Problems

2024 Paper 3 Q5

D: 1500.0 B: 1500.0

trace determinant matrix differential equation 2x2 matrices commutator inverse matrix proof

In this question, $\mathbf{M}$ and $\mathbf{N}$ are non-singular $2 \times 2$ matrices. The \emph{trace} of the matrix $\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is defined as $\operatorname{tr}(\mathbf{M}) = a + d$.

Prove that, for any two matrices $\mathbf{M}$ and $\mathbf{N}$, $\operatorname{tr}(\mathbf{MN}) = \operatorname{tr}(\mathbf{NM})$ and derive an expression for $\operatorname{tr}(\mathbf{M}+\mathbf{N})$ in terms of $\operatorname{tr}(\mathbf{M})$ and $\operatorname{tr}(\mathbf{N})$.

The entries in matrix $\mathbf{M}$ are functions of $t$ and $\dfrac{\mathrm{d}\mathbf{M}}{\mathrm{d}t}$ denotes the matrix whose entries are the derivatives of the corresponding entries in $\mathbf{M}$.

Show that \[\frac{1}{\det \mathbf{M}} \frac{\mathrm{d}}{\mathrm{d}t}(\det \mathbf{M}) = \operatorname{tr}\!\left(\mathbf{M}^{-1} \frac{\mathrm{d}\mathbf{M}}{\mathrm{d}t}\right).\]
In this part, matrix $\mathbf{M}$ satisfies the differential equation \[\frac{\mathrm{d}\mathbf{M}}{\mathrm{d}t} = \mathbf{MN} - \mathbf{NM},\] where the entries in matrix $\mathbf{N}$ are also functions of $t$. Show that $\det \mathbf{M}$, $\operatorname{tr}(\mathbf{M})$ and $\operatorname{tr}(\mathbf{M}^2)$ are independent of $t$. In the case $\mathbf{N} = \begin{pmatrix} t & t \\ 0 & t \end{pmatrix}$, and given that $\mathbf{M} = \begin{pmatrix} A & B \\ C & D \end{pmatrix}$ when $t = 0$, find $\mathbf{M}$ as a function of $t$.
In this part, matrix $\mathbf{M}$ satisfies the differential equation \[\frac{\mathrm{d}\mathbf{M}}{\mathrm{d}t} = \mathbf{MN},\] where the entries in matrix $\mathbf{N}$ are again functions of $t$. The trace of $\mathbf{M}$ is non-zero and independent of $t$. Is it necessarily true that $\operatorname{tr}(\mathbf{N}) = 0$?

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2023 Paper 2 Q6

D: 1500.0 B: 1500.0

proof by induction fibonacci sequence matrix exponentiation recurrence relation binomial theorem determinant 2x2 matrices

The sequence $F_n$, for $n = 0, 1, 2, \ldots$, is defined by $F_0 = 0$, $F_1 = 1$ and by $F_{n+2} = F_{n+1} + F_n$ for $n \geqslant 0$. Prove by induction that, for all positive integers $n$, \[\begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix} = \mathbf{Q}^n,\] where the matrix $\mathbf{Q}$ is given by \[\mathbf{Q} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}.\]

By considering the matrix $\mathbf{Q}^n$, show that $F_{n+1}F_{n-1} - F_n^2 = (-1)^n$ for all positive integers $n$.
By considering the matrix $\mathbf{Q}^{m+n}$, show that $F_{m+n} = F_{m+1}F_n + F_m F_{n-1}$ for all positive integers $m$ and $n$.
Show that $\mathbf{Q}^2 = \mathbf{I} + \mathbf{Q}$. In the following parts, you may use without proof the Binomial Theorem for matrices: \[(\mathbf{I} + \mathbf{A})^n = \sum_{k=0}^{n} \binom{n}{k} \mathbf{A}^k.\]
1. Show that, for all positive integers $n$, \[F_{2n} = \sum_{k=0}^{n} \binom{n}{k} F_k\,.\]
2. Show that, for all positive integers $n$, \[F_{3n} = \sum_{k=0}^{n} \binom{n}{k} 2^k F_k\] and also that \[F_{3n} = \sum_{k=0}^{n} \binom{n}{k} F_{n+k}\,.\]
3. Show that, for all positive integers $n$, \[\sum_{k=0}^{n} (-1)^{n+k} \binom{n}{k} F_{n+k} = 0\,.\]

