| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2024 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find eigenvalues of 3×3 matrix |
| Difficulty | Standard +0.3 This is a structured, multi-part eigenvalue question with significant scaffolding. Part (a) is routine verification by matrix multiplication. Part (b) requires computing det(A - λI) to get the characteristic equation (which is given to verify) then factoring a cubic using the known eigenvalue from (a). Part (c) applies the Cayley-Hamilton theorem, a standard Further Maths technique. While this covers several concepts, each step is guided and uses well-practiced methods, making it slightly easier than the average A-level maths question overall. |
| Spec | 4.03b Matrix operations: addition, multiplication, scalar4.03i Determinant: area scale factor and orientation4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}-11&1&8\\0&-2&0\\-16&1&13\end{pmatrix}\begin{pmatrix}1\\1\\1\end{pmatrix} = \begin{pmatrix}-2\\-2\\-2\end{pmatrix}\) | M1 | Multiplies matrix with eigenvector |
| \(\lambda = -2\) | A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix}-11-\lambda&1&8\\0&-2-\lambda&0\\-16&1&13-\lambda\end{vmatrix} = 0\) | M1 | Sets determinant equal to zero |
| \((-11-\lambda)(-2-\lambda)(13-\lambda)+128(-2-\lambda)=0\) | A1 | Expands determinant, AG |
| \((-2-\lambda)(\lambda^2-2\lambda-15)=0\) | ||
| \(\lambda^3-19\lambda-30=0\) | ||
| \(\lambda = 5, -3\) | B1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{A}^3 - 19\mathbf{A} - 30\mathbf{I} = 0\) | B1 | States that A satisfies its characteristic equation |
| \(30\mathbf{A}^{-1} = \mathbf{A}^2 - 19\mathbf{I}\) | M1 | Multiplies through by \(\mathbf{A}^{-1}\) |
| \(\mathbf{A}^2 = \begin{pmatrix}-7&-5&16\\0&4&0\\-32&-5&41\end{pmatrix} \Rightarrow \mathbf{A}^{-1} = \frac{1}{30}\begin{pmatrix}-26&-5&16\\0&-15&0\\-32&-5&22\end{pmatrix}\) | M1A1 | M1 for substituting \(\mathbf{A}^2\) |
| 4 |
# Question 4:
## Part 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}-11&1&8\\0&-2&0\\-16&1&13\end{pmatrix}\begin{pmatrix}1\\1\\1\end{pmatrix} = \begin{pmatrix}-2\\-2\\-2\end{pmatrix}$ | M1 | Multiplies matrix with eigenvector |
| $\lambda = -2$ | A1 | |
| | **2** | |
## Part 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix}-11-\lambda&1&8\\0&-2-\lambda&0\\-16&1&13-\lambda\end{vmatrix} = 0$ | M1 | Sets determinant equal to zero |
| $(-11-\lambda)(-2-\lambda)(13-\lambda)+128(-2-\lambda)=0$ | A1 | Expands determinant, AG |
| $(-2-\lambda)(\lambda^2-2\lambda-15)=0$ | | |
| $\lambda^3-19\lambda-30=0$ | | |
| $\lambda = 5, -3$ | B1 | |
| | **3** | |
## Part 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{A}^3 - 19\mathbf{A} - 30\mathbf{I} = 0$ | B1 | States that **A** satisfies its characteristic equation |
| $30\mathbf{A}^{-1} = \mathbf{A}^2 - 19\mathbf{I}$ | M1 | Multiplies through by $\mathbf{A}^{-1}$ |
| $\mathbf{A}^2 = \begin{pmatrix}-7&-5&16\\0&4&0\\-32&-5&41\end{pmatrix} \Rightarrow \mathbf{A}^{-1} = \frac{1}{30}\begin{pmatrix}-26&-5&16\\0&-15&0\\-32&-5&22\end{pmatrix}$ | M1A1 | M1 for substituting $\mathbf{A}^2$ |
| | **4** | |
---4 The matrix $\mathbf { A }$ is given by
$$\mathbf { A } = \left( \begin{array} { r r r }
- 11 & 1 & 8 \\
0 & - 2 & 0 \\
- 16 & 1 & 13
\end{array} \right)$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\left( \begin{array} { l } 1 \\ 1 \\ 1 \end{array} \right)$ is an eigenvector of $\mathbf { A }$ and state the corresponding eigenvalue.
\item Show that the characteristic equation of $\mathbf { A }$ is $\lambda ^ { 3 } - 19 \lambda - 30 = 0$ and hence find the other eigenvalues of $\mathbf { A }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{374b91df-926d-4f7f-a1d3-a54c70e8ff0e-08_2717_35_106_2015}\\
\includegraphics[max width=\textwidth, alt={}, center]{374b91df-926d-4f7f-a1d3-a54c70e8ff0e-09_2726_33_97_22}
\item Use the characteristic equation of $\mathbf { A }$ to find $\mathbf { A } ^ { - 1 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q4 [9]}}