| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2024 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find invariant lines through origin |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring matrix multiplication (routine), finding invariant lines via eigenvalues (standard FM technique), and matrix equation solving. Part (b) is the key challenge requiring eigenvalue calculation and interpretation, which is standard FM content but more demanding than typical A-level. The multi-step nature and FM-specific techniques place it moderately above average difficulty. |
| Spec | 4.03a Matrix language: terminology and notation4.03b Matrix operations: addition, multiplication, scalar4.03d Linear transformations 2D: reflection, rotation, enlargement, shear4.03g Invariant points and lines |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}-2&-1&1\\1&1&3\end{pmatrix}\begin{pmatrix}1&2&3\\2&1&3\\3&2&5\end{pmatrix}\begin{pmatrix}0&-2\\-1&3\\0&0\end{pmatrix}=\begin{pmatrix}-2&-1&1\\1&1&3\end{pmatrix}\begin{pmatrix}-2&4\\-1&-1\\-2&0\end{pmatrix}\) | M1 A1 | Multiplying two matrices correctly, correct dimensions |
| \(=\begin{pmatrix}3&-7\\-9&3\end{pmatrix}\) | B1 | Convincingly completing matrix multiplication, AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}3&-7\\-9&3\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}3x-7y\\-9x+3y\end{pmatrix}\) | B1 | Transforms \(\begin{pmatrix}x\\y\end{pmatrix}\) to \(\begin{pmatrix}X\\Y\end{pmatrix}\) |
| \(-9x+3mx=m(3x-7mx)\) | M1 A1 | Uses \(y=mx\) and \(Y=mX\) |
| \(-9+3m=3m-7m^2 \Rightarrow 7m^2=9\) | A1 | |
| \(y=\frac{3}{\sqrt{7}}x\) and \(y=-\frac{3}{\sqrt{7}}x\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Stretch | B1 | |
| parallel to the \(x\)-axis, scale factor 3 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{M}^{-1}=\frac{1}{3}\begin{pmatrix}1&0\\0&3\end{pmatrix}\) | B1 | |
| \(\mathbf{N}=\frac{1}{3}\begin{pmatrix}3&-7\\-9&3\end{pmatrix}\begin{pmatrix}1&0\\0&3\end{pmatrix}\) or \(\mathbf{N}=\begin{pmatrix}3&-7\\-9&3\end{pmatrix}\begin{pmatrix}\frac{1}{3}&0\\0&1\end{pmatrix}\) | M1 | Correct order |
| \(=\begin{pmatrix}1&-7\\-3&3\end{pmatrix}\) | A1 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}-2&-1&1\\1&1&3\end{pmatrix}\begin{pmatrix}1&2&3\\2&1&3\\3&2&5\end{pmatrix}\begin{pmatrix}0&-2\\-1&3\\0&0\end{pmatrix}=\begin{pmatrix}-2&-1&1\\1&1&3\end{pmatrix}\begin{pmatrix}-2&4\\-1&-1\\-2&0\end{pmatrix}$ | M1 A1 | Multiplying two matrices correctly, correct dimensions |
| $=\begin{pmatrix}3&-7\\-9&3\end{pmatrix}$ | B1 | Convincingly completing matrix multiplication, AG |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}3&-7\\-9&3\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}3x-7y\\-9x+3y\end{pmatrix}$ | B1 | Transforms $\begin{pmatrix}x\\y\end{pmatrix}$ to $\begin{pmatrix}X\\Y\end{pmatrix}$ |
| $-9x+3mx=m(3x-7mx)$ | M1 A1 | Uses $y=mx$ and $Y=mX$ |
| $-9+3m=3m-7m^2 \Rightarrow 7m^2=9$ | A1 | |
| $y=\frac{3}{\sqrt{7}}x$ and $y=-\frac{3}{\sqrt{7}}x$ | A1 | |
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Stretch | B1 | |
| parallel to the $x$-axis, scale factor 3 | B1 | |
## Question 4(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{M}^{-1}=\frac{1}{3}\begin{pmatrix}1&0\\0&3\end{pmatrix}$ | B1 | |
| $\mathbf{N}=\frac{1}{3}\begin{pmatrix}3&-7\\-9&3\end{pmatrix}\begin{pmatrix}1&0\\0&3\end{pmatrix}$ or $\mathbf{N}=\begin{pmatrix}3&-7\\-9&3\end{pmatrix}\begin{pmatrix}\frac{1}{3}&0\\0&1\end{pmatrix}$ | M1 | Correct order |
| $=\begin{pmatrix}1&-7\\-3&3\end{pmatrix}$ | A1 | |4 The matrices $\mathbf { A } , \mathbf { B }$ and $\mathbf { C }$ are given by
$$\mathbf { A } = \left( \begin{array} { l l l }
1 & 2 & 3 \\
2 & 1 & 3 \\
3 & 2 & 5
\end{array} \right) , \mathbf { B } = \left( \begin{array} { r r }
0 & - 2 \\
- 1 & 3 \\
0 & 0
\end{array} \right) \text { and } \mathbf { C } = \left( \begin{array} { r r r }
- 2 & - 1 & 1 \\
1 & 1 & 3
\end{array} \right)$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathbf { C A B } = \left( \begin{array} { r r } 3 & - 7 \\ - 9 & 3 \end{array} \right)$.
\item Find the equations of the invariant lines, through the origin, of the transformation in the $x - y$ plane represented by $\mathbf { C A B }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{beb9c1f1-1676-4432-a42a-c418ff9f45d8-08_2715_31_106_2016}
Let $\mathbf { M } = \left( \begin{array} { l l } 3 & 0 \\ 0 & 1 \end{array} \right)$.
\item Give full details of the transformation represented by $\mathbf { M }$.
\item Find the matrix $\mathbf { N }$ such that $\mathbf { N M } = \mathbf { C A B }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2024 Q4 [13]}}