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2021 Paper 2 Q7

D: 1500.0 B: 1500.0

rotation matrices matrix algebra determinant cayley-hamilton theorem 2x2 matrices trigonometric identities proof

The matrix $\mathbf{R}$ represents an anticlockwise rotation through angle $\varphi$ ($0^\circ \leqslant \varphi < 360^\circ$) in two dimensions, and the matrix $\mathbf{R} + \mathbf{I}$ also represents a rotation in two dimensions. Determine the possible values of $\varphi$ and deduce that $\mathbf{R}^3 = \mathbf{I}$.
Let $\mathbf{S}$ be a real matrix with $\mathbf{S}^3 = \mathbf{I}$, but $\mathbf{S} \neq \mathbf{I}$. Show that $\det(\mathbf{S}) = 1$. Given that \[ \mathbf{S} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \] show that $\mathbf{S}^2 = (a+d)\mathbf{S} - \mathbf{I}$. Hence prove that $a + d = -1$.
Let $\mathbf{S}$ be a real $2 \times 2$ matrix. Show that if $\mathbf{S}^3 = \mathbf{I}$ and $\mathbf{S} + \mathbf{I}$ represents a rotation, then $\mathbf{S}$ also represents a rotation. What are the possible angles of the rotation represented by $\mathbf{S}$?

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2021 Paper 3 Q2

D: 1500.0 B: 1500.0

3x3 matrices determinant inequality algebraic manipulation matrix equation am-gm inequality

Let \[ x = \frac{a}{b - c}, \qquad y = \frac{b}{c - a} \qquad \text{and} \qquad z = \frac{c}{a - b}, \] where $a$, $b$ and $c$ are distinct real numbers. Show that \[ \begin{pmatrix} 1 & -x & x \\ y & 1 & -y \\ -z & z & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \] and use this result to deduce that $yz + zx + xy = -1$. Hence show that \[ \frac{a^2}{(b-c)^2} + \frac{b^2}{(c-a)^2} + \frac{c^2}{(a-b)^2} \geqslant 2. \]
Let \[ x = \frac{2a}{b+c}, \qquad y = \frac{2b}{c+a} \qquad \text{and} \qquad z = \frac{2c}{a+b}, \] where $a$, $b$ and $c$ are positive real numbers. Using a suitable matrix, show that $xyz + yz + zx + xy = 4$. Hence show that \[ (2a + b + c)(a + 2b + c)(a + b + 2c) > 5(b+c)(c+a)(a+b). \] Show further that \[ (2a + b + c)(a + 2b + c)(a + b + 2c) > 7(b+c)(c+a)(a+b). \]

Solution:

$\,$ \begin{align*} && \begin{pmatrix} 1 & -x & x \\ y & 1 & -y \\ -z & z & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} &= \begin{pmatrix} a-xb+xc \\ ay+b-yc \\ -za+zb+c \end{pmatrix} \\ &&&= \begin{pmatrix} a-x(b-c) \\ b-y(c-a) \\ c-z(a-b) \end{pmatrix} \\ &&&= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \end{align*} Notice since $a,b$ and $c$ are distinct real numbers the vector $\langle a,b,c \rangle$ cannot be the zero vector, so the determinant of the matrix is zero, ie $0= 1(1+yz)+x(y-yz)+x(yz+z) = 1 +yz+yx+zx$. Notice also then that \begin{align*} && \frac{a^2}{(b-c)^2} + \frac{b^2}{(c-a)^2} + \frac{c^2}{(a-b)^2} &= x^2+y^2+z^2 \\ &&&= (x+y+z)^2 - 2(xy+yz+zx) \\ &&&= 2 + (x+y+z)^2 \geq 2 \end{align*}
$\,$ \begin{align*} && \begin{pmatrix} 2 & -x & -x \\ -y & 2 & -y \\ -z & -z & 2 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} &= \begin{pmatrix} 2a-xb-xc \\ -ay+2b-yc \\ -za-zb+2c \end{pmatrix} \\ &&&= \begin{pmatrix} 2a-x(b+c) \\ 2b-y(c+a) \\ 2c-z(a+b) \end{pmatrix} \\ &&&= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \\ \Rightarrow && 0 &= \det \begin{pmatrix} 2 & -x & -x \\ -y & 2 & -y \\ -z & -z & 2 \end{pmatrix} \\ &&&= 2(4 -yz)+x(-2y-yz)-x(yz+2z) \\ &&&= 8 - 2yz-2yx-2xyz-2zx\\ \Rightarrow && 4 &= xyz+xy+yz+zx \end{align*} \begin{align*} && (2a + b + c)(a + 2b + c)(a + b + 2c) &> 5(b+c)(c+a)(a+b) \\ \Leftrightarrow && \left ( \frac{2a}{b+c}+1 \right)\left ( \frac{2b}{c+a}+1 \right)\left ( \frac{2c}{a+b}+1 \right) &> 5 \\ \Leftrightarrow && \left ( x+1 \right)\left ( y+1 \right)\left ( x+1 \right) &> 5 \\ \Leftrightarrow && xyz+xy+yz+zx+x+y+z+1 &> 5 \\ \Leftrightarrow && 5+x+y+z&> 5 \\ \end{align*} Which is clearly true since if $a,b,c$ are positve real numbers so are $x,y,z$. This final inequality is equivalent to showing $x+y+z > 2$ ie \begin{align*} && x+y+z &> 2 \\ \Leftrightarrow && \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} &> 1 \\ \\ && \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} & > \frac{a}{a+b+c} + \frac{b}{a+b+c} + \frac{c}{a+b+c} = 1 \end{align*} So we're done.

