| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find invariant lines through origin |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths question involving matrix multiplication, singularity conditions, and invariant lines. Part (a) is routine computation, part (b) requires setting det(A)=0 and solving, and part (c) requires finding eigenvectors of a 2×2 matrix. While it involves several Further Maths concepts, each step follows standard procedures without requiring novel insight or particularly complex reasoning. |
| Spec | 4.03b Matrix operations: addition, multiplication, scalar4.03g Invariant points and lines4.03l Singular/non-singular matrices |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix}0&1&1\\-1&2&0\end{pmatrix}\begin{pmatrix}2&k&k\\5&-1&3\\1&0&1\end{pmatrix}\begin{pmatrix}1&0\\0&1\\1&0\end{pmatrix} = \begin{pmatrix}0&1&1\\-1&2&0\end{pmatrix}\begin{pmatrix}2+k&k\\8&-1\\2&0\end{pmatrix}\) | M1 A1 | Multiplies two matrices correctly. |
| \(\begin{pmatrix}10&-1\\-k+14&-k-2\end{pmatrix}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(2\begin{vmatrix}-1&3\\0&1\end{vmatrix} - k\begin{vmatrix}5&3\\1&1\end{vmatrix} + k\begin{vmatrix}5&-1\\1&0\end{vmatrix} = 0\) leading to \(-2 - 2k + k = 0\) | M1 A1 | Sets determinant equal to zero and forms linear equation. |
| \(k = -2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix}10&-1\\16&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}10x-y\\16x\end{pmatrix}\) | M1 | Transforms \(\begin{pmatrix}x\\y\end{pmatrix}\). Allow \(q\begin{pmatrix}1\\m\end{pmatrix}\) where \(q\) is \(x\), \(t\) or a nonzero number. |
| \(10x - mx = X\) and \(16x = mX\) | M1 A1 | Uses \(y = mx\) and \(Y = mX\). Expect \(16x = m(10x - mx)\). |
| \(16 = 10m - m^2 \quad [m^2 - 10m + 16 = 0]\) | A1 | OE |
| \(y = 2x\) and \(y = 8x\) | A1 |
# Question 4:
## Part (a):
$\begin{pmatrix}0&1&1\\-1&2&0\end{pmatrix}\begin{pmatrix}2&k&k\\5&-1&3\\1&0&1\end{pmatrix}\begin{pmatrix}1&0\\0&1\\1&0\end{pmatrix} = \begin{pmatrix}0&1&1\\-1&2&0\end{pmatrix}\begin{pmatrix}2+k&k\\8&-1\\2&0\end{pmatrix}$ | **M1 A1** | Multiplies two matrices correctly.
$\begin{pmatrix}10&-1\\-k+14&-k-2\end{pmatrix}$ | **A1** |
## Part (b):
$2\begin{vmatrix}-1&3\\0&1\end{vmatrix} - k\begin{vmatrix}5&3\\1&1\end{vmatrix} + k\begin{vmatrix}5&-1\\1&0\end{vmatrix} = 0$ leading to $-2 - 2k + k = 0$ | **M1 A1** | Sets determinant equal to zero and forms linear equation.
$k = -2$ | **A1** |
## Part (c):
$\begin{pmatrix}10&-1\\16&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}10x-y\\16x\end{pmatrix}$ | **M1** | Transforms $\begin{pmatrix}x\\y\end{pmatrix}$. Allow $q\begin{pmatrix}1\\m\end{pmatrix}$ where $q$ is $x$, $t$ or a nonzero number.
$10x - mx = X$ and $16x = mX$ | **M1 A1** | Uses $y = mx$ and $Y = mX$. Expect $16x = m(10x - mx)$.
$16 = 10m - m^2 \quad [m^2 - 10m + 16 = 0]$ | **A1** | OE
$y = 2x$ and $y = 8x$ | **A1** |
---4 The matrices $\mathbf { A } , \mathbf { B }$ and $\mathbf { C }$ are given by
$$\mathbf { A } = \left( \begin{array} { c c c }
2 & k & k \\
5 & - 1 & 3 \\
1 & 0 & 1
\end{array} \right) , \quad \mathbf { B } = \left( \begin{array} { c c }
1 & 0 \\
0 & 1 \\
1 & 0
\end{array} \right) \text { and } \quad \mathbf { C } = \left( \begin{array} { r c c }
0 & 1 & 1 \\
- 1 & 2 & 0
\end{array} \right)$$
where $k$ is a real constant.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathbf { C A B }$.
\item Given that $\mathbf { A }$ is singular, find the value of $k$.
\item Using the value of $k$ from part (b), find the equations of the invariant lines, through the origin, of the transformation in the $x - y$ plane represented by $\mathbf { C A B }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2021 Q4 [11]}}