| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2023 |
| Session | November |
| Marks | 15 |
| 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, finding invariant lines via eigenvalues/eigenvectors, and decomposing a transformation into geometric components. Part (a) is routine computation, part (b) requires standard eigenvalue techniques (characteristic equation, eigenvectors), and part (c) involves recognizing standard transformation matrices and solving a matrix equation. While it covers several techniques, each step follows established procedures without requiring novel insight. The Further Maths context and multi-step nature place it moderately above average difficulty. |
| Spec | 4.03a Matrix language: terminology and notation4.03d Linear transformations 2D: reflection, rotation, enlargement, shear4.03g Invariant points and lines4.03l Singular/non-singular matrices |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix}1&3\\2&5\end{vmatrix} - k\begin{vmatrix}2&3\\3&5\end{vmatrix} + 3\begin{vmatrix}2&1\\3&2\end{vmatrix} = 0 \Rightarrow -1-k+3=0 \Rightarrow k=2\) | M1 A1 | Sets determinant of A equal to zero |
| \(\begin{pmatrix}-2&-1&1\\1&1&3\end{pmatrix}\begin{pmatrix}1&2&3\\2&1&3\\3&2&5\end{pmatrix}\begin{pmatrix}0&-2\\-1&3\\0&0\end{pmatrix} = \begin{pmatrix}-2&-1&1\\1&1&3\end{pmatrix}\begin{pmatrix}-2&4\\-1&-1\\-2&0\end{pmatrix}\) | M1 | Multiplying two matrices correctly, correct dimensions |
| \(\begin{pmatrix}3&-7\\-9&3\end{pmatrix}\) | M1 A1 | Completing matrix multiplication, AG |
| Total: 5 |
| 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 | |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{D} = \begin{pmatrix}\alpha&0\\0&\alpha\end{pmatrix}\) | B1 | |
| \(\mathbf{E} = \begin{pmatrix}\beta&0\\0&1\end{pmatrix}\) | B1 | |
| \(\mathbf{F} = \begin{pmatrix}0&1\\1&0\end{pmatrix}\) | B1 | |
| \(\begin{pmatrix}3&-7\\-9&3\end{pmatrix} = \begin{pmatrix}\alpha&0\\0&\alpha\end{pmatrix} - 9\begin{pmatrix}0&\beta\\1&0\end{pmatrix}\) | M1 | Setting up simultaneous equations using their D and E |
| \(\mathbf{D} = \begin{pmatrix}3&0\\0&3\end{pmatrix}\), \(\mathbf{E} = \begin{pmatrix}\frac{7}{9}&0\\0&1\end{pmatrix}\) | A1 | Condone \(\alpha=3\), \(\beta=\frac{7}{9}\) if it is clear that they refer to the correct matrices |
| Total: 5 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix}1&3\\2&5\end{vmatrix} - k\begin{vmatrix}2&3\\3&5\end{vmatrix} + 3\begin{vmatrix}2&1\\3&2\end{vmatrix} = 0 \Rightarrow -1-k+3=0 \Rightarrow k=2$ | M1 A1 | Sets determinant of **A** equal to zero |
| $\begin{pmatrix}-2&-1&1\\1&1&3\end{pmatrix}\begin{pmatrix}1&2&3\\2&1&3\\3&2&5\end{pmatrix}\begin{pmatrix}0&-2\\-1&3\\0&0\end{pmatrix} = \begin{pmatrix}-2&-1&1\\1&1&3\end{pmatrix}\begin{pmatrix}-2&4\\-1&-1\\-2&0\end{pmatrix}$ | M1 | Multiplying two matrices correctly, correct dimensions |
| $\begin{pmatrix}3&-7\\-9&3\end{pmatrix}$ | M1 A1 | Completing matrix multiplication, AG |
| **Total: 5** | | |
---
## Question 5(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 | |
| **Total: 5** | | |
---
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{D} = \begin{pmatrix}\alpha&0\\0&\alpha\end{pmatrix}$ | B1 | |
| $\mathbf{E} = \begin{pmatrix}\beta&0\\0&1\end{pmatrix}$ | B1 | |
| $\mathbf{F} = \begin{pmatrix}0&1\\1&0\end{pmatrix}$ | B1 | |
| $\begin{pmatrix}3&-7\\-9&3\end{pmatrix} = \begin{pmatrix}\alpha&0\\0&\alpha\end{pmatrix} - 9\begin{pmatrix}0&\beta\\1&0\end{pmatrix}$ | M1 | Setting up simultaneous equations using their **D** and **E** |
| $\mathbf{D} = \begin{pmatrix}3&0\\0&3\end{pmatrix}$, $\mathbf{E} = \begin{pmatrix}\frac{7}{9}&0\\0&1\end{pmatrix}$ | A1 | Condone $\alpha=3$, $\beta=\frac{7}{9}$ if it is clear that they refer to the correct matrices |
| **Total: 5** | | |
---5 Let $k$ be a constant. The matrices $\mathbf { A } , \mathbf { B }$ and $\mathbf { C }$ are given by
$$\mathbf { A } = \left( \begin{array} { l l l }
1 & k & 3 \\
2 & 1 & 3 \\
3 & 2 & 5
\end{array} \right) , \quad \mathbf { B } = \left( \begin{array} { r r }
0 & - 2 \\
- 1 & 3 \\
0 & 0
\end{array} \right) \quad \text { and } \quad \mathbf { C } = \left( \begin{array} { r r r }
- 2 & - 1 & 1 \\
1 & 1 & 3
\end{array} \right)$$
It is given that $\mathbf { A }$ is singular.
\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 $\mathbf { C A B }$.
\item The matrices $\mathbf { D } , \mathbf { E }$ and $\mathbf { F }$ represent geometrical transformations in the $x - y$ plane.
\begin{itemize}
\item D represents an enlargement, centre the origin.
\item E represents a stretch parallel to the $x$-axis.
\item F represents a reflection in the line $y = x$.
\end{itemize}
Given that $\mathbf { C A B } = \mathbf { D } - 9 \mathbf { E F }$, find $\mathbf { D } , \mathbf { E }$ and $\mathbf { F }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q5 [15]}}