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2020 Paper 2 Q6

D: 1500.0 B: 1500.0

matrices trace determinant proof matrix powers necessary and sufficient conditions involutory matrix

A $2 \times 2$ matrix $\mathbf{M}$ is real if it can be written as $\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, where $a$, $b$, $c$ and $d$ are real. In this case, the \emph{trace} of matrix $\mathbf{M}$ is defined to be $\mathrm{tr}(\mathbf{M}) = a + d$ and $\det(\mathbf{M})$ is the determinant of matrix $\mathbf{M}$. In this question, $\mathbf{M}$ is a real $2 \times 2$ matrix.

Prove that \[\mathrm{tr}(\mathbf{M}^2) = \mathrm{tr}(\mathbf{M})^2 - 2\det(\mathbf{M}).\]
Prove that \[\mathbf{M}^2 = \mathbf{I} \text{ but } \mathbf{M} \neq \pm\mathbf{I} \iff \mathrm{tr}(\mathbf{M}) = 0 \text{ and } \det(\mathbf{M}) = -1,\] and that \[\mathbf{M}^2 = -\mathbf{I} \iff \mathrm{tr}(\mathbf{M}) = 0 \text{ and } \det(\mathbf{M}) = 1.\]
Use part (ii) to prove that \[\mathbf{M}^4 = \mathbf{I} \iff \mathbf{M}^2 = \pm\mathbf{I}.\] Find a necessary and sufficient condition on $\det(\mathbf{M})$ and $\mathrm{tr}(\mathbf{M})$ so that $\mathbf{M}^4 = -\mathbf{I}$.
Give an example of a matrix $\mathbf{M}$ for which $\mathbf{M}^8 = \mathbf{I}$, but which does not represent a rotation or reflection. [Note that the matrices $\pm\mathbf{I}$ are both rotations.]

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2019 Paper 2 Q8

D: 1500.0 B: 1638.7

matrices multiplicative function determinant 2x2 matrix identity matrix matrix properties proof function properties

The domain of the function f is the set of all $2 \times 2$ matrices and its range is the set of real numbers. Thus, if $M$ is a $2 \times 2$ matrix, then $f(M) \in \mathbb{R}$. The function f has the property that $f(MN) = f(M)f(N)$ for any $2 \times 2$ matrices $M$ and $N$.

