| 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 Further Maths question on eigenvalues with significant scaffolding. Part (a) is routine verification by matrix multiplication. Part (b) involves computing det(A - λI) for a 3×3 matrix (standard but computational) with the answer given to verify, then factoring a cubic where one root is known. Part (c) applies Cayley-Hamilton theorem, which is a standard technique once learned. While this is Further Maths content, the heavy scaffolding and routine application of techniques makes it slightly easier than average for A-level overall. |
| Spec | 4.03j Determinant 3x3: calculation4.03o Inverse 3x3 matrix |
| 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 |
| 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. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\lambda^3-19\lambda-30=0\) | A1 | Expands determinant, AG. |
| \(\lambda = 5, -3\) | B1 |
| 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 \(\mathbf{A}^2\). |
## Question 4(a):
$\begin{pmatrix}-11 & 1 & 8\\0 & -2 & 0\\-16 & 1 & 13\end{pmatrix}\begin{pmatrix}1\\1\\1\end{pmatrix} = \begin{pmatrix}-2\\-2\\-2\end{pmatrix}$ | M1 | Multiplies matrix with eigenvector.
$\lambda = -2$ | A1 |
**Total: 2**
---
## Question 4(b):
$\begin{vmatrix}-11-\lambda & 1 & 8\\0 & -2-\lambda & 0\\-16 & 1 & 13-\lambda\end{vmatrix} = 0$ | M1 | Sets determinant equal to zero.
$(-11-\lambda)(-2-\lambda)(13-\lambda)+128(-2-\lambda)=0$
$(-2-\lambda)(\lambda^2-2\lambda-15)=0$
$\lambda^3-19\lambda-30=0$ | A1 | Expands determinant, AG.
$\lambda = 5, -3$ | B1 |
**Total: 3**
---
## Question 4(c):
$\mathbf{A}^3 - 19\mathbf{A} - 30\mathbf{I} = 0$ | 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 $\mathbf{A}^2$.
**Total: 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]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-08_2715_44_110_2006}\\
\includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-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]}}