Problems

2025 Paper 3 Q4

D: 1500.0 B: 1500.0

linear transformations rotation matrix ellipse reflection invariant line curve sketching implicit differentiation conics

$x_2$ and $y_2$ are defined in terms of $x_1$ and $y_1$ by the equation $$\begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$$ $G_1$ is the graph with equation $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$ and $G_2$ is the graph with equation $$\frac{\left(\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}}\right)^2}{9} + \frac{\left(-\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}}\right)^2}{4} = 1$$ Show that, if $(x_1, y_1)$ is a point on $G_1$, then $(x_2, y_2)$ is a point on $G_2$. Show that $G_2$ is an anti-clockwise rotation of $G_1$ through $45^\circ$ about the origin.
1. The matrix $$\begin{pmatrix} -0.6 & 0.8 \\ 0.8 & 0.6 \end{pmatrix}$$ represents a reflection. Find the line of invariant points of this matrix.
2. Sketch, on the same axes, the graphs with equations $$y = 2^x \text{ and } 0.8x + 0.6y = 2^{-0.6x+0.8y}$$
Sketch, on the same axes, for $0 \leq x \leq 2\pi$, the graphs with equations $$y = \sin x \text{ and } y = \sin(x - 2y)$$ You should determine the exact co-ordinates of the points on the graph with equation $y = \sin(x - 2y)$ where the tangent is horizontal and those where it is vertical.

Solution:

Suppose \begin{align*} && \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} &= \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} \\ \Rightarrow && \binom{x_1}{y_1} &= \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \binom{x_2}{y_2} \end{align*} Therefore if $\frac{x_1^2}9+\frac{y_1^2}{4} = 1$ we must have \begin{align*} \frac{(\frac{x_2}{\sqrt{2}}+\frac{y_2}{\sqrt{2}})^2 }{9} + \frac{(-\frac{x_2}{\sqrt{2}}+\frac{y_2}{\sqrt{2}})^2}{4} = 1 \end{align*} but this is precisely the statement that $(x_1, y_1)$ is on $G_1$ is equivalent to $(x_2,y_2)$ being on the $G_2$. Since the point $(x_2,y_2)$ is a $45^{\circ}$ rotation of $(x_1,y_1)$ anticlockwise about the origin, this means $G_2$ is a $45^{\circ}$ anticlockwise rotation of $G_1$.
1. \begin{align*} && \begin{pmatrix} -0.6 & 0.8 \\ 0.8 & 0.6 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} x \\ y \end{pmatrix} \\ \Rightarrow && \begin{pmatrix} -0.6 x + 0.8y \\ 0.8x + 0.6y \end{pmatrix} &= \begin{pmatrix} x \\ y \end{pmatrix} \\ \Rightarrow && \begin{pmatrix} -1.6 x + 0.8y \\ 0.8x -0.4y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \Rightarrow && y &=2 x \end{align*}
Consider the transformation $\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ which is a shear, leaving the $x$-axis invariant. Then we must have:

Since the shear leaves lines of the form $y = k$ invariant, the points where $\frac{\d y}{\d x} = 0$ must also map to points where this is true, ie $(\tfrac{\pi}{2}, 1), (\tfrac{3\pi}{2}, -1)$ map to points $(\tfrac{\pi}{2}+2,1), (\tfrac{3\pi}{2} -2,-1)$ where the tangent is horizontal. The line $x = c$ map back to lines $\begin{pmatrix} 1 & -2 \\ 0 & 1\end{pmatrix} \begin{pmatrix} c \\ t\end{pmatrix} = \begin{pmatrix}c - 2t \\ t \end{pmatrix}$, ie $y = -\frac12 x- \frac{c}{2}$. Therefore we are interested in points on the original curve where the gradient is $-\frac12$, ie $(\frac{2\pi}{3}, \frac{\sqrt{3}}{2}), (\frac{4\pi}{3}, -\frac{\sqrt{3}}{2})$, these map to $(\frac{2\pi}{3}+\sqrt{3},\frac{\sqrt{3}}{2}), (\frac{4\pi}{3}-\sqrt{3}, -\frac{\sqrt{3}}{2})$

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2010 Paper 3 Q6

D: 1700.0 B: 1484.0

trigonometry 3D rotation rotation matrix sphere vectors coordinate geometry dot product axis of rotation

The points $P$, $Q$ and $R$ lie on a sphere of unit radius centred at the origin, $O$, which is fixed. Initially, $P$ is at $P_0(1, 0, 0)$, $Q$ is at $Q_0(0, 1, 0)$ and $R$ is at $R_0(0, 0, 1)$.