You are given that there is a matrix $M$ such that $f(M) \neq 0$. Let $I$ be the $2 \times 2$ identity matrix. By considering $f(MI)$, show that $f(I) = 1$.
Let $J = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. You are given that $f(J) \neq 1$. By considering $J^2$, evaluate $f(J)$. Using $J$, show that, for any real numbers $a$, $b$, $c$ and $d$, $$.f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right) = -f\left(\begin{pmatrix} c & d \\ a & b \end{pmatrix}\right) = f\left(\begin{pmatrix} d & c \\ b & a \end{pmatrix}\right)$$
Let $K = \begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}$ where $k \in \mathbb{R}$. Use $K$ to show that, if the second row of the matrix $A$ is a multiple of the first row, then $f(A) = 0$.
Let $P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. By considering the matrices $P^2$, $P^{-1}$, and $K^{-1}PK$ for suitable values of $k$, evaluate $f(P)$.

Solution:

Consider $f(M) = f(MI) = f(M)f(I)$. Since $f(M) \neq 0$ we can divide by $f(M)$ to obtain $f(I) = 1$
Let $J = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, then $J^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$. Therefore $1 = f(I) = f(J^2) = f(J)f(J) \Rightarrow f(J) = \pm 1 \Rightarrow f(J) = -1$ since $f(J) \neq 1$. \begin{align*} \begin{pmatrix} a & b \\ c & d \end{pmatrix}J &= \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\ &= \begin{pmatrix} b & a \\ d & c \end{pmatrix} \\ J\begin{pmatrix} a & b \\ c & d \end{pmatrix} &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} \\ &= \begin{pmatrix} c & d \\ a & b \end{pmatrix} \\ J\begin{pmatrix} a & b \\ c & d \end{pmatrix}J &=\begin{pmatrix} d & c \\ b & a \end{pmatrix} \end{align*} Therefore $f\left(\begin{pmatrix} c & d \\ a & b \end{pmatrix}\right) = f \left (J\begin{pmatrix} a & b \\ c & d \end{pmatrix} \right) = f(J) f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right) = -f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right)$ and $f\left(\begin{pmatrix} d & c \\ b & a \end{pmatrix}\right) = f\left(J\begin{pmatrix} a & b \\ c & d \end{pmatrix}J \right) = f(J)f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right)f(J) = f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right)$ as required.
First consider $O$ the matrix of $0$, then \begin{align*} && JO &= O \\ \Rightarrow && f(JO) &= f(O) \\ \Rightarrow && f(J)f(O) &= f(O) \\ \Rightarrow && -f(O) &= f(O) \\ \Rightarrow && f(O) &= 0 \end{align*} Now consider $K_{k} = \begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}$. Suppose $A = \begin{pmatrix} a & b \\ ka & kb \end{pmatrix}$ then \begin{align*} K_{\frac1k}A &= \begin{pmatrix} 1 & 0 \\ 0 & \frac1k \end{pmatrix} \begin{pmatrix} a & b \\ ka & kb \end{pmatrix} \\ &= \begin{pmatrix} a & b \\ a & b \end{pmatrix} \end{align*} And so $f(K_{\frac1k}A) = f\left ( \begin{pmatrix} a & b \\ a & b \end{pmatrix} \right) = - f \left ( \begin{pmatrix} a & b \\ a & b \end{pmatrix} \right) = 0$, therefore either $f(K_{\frac1k}) = 0$ or $f(A) = 0$, but we know that $f(I) \neq 0$ therefore $f(A) = 0$.
Let $P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, then $P^2 = \begin{pmatrix} 1 &2 \\ 0 & 1 \end{pmatrix}$, $P^{-1} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$, $K_k^{-1}PK_k = K_k^{-1}\begin{pmatrix} 1 & k \\ 0 & k \end{pmatrix} = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$. If $A$ has an inverse then $f(A) \neq 0$ since $1 = f(I) = f(A)f(A^{-1})$, in particular, $f(A)f(A^{-1}) = 1$. Using this for $K_2$ we have: $f(P)^2 = f(P^2) = f(K_2^{-1}PK_2) = f(P)$ therefore $f(P) = 0, 1$, but since $f(P)$ has an inverse, $f(P) \neq 0$ so $f(P) = 1$

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1996 Paper 2 Q2

D: 1600.0 B: 1500.0

3x3 matrices simultaneous equations determinant substitution nonlinear equations Cramer's rule symmetric functions

Consider the system of equations \begin{alignat*}{1} 2yz+zx-5xy & =2\\ yz-zx+2xy & =1\\ yz-2zx+6xy & =3 \end{alignat*} Show that \[xyz=\pm 6\] and find the possible values of $x$, $y$ and $z$.

