Question 5 - A-Level Maths

CAIE Further Paper 1 2022 June — Question 5 12 marks

Exam Board	CAIE
Module	Further Paper 1 (Further Paper 1)
Year	2022
Session	June
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Proof by induction
Type	Prove matrix power formula
Difficulty	Standard +0.3 This is a straightforward Further Maths induction question with a standard matrix power formula. Part (b) requires routine application of the induction framework with simple matrix multiplication. Parts (a) and (c) add context but are also standard techniques (identifying shears and finding invariant lines via eigenvectors). The formula pattern is obvious and verification is mechanical, making this easier than average even for Further Maths.
Spec	4.01a Mathematical induction: construct proofs 4.03d Linear transformations 2D: reflection, rotation, enlargement, shear 4.03g Invariant points and lines

5 Let $\mathbf { A } = \left( \begin{array} { l l } 1 & a \\ 0 & 1 \end{array} \right)$, where $a$ is a positive constant.

State the type of the geometrical transformation in the $x - y$ plane represented by $\mathbf { A }$.
Prove by mathematical induction that, for all positive integers $n$, $$\mathbf { A } ^ { \mathrm { n } } = \left( \begin{array} { c c } 1 & \mathrm { na } \\ 0 & 1 \end{array} \right)$$ Let $\mathbf { B } = \left( \begin{array} { c c } b & b \\ a ^ { - 1 } & a ^ { - 1 } \end{array} \right)$, where $b$ is a positive constant.
Find the equations of the invariant lines, through the origin, of the transformation in the $x - y$ plane represented by $\mathbf { A } ^ { n } \mathbf { B }$.

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Question 5(a):

Answer	Marks	Guidance
A shear in the $x$-direction	B1	Accept 'shear'

Question 5(b):

Answer	Marks	Guidance
$\mathbf{A} = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}$ so true when $n=1$	B1	States base case
Assume true for $n=k$, so $\mathbf{A}^k = \begin{pmatrix} 1 & ka \\ 0 & 1 \end{pmatrix}$	B1	States inductive hypothesis
$\mathbf{A}^{k+1} = \begin{pmatrix} 1 & ka \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & a+ak \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & (k+1)a \\ 0 & 1 \end{pmatrix}$	M1 A1	Multiplies $\mathbf{A}^k$ with $\mathbf{A}$
If true for $[n=1$ and$]$ $n=k$ then also true for $n=k+1$. Hence by induction, true for all positive integers.	A1	Everything correct and states conclusion

Question 5(c):

Answer	Marks	Guidance
$\mathbf{A}^n\mathbf{B} = \begin{pmatrix} 1 & na \\ 0 & 1 \end{pmatrix}\begin{pmatrix} b & b \\ \frac{1}{a} & \frac{1}{a} \end{pmatrix} = \begin{pmatrix} b+n & b+n \\ \frac{1}{a} & \frac{1}{a} \end{pmatrix}$	M1 A1	Uses formula given in part (b)
$\begin{pmatrix} b+n & b+n \\ \frac{1}{a} & \frac{1}{a} \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} (b+n)(x+y) \\ \frac{1}{a}(x+y) \end{pmatrix}$	B1	Transforms $\begin{pmatrix} x \\ y \end{pmatrix}$ by multiplying matrices to find $\begin{pmatrix} X \\ Y \end{pmatrix}$
$\frac{1}{a}(1+m) = m(b+n)(1+m)$	M1	Uses $y = mx$ and $Y = mX$
$y = -x$	A1
$y = \frac{1}{a(b+n)}x$	A1

## Question 5(a):

A shear in the $x$-direction | **B1** | Accept 'shear'

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## Question 5(b):

$\mathbf{A} = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}$ so true when $n=1$ | **B1** | States base case

Assume true for $n=k$, so $\mathbf{A}^k = \begin{pmatrix} 1 & ka \\ 0 & 1 \end{pmatrix}$ | **B1** | States inductive hypothesis

$\mathbf{A}^{k+1} = \begin{pmatrix} 1 & ka \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & a+ak \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & (k+1)a \\ 0 & 1 \end{pmatrix}$ | **M1 A1** | Multiplies $\mathbf{A}^k$ with $\mathbf{A}$

