}_1)\ \mid\ T(\mathbf{e}_2)\ \mid\ T(\mathbf{e}_3)\,\bigr), \quad \mathbf{e}_1=\begin{pmatrix}1\\0\\0\end{pmatrix},\ \mathbf{e}_2=\begin{pmatrix}0\\1\\0\end{pmatrix},\ \mathbf{e}_3=\begin{pmatrix}0\\0\\1\end{pmatrix}.$$M=(T(e1​) ∣ T(e2​) ∣ T(e3​)),e1​=​100​​, e2​=​010​​, e3​=​001​​.

The derivation is identical to 2D: write $\mathbf{x}=x\mathbf{e}_1+y\mathbf{e}_2+z\mathbf{e}_3$x=xe1​+ye2​+ze3​, apply linearity, and the images of the basis vectors fall out as the columns. So every standard 3D matrix is built by asking where each of $\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3$e1​,e2​,e3​ goes.

A useful mental picture is the unit cube spanned by $\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3$e1​,e2​,e3​. The transformation sends this cube to the parallelepiped spanned by the three columns of $M$M; reading the columns tells you the new edge vectors, and as we shall see in §4 the volume of that parallelepiped is exactly $\lvert\det M\rvert$∣detM∣. The same two directions of reasoning available in 2D apply: to find a matrix, track where the three basis vectors go and write their images as columns; to read a matrix, the $j$j-th column is the image of $\mathbf{e}_j$ej​, so the first column shows the fate of the $x$x-direction, the second of the $y$y-direction, the third of the $z$z-direction. This single habit — "columns are images of basis vectors" — is the workhorse of the entire lesson, used both to construct the standard library and to recognise an unknown matrix as a named transformation.

---

3. The standard 3D library

Reflections in coordinate planes fix two coordinates and flip the third.

Reflection in plane	Matrix	$\det$det
$xy$xy-plane ($z=0$z=0)	$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$​100​010​00−1​​	$-1$−1
$xz$xz-plane ($y=0$y=0)	$\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$​100​0−10​001​​	$-1$−1
$yz$yz-plane ($x=0$x=0)	$\begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$​−100​010​001​​	$-1$−1

Rotations about the coordinate axes. A rotation by $\theta$θ (anticlockwise looking back along the positive axis towards the origin) leaves that axis's coordinate fixed and rotates the other two as a 2D rotation:

$$R_z(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix},\quad R_x(\theta) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \end{pmatrix},\quad R_y(\theta) = \begin{pmatrix} \cos\theta & 0 & \sin\theta \\ 0 & 1 & 0 \\ -\sin\theta & 0 & \cos\theta \end{pmatrix}.$$Rz​(θ)=​cosθsinθ0​−sinθcosθ0​001​​,Rx​(θ)=​100​0cosθsinθ​0−sinθcosθ​​,Ry​(θ)=​cosθ0−sinθ​010​sinθ0cosθ​​.

Note the sign pattern of $R_y$Ry​ is the odd one out: the $+\sin\theta$+sinθ sits in the top-right. This is not a typo — it follows from the cyclic order $x\to y\to z\to x$x→y→z→x; the $y$y-rotation "wraps around" and the signs flip relative to $R_x$Rx​ and $R_z$Rz​. Getting this sign right is a classic discriminator.

Other standard transformations.

Transformation	Matrix	$\det$det
Enlargement, factor $k$k	$kI = \begin{pmatrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \end{pmatrix}$kI=​k00​0k0​00k​​	$k^3$k3
Stretch in $z$z, factor $c$c	$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & c \end{pmatrix}$​100​010​00c​​	$c$c
Projection onto $xy$xy-plane	$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$​100​010​000​​	$0$0

---

4. The determinant as the volume scale factor

For a $3\times3$3×3 transformation matrix $M$M, the unit cube (volume 1) maps to a parallelepiped of volume $\lvert\det M\rvert$∣detM∣, and every region's volume is multiplied by $\lvert\det M\rvert$∣detM∣. The sign records orientation (handedness):

$\det M$detM	Geometric meaning
$>0$>0	orientation-preserving (rotation, enlargement)
$<0$<0	orientation-reversing (a reflection is involved)
$=1$=1	volume-preserving and orientation-preserving (a pure rotation)
$=-1$=−1	volume-preserving but orientation-reversing (a pure reflection)
$=0$=0	a dimension collapses (e.g. projection onto a plane) — no inverse

This is why an enlargement of factor $k$k multiplies volume by $k^3$k3 (determinant $k^3$k3), while any rotation has $\det = 1$det=1: the rotation matrices above all satisfy $\det R = \cos^2\theta+\sin^2\theta = 1$detR=cos2θ+sin2θ=1 on expansion.

Worked Example 1 — images under a reflection and a rotation (with mark scheme)

(a) Find the image of $(1,2,3)$(1,2,3) under reflection in the $xz$xz-plane. (b) Find the image of $(1,0,0)$(1,0,0) under rotation by $90^\circ$90∘ about the $z$z-axis.

$$\text{(a)}\quad \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}, \qquad \text{(b)}\quad \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}.$$(a)​100​0−10​001​​​123​​=​1−23​​,(b)​010​−100​001​​​100​​=​010​​.