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1996 Paper 3 Q6

D: 1674.0 B: 1529.9

groups matrix group axioms closure singular matrix determinant proof matrix multiplication

Let $S$ be the set of matrices of the form \[ \begin{pmatrix}a & a\\ a & a \end{pmatrix}, \] where $a$ is any real non-zero number. Show that $S$ is closed under matrix multiplication and, further, that $S$ is a group under matrix multiplication.
Let $G$ be a set of $n\times n$ matrices which is a group under matrix multiplication, with identity element $\mathbf{E}.$ By considering equations of the form $\mathbf{BC=D}$ for suitable elements $\mathbf{B},$ $\mathbf{C}$ and $\mathbf{D}$ of $G$, show that if a given element $\mathbf{A}$ of $G$ is a singular matrix (i.e. $\det\mathbf{A}=0$), then all elements of $G$ are singular. Give, with justification, an example of such a group of singular matrices in the case $n=3.$

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1994 Paper 3 Q8

D: 1700.0 B: 1516.0

complex numbers matrix determinant sum of squares bilinear forms complex conjugate matrix multiplication

Let $a,b,c,d,p,q,r$ and $s$ be real numbers. By considering the determinant of the matrix product \[ \begin{pmatrix}z_{1} & z_{2}\\ -z_{2}^{*} & z_{1}^{*} \end{pmatrix}\begin{pmatrix}z_{3} & z_{4}\\ -z_{4}^{*} & z_{3}^{*} \end{pmatrix}, \] where $z_{1},z_{2},z_{3}$ and $z_{4}$ are suitably chosen complex numbers, find expressions $L_{1},L_{2},L_{3}$ and $L_{4},$ each of which is linear in $a,b,c$ and $d$ and also linear in $p,q,r$ and $s,$ such that \[ (a^{2}+b^{2}+c^{2}+d^{2})(p^{2}+q^{2}+r^{2}+s^{2})=L_{1}^{2}+L_{2}^{2}+L_{3}^{2}+L_{4}^{2}. \]

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1993 Paper 2 Q6

D: 1600.0 B: 1516.0

matrices 2x2 matrix matrix multiplication determinant counterexample proof true or false commutativity singular matrix

In this question, $\mathbf{A,\mathbf{B$ }}and $\mathbf{X$ are non-zero $2\times2$ real matrices.} Are the following assertions true or false? You must provide a proof or a counterexample in each case.

If $\mathbf{AB=0}$ then $\mathbf{BA=0}.$
$(\mathbf{A-B)(A+B)=}\mathbf{A}^{2}-\mathbf{B}^{2}.$
The equation $\mathbf{AX=0}$ has a non-zero solution $\mathbf{X}$ if and only if $\det\mathbf{A}=0.$
For any $\mathbf{A}$ and $\mathbf{B}$ there are at most two matrices $\mathbf{X}$ such that $\mathbf{X}^{2}+\mathbf{AX}+\mathbf{B}=\mathbf{0}.$

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1993 Paper 3 Q10

D: 1700.0 B: 1484.0

linear transformations matrix parallelogram coordinate geometry area preservation determinant parallel lines

The transformation $T$ of the point $P$ in the $x$,$y$ plane to the point $P'$ is constructed as follows: \hfil\break Lines are drawn through $P$ parallel to the lines $y=mx$ and $y=-mx$ to cut the line $y=kx$ at $Q$ and $R$ respectively, $m$ and $k$ being given constants. $P'$ is the fourth vertex of the parallelogram $PQP'R$. Show that if $P$ is $(x_1,y_1)$ then $Q$ is $$ \left( {mx_1-y_1 \over m-k}, {k(mx_1-y_1)\over m-k}\right). $$ Obtain the coordinates of $P'$ in terms of $x_1$, $y_1$, $m$ and $k$, and express $T$ as a matrix transformation. Show that areas are transformed under $T$ into areas of the same magnitude.