The sphere is then rotated about the $z$-axis, so that the line $OP$ turns directly towards the positive $y$-axis through an angle $\phi$. The position of $P$ after this rotation is denoted by $P_1$. Write down the coordinates of $P_1$.
The sphere is now rotated about the line in the $x$-$y$ plane perpendicular to $OP_1$, so that the line $OP$ turns directly towards the positive $z$-axis through an angle $\lambda$. The position of $P$ after this rotation is denoted by $P_2$. Find the coordinates of $P_2$. Find also the coordinates of the points $Q_2$ and $R_2$, which are the positions of $Q$ and $R$ after the two rotations.
The sphere is now rotated for a third time, so that $P$ returns from $P_2$ to its original position~$P_0$. During the rotation, $P$ remains in the plane containing $P_0$, $P_2$ and $O$. Show that the angle of this rotation, $\theta$, satisfies \[ \cos\theta = \cos\phi \cos\lambda\,, \] and find a vector in the direction of the axis about which this rotation takes place.

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1997 Paper 3 Q8

D: 1700.0 B: 1484.0

linear transformations rotation matrix diagonalisation symmetric matrix ellipse quadratic form matrix multiplication conics optimisation

Let $R_{\alpha}$ be the $2\times2$ matrix that represents a rotation through the angle $\alpha$ and let $$A=\begin{pmatrix}a&b\\b&c\end{pmatrix}.$$

Find in terms of $a$, $b$ and $c$ an angle $\alpha$ such that $R_{-\alpha}AR_{\alpha}$ is a diagonal matrix (i.e. has the value zero in top-right and bottom-left positions).
Find values of $a$, $b$ and $c$ such that the equation of the ellipse \[x^2+(y+2x\cot2\theta)^2=1\qquad(0 < \theta < \tfrac{1}{4}\pi)\] can be expressed in the form \[\begin{pmatrix}x&y\end{pmatrix}A\begin{pmatrix}x\\y\end{pmatrix}=1.\] Show that, for this $A$, $R_{-\alpha}AR_{\alpha}$ is diagonal if $\alpha=\theta$. Express the non--zero elements of this matrix in terms of $\theta$.
Deduce, or show otherwise, that the minimum and maximum distances from the centre to the circumference of this ellipse are $\tan\theta$ and $\cot\theta$.