So (a) $(1,-2,3)$(1,−2,3) — only the $y$y-coordinate flips, as reflection in $y=0$y=0 should; (b) $(0,1,0)$(0,1,0) — the point on the positive $x$x-axis swings to the positive $y$y-axis under a $90^\circ$90∘ turn about $z$z.

Mark scheme: (a) M1 correct reflection matrix; A1 $(1,-2,3)$(1,−2,3). (b) M1 $R_z(90^\circ)$Rz​(90∘) with $\cos90^\circ=0,\sin90^\circ=1$cos90∘=0,sin90∘=1; A1 $(0,1,0)$(0,1,0).

Worked Example 2 — composition (with mark scheme)

Find the single matrix for a reflection in the $xy$xy-plane followed by a rotation of $90^\circ$90∘ about the $z$z-axis.

Reflection $A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$A=​100​010​00−1​​; rotation $B = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$B=​010​−100​001​​. "Reflection then rotation" applies $A$A first, so the matrix is $BA$BA:

$$BA = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}.$$BA=​010​−100​001​​​100​010​00−1​​=​010​−100​00−1​​.

(Check: $\det(BA) = \det B\cdot\det A = (1)(-1) = -1$det(BA)=detB⋅detA=(1)(−1)=−1, so the composite is orientation-reversing — correct, since exactly one reflection is involved.)

Mark scheme: M1 both matrices; M1 correct order $BA$BA; A1 the $3\times3$3×3 product; B1 the determinant check or a statement that the result is orientation-reversing.

Worked Example 3 — recognising an unknown matrix (with mark scheme)

Describe fully the transformation with matrix $M = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}$M=​100​001​0−10​​.

Read the columns: $\mathbf{e}_1\mapsto\mathbf{e}_1$e1​↦e1​ (the $x$x-axis is fixed), $\mathbf{e}_2\mapsto\begin{pmatrix}0\\0\\1\end{pmatrix}=\mathbf{e}_3$e2​↦​001​​=e3​, $\mathbf{e}_3\mapsto\begin{pmatrix}0\\-1\\0\end{pmatrix}=-\mathbf{e}_2$e3​↦​0−10​​=−e2​. The $x$x-axis is unmoved while $y$y and $z$z rotate, with $\mathbf{e}_2\to\mathbf{e}_3$e2​→e3​ and $\mathbf{e}_3\to-\mathbf{e}_2$e3​→−e2​: comparing with $R_x(\theta)$Rx​(θ), this is $R_x(90^\circ)$Rx​(90∘) — a rotation of $90^\circ$90∘ about the $x$x-axis. Confirm: $\det M = 1$detM=1 (a pure rotation), and $M$M fixes the $x$x-axis (the rotation axis) ✓.

Mark scheme: M1 read images of the basis vectors / state the $x$x-axis is fixed; A1 identify a rotation about the $x$x-axis; A1 angle $90^\circ$90∘; B1 confirm with $\det M = 1$detM=1. Stating "axis = the $x$x-axis" earns the full-description credit.

The picture: under $R_x(90^\circ)$Rx​(90∘) the $x$x-axis (red) is the fixed rotation axis; the unit vector $\mathbf{e}_2$e2​ along $y$y swings to $\mathbf{e}_3$e3​ along $z$z.

---

5. Rotations have determinant 1 — and an invariant axis

Two structural facts about 3D rotations earn marks and aid recognition. First, every rotation matrix has $\det = 1$det=1 (orientation- and volume-preserving). Second, a rotation about an axis fixes that axis: the axis direction is an eigenvector with eigenvalue $1$1 (it is the one direction left unmoved). For the coordinate-axis rotations this is obvious — $R_z$Rz​ fixes $\mathbf{e}_3$e3​, and so on — but it is the general principle by which you locate the axis of an unknown rotation: solve $M\mathbf{v}=\mathbf{v}$Mv=v, i.e. $(M-I)\mathbf{v}=\mathbf{0}$(M−I)v=0, and the solution direction is the rotation axis (the link to Lesson 8's eigenvectors). A reflection in a plane, by contrast, has $\det = -1$det=−1 and fixes a plane of points (eigenvalue $1$1, two-dimensional) while flipping the normal direction (eigenvalue $-1$−1).

How do you also recover the angle of an unknown rotation? Use the trace. For any rotation by $\theta$θ about any axis, the trace of the matrix is $1 + 2\cos\theta$1+2cosθ. You can check this against the coordinate rotations: $\operatorname{tr} R_z(\theta) = \cos\theta + \cos\theta + 1 = 1 + 2\cos\theta$trRz​(θ)=cosθ+cosθ+1=1+2cosθ, and the same for $R_x,R_y$Rx​,Ry​. The trace is unchanged by the choice of coordinate axes (a property called similarity-invariance), so the formula holds for a rotation about any axis. Thus the recipe to fully describe an unknown rotation matrix $M$M is: (i) confirm $\det M = 1$detM=1 so it really is a rotation; (ii) solve $(M-I)\mathbf{v}=\mathbf{0}$(M−I)v=0 for the axis $\mathbf{v}$v; (iii) read off the angle from $\operatorname{tr} M = 1 + 2\cos\theta$trM=1+2cosθ. Three short steps turn a grid of numbers into a complete geometric description.

Worked Example 4 — axis and angle of a rotation (with mark scheme)