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1992 Paper 2 Q5

D: 1577.1 B: 1470.1

groups matrix order of element inverse determinant group axioms proof

Explain what is meant by the order of an element $g$ of a group $G$. The set $S$ consists of all $2\times2$ matrices whose determinant is $1$. Find the inverse of the element $\mathbf{A}$ of $S$, where \[ \mathbf{A}=\begin{pmatrix}w & x\\ y & z \end{pmatrix}. \] Show that $S$ is a group under matrix multiplication (you may assume that matrix multiplication is associative). For which elements $\mathbf{A}$ is $\mathbf{A}^{-1}=\mathbf{A}$? Which element or elements have order 2? Show that the element $\mathbf{A}$ of $S$ has order 3 if, and only if, $w+z+1=0.$ Write down one such element.

Solution: The order of an element $g$ is the smallest positive number $k$ such that $g^k = e$. $\mathbf{A}^{-1} = \begin{pmatrix}z & -x\\ -y & w \end{pmatrix}$. Claim, $S$ is a group. \begin{enumerate} \item (Closure) The product of two $2\times2$ matrices is always a $2\times 2$ matrix so we only need to check the determinant. Suppose $\det(\mathbf{A}) = \det (\mathbf{B}) = 1$, then $\det(AB) = \det(A)\det(B) = 1$, so our operation is closed \item (Associativity) Inherited from matrix multiplication \item (Identity) $\mathbf{I} =\begin{pmatrix}1 & 0\\ 1 & 1 \end{pmatrix}$ has determinant $1$. \item (Inverses) The inverse is always fine since the matrix of cofactors always contains integers and the determinant is one, so we never end up with anything which isn't an integer. \end{itemize} If $\mathbf{A}^-1 = \mathbf{A}$ then assuming $\mathbf{A} = \begin{pmatrix}a & b\\ c & d \end{pmatrix}$ then $\mathbf{A}^{-1} = \begin{pmatrix}d & - b\\ -c & a \end{pmatrix}$ so we must have $a=d, -b=b, -c=c$, so $b = c = 0$ and $a = d$. For the determinant to be $1$ we must have $ad = a^2 = 1$, ie $a = \pm 1$. Therefore we must have $\mathbf{A} = \begin{pmatrix}1 & 0\\ 0 & 1 \end{pmatrix}$ or $\mathbf{A} = \begin{pmatrix}-1 & 0\\ 0 & -1 \end{pmatrix}$. For an element to have order $2$ then $\mathbf{A}^2 = \mathbf{I}$ ie, $\mathbf{A} = \mathbf{A}^{-1}$ and $\mathbf{A} \neq \mathbf{I}$ therefore the only element of order $2$ is $\begin{pmatrix}-1 & 0\\ 0 & -1 \end{pmatrix}$. For an element to have order $3$ we must have $\mathbf{A}^2 = \mathbf{A}^{-1}$, ie $\begin{pmatrix}w^2 + xy & x(w+z)\\ y(w+z) & z^2 + xy \end{pmatrix} = \begin{pmatrix}z & -x\\ -y & w \end{pmatrix}$. Therefore $w^2 + xy = z, x(w+z) = -x, y(w+z) = -y, z^2+xy = w$. The second and third equations are satisfied iff $w+z+1 = 0$ or $x = 0$ and $y = 0$, but if $x = 0$ and $y = 0$ then we aren't order $3$, so we just need to check this is sufficient for the first and last equations. Since $\det(\mathbf{A}) = 1$ we have $wz =xy +1$, so the first and last equations are equivalent to $w^2 + wz - 1 = z$ and $x^2 + wz-1 = w$ which are equivalent to $w(w+z) = z+1$ or $w + z+ 1 = 0$ as required

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1991 Paper 3 Q2

D: 1700.0 B: 1516.0

complex numbers similar triangles 3x3 matrices determinant Argand diagram geometric proof centroid equilateral triangle factorisation

The distinct points $P_{1},P_{2},P_{3},Q_{1},Q_{2}$ and $Q_{3}$ in the Argand diagram are represented by the complex numbers $z_{1},z_{2},z_{3},w_{1},w_{2}$ and $w_{3}$ respectively. Show that the triangles $P_{1}P_{2}P_{3}$ and $Q_{1}Q_{2}Q_{3}$ are similar, with $P_{i}$ corresponding to $Q_{i}$ ($i=1,2,3$) and the rotation from $1$ to $2$ to $3$ being in the same sense for both triangles, if and only if \[ \frac{z_{1}-z_{2}}{z_{2}-z_{3}}=\frac{w_{1}-w_{2}}{w_{1}-w_{3}}. \] Verify that this condition may be written \[ \det\begin{pmatrix}z_{1} & z_{2} & z_{3}\\ w_{1} & w_{2} & w_{3}\\ 1 & 1 & 1 \end{pmatrix}=0. \]