If true for $[n=1$ and$]$ $n=k$ then also true for $n=k+1$. Hence by induction, true for all positive integers. | **A1** | Everything correct and states conclusion

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## Question 5(c):

$\mathbf{A}^n\mathbf{B} = \begin{pmatrix} 1 & na \\ 0 & 1 \end{pmatrix}\begin{pmatrix} b & b \\ \frac{1}{a} & \frac{1}{a} \end{pmatrix} = \begin{pmatrix} b+n & b+n \\ \frac{1}{a} & \frac{1}{a} \end{pmatrix}$ | **M1 A1** | Uses formula given in part (b)

$\begin{pmatrix} b+n & b+n \\ \frac{1}{a} & \frac{1}{a} \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} (b+n)(x+y) \\ \frac{1}{a}(x+y) \end{pmatrix}$ | **B1** | Transforms $\begin{pmatrix} x \\ y \end{pmatrix}$ by multiplying matrices to find $\begin{pmatrix} X \\ Y \end{pmatrix}$

$\frac{1}{a}(1+m) = m(b+n)(1+m)$ | **M1** | Uses $y = mx$ and $Y = mX$

$y = -x$ | **A1** |

$y = \frac{1}{a(b+n)}x$ | **A1** |

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5 Let $\mathbf { A } = \left( \begin{array} { l l } 1 & a \\ 0 & 1 \end{array} \right)$, where $a$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item State the type of the geometrical transformation in the $x - y$ plane represented by $\mathbf { A }$.
\item Prove by mathematical induction that, for all positive integers $n$,

$$\mathbf { A } ^ { \mathrm { n } } = \left( \begin{array} { c c } 
1 & \mathrm { na } \\
0 & 1
\end{array} \right)$$

Let $\mathbf { B } = \left( \begin{array} { c c } b & b \\ a ^ { - 1 } & a ^ { - 1 } \end{array} \right)$, where $b$ is a positive constant.
\item Find the equations of the invariant lines, through the origin, of the transformation in the $x - y$ plane represented by $\mathbf { A } ^ { n } \mathbf { B }$.
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 1 2022 Q5 [12]}}

This paper (7 questions)

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Q1 6 Q2 7 Q3 10 Q4 9 Q5 12 Q6 13 Q7 18

Answer	Marks	Guidance
\(\mathbf{A} = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}\) so true when \(n=1\)	B1	States base case
Assume true for \(n=k\), so \(\mathbf{A}^k = \begin{pmatrix} 1 & ka \\ 0 & 1 \end{pmatrix}\)	B1	States inductive hypothesis
\(\mathbf{A}^{k+1} = \begin{pmatrix} 1 & ka \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & a+ak \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & (k+1)a \\ 0 & 1 \end{pmatrix}\)	M1 A1	Multiplies \(\mathbf{A}^k\) with \(\mathbf{A}\)
If true for \([n=1\) and\(]\) \(n=k\) then also true for \(n=k+1\). Hence by induction, true for all positive integers.	A1	Everything correct and states conclusion

Answer	Marks	Guidance
\(\mathbf{A}^n\mathbf{B} = \begin{pmatrix} 1 & na \\ 0 & 1 \end{pmatrix}\begin{pmatrix} b & b \\ \frac{1}{a} & \frac{1}{a} \end{pmatrix} = \begin{pmatrix} b+n & b+n \\ \frac{1}{a} & \frac{1}{a} \end{pmatrix}\)	M1 A1	Uses formula given in part (b)
\(\begin{pmatrix} b+n & b+n \\ \frac{1}{a} & \frac{1}{a} \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} (b+n)(x+y) \\ \frac{1}{a}(x+y) \end{pmatrix}\)	B1	Transforms \(\begin{pmatrix} x \\ y \end{pmatrix}\) by multiplying matrices to find \(\begin{pmatrix} X \\ Y \end{pmatrix}\)
\(\frac{1}{a}(1+m) = m(b+n)(1+m)\)	M1	Uses \(y = mx\) and \(Y = mX\)
\(y = -x\)	A1
\(y = \frac{1}{a(b+n)}x\)	A1