Solution: \begin{questionparts} \item \begin{align*} R_{-\alpha}AR_{\alpha} &= \begin{pmatrix} \cos \alpha & \sin\alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix}\begin{pmatrix} a & b \\ b & c \end{pmatrix} \begin{pmatrix} \cos \alpha & -\sin\alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} \\ &= \begin{pmatrix} \cos \alpha & \sin\alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix} \begin{pmatrix} a\cos \alpha + b \sin \alpha & -a\sin\alpha + b \cos\alpha \\ b\cos\alpha + c \sin\alpha & c\cos\alpha-b\sin\alpha \end{pmatrix} \\ &= \begin{pmatrix} a\cos^2\alpha+2b\sin\alpha\cos\alpha+c\sin^2\alpha & -a\sin\alpha\cos \alpha+b\cos^2\alpha +c\sin\alpha\cos\alpha-b\sin^2 \alpha\\ (c-a)\sin\alpha\cos \alpha +b(\cos^2\alpha-\sin^2 \alpha) & a\sin^2 \alpha -2b\sin\alpha\cos\alpha+c\cos^2\alpha \end{pmatrix} \\ &= \begin{pmatrix} * & \frac{c-a}{2}\sin2\alpha+b \cos 2\alpha\\\frac{c-a}{2}\sin2\alpha+b \cos 2\alpha & * \end{pmatrix} \end{align*} Therefore this will be diagonal if $\tan 2\alpha = \frac{2b}{a-c} \Rightarrow \alpha = \frac12 \tan^{-1} \l \frac{2b}{a-c} \r$ \item \begin{align*} x^2+(y+2x\cot2\theta)^2 &= x^2(1 + 4\cot^22\theta) + 4\cot2\theta xy + y^2 \\ &= \begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix} 1 + 4\cot^22\theta & 2\cot 2\theta \\ 2\cot 2\theta & 1 \end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} \end{align*} Plugging this $\mathbf{A}$ in our result from before we discover \begin{align*} \frac12 \tan^{-1} \l \frac{2b}{a-c} \r &= \frac12 \tan^{-1} \l \frac{4\cot 2\theta}{1 + 4\cot^22\theta-1} \r \\ &= \frac12 \tan^{-1} \l \tan 2 \theta \r \\ &= \theta \end{align*} Therefore, the matrix will be: \begin{align*} & \textrm{diag}\begin{pmatrix} (1+4\cot^2 2\theta)\cos^2 \theta + 4\cot2\theta \sin\theta\cos\theta + \sin^2 \theta \\ (1+4\cot^2 2\theta)\sin^2 \theta - 4\cot2\theta \sin\theta\cos\theta + \cos^2 \theta \end{pmatrix} \\ =& \textrm{diag}\begin{pmatrix} \cos^2\theta + \frac{\cos^2 2\theta}{\sin^2 \theta} + 2\cos 2\theta + \sin^2 \theta \\ \sin^2\theta + \frac{\cos^2 2\theta}{\cos^2 \theta} - 2\cos 2\theta + \cos^2 \theta \end{pmatrix} \\ =& \textrm{diag}\begin{pmatrix} 1 + \cos 2\theta \l \frac{\cos2\theta}{\sin^2 \theta} + 2\r \\ 1 + \cos 2\theta \l \frac{\cos2\theta}{\cos^2 \theta} - 2\r \\ \end{pmatrix} \\ =& \textrm{diag}\begin{pmatrix} 1 + \cos 2\theta \l \frac{\cos^2 \theta + \sin^2 \theta}{\sin^2 \theta}\r \\ 1 -\cos 2\theta \l \frac{-\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta}\r \\ \end{pmatrix} \\ =& \textrm{diag}\begin{pmatrix} 1 + (\cos^2\theta - \sin^2 \theta) \cosec^2 \theta \\ 1 - (\cos^2\theta - \sin^2 \theta) \sec^2 \theta \\ \end{pmatrix} \\ =& \textrm{diag}\begin{pmatrix} \cot^2 \theta \\ \tan^2 \theta \\ \end{pmatrix} \\ \end{align*} Therefore this is a rotation of an ellipse with equation: $(\cot \theta x)^2 + (\tan \theta y)^2 = 1$, ie the shortest side and longest side are $\cot \theta$ and $\tan \theta$ respectively, but we know since $0 < \theta < \tfrac{1}{4}\pi$ the shortest will be $\tan \theta$ and the longest $\cot \theta$.

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1995 Paper 3 Q7

D: 1654.7 B: 1516.0

groups group axioms closure identity element inverse isomorphism complex numbers modular arithmetic matrix rotation matrix unit modulus

Consider the following sets with the usual definition of multiplication appropriate to each. In each case you may assume that the multiplication is associative. In each case state, giving adequate reasons, whether or not the set is a group.

the complex numbers of unit modulus;
the integers modulo 4;
the matrices \[ \mathrm{M}(\theta)=\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix}, \] where $0\leqslant\theta<2\pi$;
the integers $1,3,5,7$ modulo 8;
the $2\times2$ matrices all of whose entries are integers;
the integers $1,2,3,4$ modulo 5.

In the case of each pair of groups above state, with reasons, whether or not they are isomorphic.

Solution:

$\{ z \in \mathbb{C} : |z| = 1\}$ is a group.
1. (Closure) $|z_1z_2| = |z_1||z_2| = 1$. Set is closed under multiplication
2. (Associativity) Multiplication of complex numbers is associative
3. (Identity) $|1| = 1$
4. (Inverses) $| \frac{1}{z} | = \frac{1}{|z|} = \frac{1}{1} = 1$, the set contains inverses
the integers $\pmod{4}$ are not a group under multiplication, $2$ has no inverse, since $0 \times k \equiv 0 \pmod{4}$
The set of rotation matrices is a group:
1. (Closure) \begin{align*} \begin{pmatrix}\cos\theta_1 & -\sin\theta_1\\ \sin\theta_1 & \cos\theta_1 \end{pmatrix} \begin{pmatrix}\cos\theta_2 & -\sin\theta_2\\ \sin\theta_2 & \cos\theta_2 \end{pmatrix} &= {\scriptscriptstyle\begin{pmatrix}\cos\theta_1 \cos \theta_2 - \sin \theta_1\sin \theta_2 & -\sin\theta_1\ \cos \theta_1 - \sin \theta_2\cos\theta_1\\ \sin\theta_1\ \cos \theta_1 + \sin \theta_2\cos\theta_1 & \cos\theta_1 \cos \theta_2 - \sin \theta_1\sin \theta_2 \end{pmatrix}} \\ &= \begin{pmatrix}\cos(\theta_1+\theta_2) & -\sin(\theta_1+\theta_2)\\ \sin(\theta_1+\theta_2) & \cos(\theta_1+\theta_2) \end{pmatrix} \end{align*} Since $\cos, \sin$ are periodic with period $2\pi$, we can find $\theta_3 = \theta_1 + \theta_2 + 2k\pi$ such that $0 \leq \theta_3 < 2 \pi$, so our set is closed
2. (Associativity) Matrix multiplication is associative
3. (Identity) Consider $\theta = 0$
4. (Inverses) Consider $2\pi - \theta$
$\{1, 3, 5, 7\} \pmod{8}$ is a group:
1 3 5 7
1 1 3 5 7
3 3 1 7 5
5 5 7 1 3
7 7 5 3 1
1. (Closure) See Cayley table
2. (Associativity) Integer multiplication is associative
3. (Identity) $1$
4. (Inverses) $x \mapsto x$ (See Cayley table)
$2\times2$ matrices are not a group, consider $0 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$, then $\mathbf{0}\mathbf{M} = \mathbf{0}$ for all other matrices.
1 2 3 4
1 1 2 3 4
2 2 4 1 3
3 3 1 4 2
4 4 3 2 1
1. (Closure) See Cayley table
2. (Associativity) Integer multiplication is associative
3. (Identity) $1$
4. (Inverses) $1 \mapsto 1, 2 \mapsto 3, 3 \mapsto 2, 4 \mapsto 4$ (See Cayley table)

	(i)	(iii)	(iv)	(vi)
(i)	$\checkmark$	$\checkmark$ consider \(z \mapsto \begin{pmatrix} \cos \arg (z)	- \sin \arg(z)
\sin \arg(z)	\cos \arg(z) \end{pmatrix}\)	not finite	not finite
(iii)		$\checkmark$	not finite	not finite
(iv)			$\checkmark$	no element order $4$
(vi)				$\checkmark$

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1989 Paper 3 Q3

D: 1675.2 B: 1469.0

groups rotation matrix matrix geometric series proof by induction recurrence relation binary operation finite group

The matrix $\mathbf{M}$ is given by \[ \mathbf{M}=\begin{pmatrix}\cos(2\pi/m) & -\sin(2\pi/m)\\ \sin(2\pi/m) & \cos(2\pi/m) \end{pmatrix}, \] where $m$ is an integer greater than $1.$ Prove that \[ \mathbf{M}^{m-1}+\mathbf{M}^{m-2}+\cdots+\mathbf{M}^{2}+\mathbf{M}+\mathbf{I}=\mathbf{O}, \] where $\mathbf{I}=\begin{pmatrix}1 & 0\\ 0 & 1 \end{pmatrix}$ and $\mathbf{O}=\begin{pmatrix}0 & 0\\ 0 & 0 \end{pmatrix}.$ The sequence $\mathbf{X}_{0},\mathbf{X}_{1},\mathbf{X}_{2},\ldots$ is defined by \[ \mathbf{X}_{k+1}=\mathbf{PX}_{k}+\mathbf{Q}, \] where $\mathbf{P,Q}$ and $\mathbf{X}_{0}$ are given $2\times2$ matrices. Suggest a suitable expression for $\mathbf{X}_{k}$ in terms of $\mathbf{P},$ $\mathbf{Q}$ and $\mathbf{X}_{0},$ and justify it by induction. The binary operation $*$ is defined as follows: \[ \mathbf{X}_{i}*\mathbf{X}_{j}\mbox{ is the result of substituting \ensuremath{\mathbf{X}_{j}}for \ensuremath{\mathbf{X}_{0}}in the expression for \ensuremath{\mathbf{X}_{i}}. } \] Show that if $\mathbf{P=M},$ the set $\{\mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3},\ldots\}$ forms a finite group under the operation $*$.

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