Show that if $w_{i}=z_{i}^{2}$ ($i=1,2,3$) then triangle $P_{1}P_{2}P_{3}$ is not similar to triangle $Q_{1}Q_{2}Q_{3}.$
Show that if $w_{i}=z_{i}^{3}$ ($i=1,2,3$) then triangle $P_{1}P_{2}P_{3}$ is similar to triangle $Q_{1}Q_{2}Q_{3}$ if and only if the centroid of triangle $P_{1}P_{2}P_{3}$ is the origin. {[}The centroid of triangle $P_{1}P_{2}P_{3}$ is represented by the complex number $\frac{1}{3}(z_{1}+z_{2}+z_{3})$.{]}
Show that the triangle $P_{1}P_{2}P_{3}$ is equilateral if and only if \[ z_{2}z_{3}+z_{3}z_{1}+z_{1}z_{2}=z_{1}^{2}+z_{2}^{2}+z_{3}^{2}. \]

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1990 Paper 2 Q10

D: 1600.0 B: 1496.1

matrices determinant matrix product zero matrix polynomial factorisation upper triangular matrix Cayley-Hamilton counterexample

Two square matrices $\mathbf{A}$ and $\mathbf{B}$ satisfies $\mathbf{AB=0}.$ Show that either $\det\mathbf{A}=0$ or $\det\mathbf{B}=0$ or $\det\mathbf{A}=\det\mathbf{B}=0$. If $\det\mathbf{B}\neq0$, what must $\mathbf{A}$ be? Give an example to show that the condition $\det\mathbf{A}=\det\mathbf{B}=0$ is not sufficient for the equation $\mathbf{AB=0}$ to hold. Find real numbers $p,q$ and $r$ such that \[ \mathbf{M}^{3}+2\mathbf{M}^{2}-5\mathbf{M}-6\mathbf{I}=(\mathbf{M}+p\mathbf{I})(\mathbf{M}+q\mathbf{I})(\mathbf{M}+r\mathbf{I}), \] where $\mathbf{M}$ is any square matrix and $\mathbf{I}$ is the appropriate identity matrix. Hence, or otherwise, find all matrices $\mathbf{M}$ of the form $\begin{pmatrix}a & c\\ 0 & b \end{pmatrix}$ which satisfy the equation \[ \mathbf{M}^{3}+2\mathbf{M}^{2}-5\mathbf{M}-6\mathbf{I}=\mathbf{0}. \]

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1988 Paper 3 Q8

D: 1700.0 B: 1500.0

conics parabola tangent-normal parametric determinant area of triangle concurrent lines

Find the equations of the tangent and normal to the parabola $y^{2}=4ax$ at the point $(at^{2},2at).$ For $i=1,2,$ and 3, let $P_{i}$ be the point $(at_{i}^{2},2at_{i}),$ where $t_{1},t_{2}$ and $t_{3}$ are all distinct. Let $A_{1}$ be the area of the triangle formed by the tangents at $P_{1},P_{2}$ and $P_{3},$ and let $A_{2}$ be the area of the triangle formed by the normals at $P_{1},P_{2}$ and $P_{3}.$ Using the fact that the area of the triangle with vertices at $(x_{1},y_{1}),(x_{2},y_{2})$ and $(x_{3},y_{3})$ is the absolute value of \[ \tfrac{1}{2}\det\begin{pmatrix}x_{1} & y_{1} & 1\\ x_{2} & y_{2} & 1\\ x_{3} & y_{3} & 1 \end{pmatrix}, \] show that $A_{3}=(t_{1}+t_{2}+t_{3})^{2}A_{1}.$ Deduce a necessary and sufficient condition in terms of $t_{1},t_{2}$ and $t_{3}$ for the normals at $P_{1},P_{2}$ and $P_{3}$ to be concurrent.

Solution: $\frac{dy}{dt} = 2a, \frac{dx}{dt} = 2at \Rightarrow \frac{dy}{dx} = \frac{1}{t}$. Therefore the equation of the tangent will be $\frac{y - 2at}{x-at^2} = \frac{1}{t} \Rightarrow y = \frac1tx +at$ and normal will be $\frac{y-2at}{x-at^2} = -t \Rightarrow y = t(at^2-x+2a)$. The tangents will meet when: \begin{align*} && \begin{cases} t_iy -x &= at_i^2 \\ t_j y - x &= at_j^2 \\ \end{cases} \\ \Rightarrow &&(t_i - t_j)y &= a(t_i-t_j)(t_i+t_j) \\ \Rightarrow && y &= a(t_i+t_j) \\ && x &= at_it_j \end{align*} The normals will meet when: \begin{align*} && \begin{cases} y +t_i x &= at_i^3+2at_i \\ y +t_j x &= at_j^3+2at_j \\ \end{cases} \\ \Rightarrow &&(t_i - t_j)x &= a(t_i-t_j)(t_i^2+t_it_j+t_j^2+2) \\ \Rightarrow && x&= a(t_i^2+t_it_j+t_j^2+2) \\ && y &= -at_it_j(t_i+t_j) \end{align*} Therefore the area of our triangles will be: \begin{align*} \tfrac{1}{2}\det\begin{pmatrix}at_1t_2 & a(t_1+t_2) & 1\\ at_2t_3 & a(t_2+t_3) & 1\\ at_3t_1 & a(t_3+t_1) & 1 \end{pmatrix} &= \frac{a^2}{2}\det\begin{pmatrix}t_1t_2 & (t_1+t_2) & 1\\ t_2t_3 & (t_2+t_3) & 1\\ t_3t_1 & (t_3+t_1) & 1 \end{pmatrix} \\ &= \frac{a^2}{2}\det\begin{pmatrix}t_1t_2 & (t_1+t_2) & 1\\ t_2(t_3-t_1) & (t_3-t_1) & 0\\ t_1(t_3-t_2) & (t_3-t_2) & 0 \end{pmatrix} \\ &= \frac{a^2}{2}|(t_2-t_1)(t_3-t_2)(t_1-t_3)| \end{align*} and \begin{align*} \tfrac{1}{2}\det\begin{pmatrix}a(t_1^2+t_1t_2+t_2^2+2) & -at_1t_2(t_1+t_2) & 1\\ a(t_2^2+t_2t_3+t_3^2+2) & -at_2t_3(t_2+t_3) & 1\\ a(t_3^2+t_3t_1+t_1^2+2) & -at_3t_1(t_3+t_1) & 1\\ \end{pmatrix} &= \frac{a^2}{2}\det\begin{pmatrix}(t_1^2+t_1t_2+t_2^2+2) & -t_1t_2(t_1+t_2) & 1\\ (t_2^2+t_2t_3+t_3^2+2) & -t_2t_3(t_2+t_3) & 1\\ (t_3^2+t_3t_1+t_1^2+2) & -t_3t_1(t_3+t_1) & 1\\ \end{pmatrix} \\ &= \frac{a^2}{2}\det\begin{pmatrix}(t_1^2+t_1t_2+t_2^2+2) & -t_1t_2(t_1+t_2) & 1\\ t_3^2-t_1^2+t_2(t_3-t_1) & t_2(t_1^2+t_1t_2-t_2t_3-t_3^2) & 0\\ t_3^2-t_2^2+t_1(t_3-t_2) & t_1(t_2^2+t_2t_1-t_1t_3-t_3^2) & 0\\ \end{pmatrix} \\ &= \frac{a^2}{2}\det\begin{pmatrix}(t_1^2+t_1t_2+t_2^2+2) & -t_1t_2(t_1+t_2) & 1\\ (t_3-t_1)(t_3+t_2+t_1) & t_2(t_1-t_3)(t_1+t_3+t_2) & 0\\ (t_3-t_2)(t_3+t_2+t_1) & t_1(t_2-t_3)(t_1+t_2+t_3)& 0\\ \end{pmatrix} \\ &= \frac{a^2}{2}(t_1+t_2+t_3)^2|(t_2-t_1)(t_3-t_2)(t_1-t_3)| \end{align*} as required. The normals will be concurrent iff the area of their triangle is $0$. This is certainly true if $t_1+t_2+t_3 = 0$. In fact the only if is also true, since no $3$ tangents can be concurrent.

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