| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Use Cayley-Hamilton for inverse |
| Difficulty | Standard +0.8 This is a Further Maths question requiring eigenvalue calculation for a 3×3 matrix, then applying Cayley-Hamilton theorem to express the inverse in terms of A² and I. While the technique is standard for Further Maths, the 3×3 characteristic polynomial and algebraic manipulation to isolate A⁻¹ requires careful execution and is more demanding than typical A-level questions. |
| Spec | 4.03a Matrix language: terminology and notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} 6-\lambda & -9 & 5 \\ 5 & -8-\lambda & 5 \\ 1 & -1 & 2-\lambda \end{vmatrix} = 0\) | M1 | Sets determinant equal to zero |
| \(\lambda^3 - 7\lambda + 6 = 0\) | M1 | Expands |
| \(\lambda = 1, 2, -3\) | A1 A1 | Award A1A0 for 2 correct solutions, A0A0 for 1 correct solution |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{A}^3 - 7\mathbf{A} + 6\mathbf{I} = 0\) | B1 | States that A satisfies its characteristic equation |
| \(\mathbf{A}^2 - 7\mathbf{I} + 6\mathbf{A}^{-1} = 0\) | M1 | Multiplies through by \(\mathbf{A}^{-1}\) |
| \(\mathbf{A}^{-1} = -\frac{1}{6}\mathbf{A}^2 + \frac{7}{6}\mathbf{I}\) | A1 | |
| Total: 3 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} 6-\lambda & -9 & 5 \\ 5 & -8-\lambda & 5 \\ 1 & -1 & 2-\lambda \end{vmatrix} = 0$ | M1 | Sets determinant equal to zero |
| $\lambda^3 - 7\lambda + 6 = 0$ | M1 | Expands |
| $\lambda = 1, 2, -3$ | A1 A1 | Award A1A0 for 2 correct solutions, A0A0 for 1 correct solution |
| **Total: 4** | | |
---
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{A}^3 - 7\mathbf{A} + 6\mathbf{I} = 0$ | B1 | States that **A** satisfies its characteristic equation |
| $\mathbf{A}^2 - 7\mathbf{I} + 6\mathbf{A}^{-1} = 0$ | M1 | Multiplies through by $\mathbf{A}^{-1}$ |
| $\mathbf{A}^{-1} = -\frac{1}{6}\mathbf{A}^2 + \frac{7}{6}\mathbf{I}$ | A1 | |
| **Total: 3** | | |
---3 The matrix $\mathbf { A }$ is given by
$$\mathbf { A } = \left( \begin{array} { l l l }
6 & - 9 & 5 \\
5 & - 8 & 5 \\
1 & - 1 & 2
\end{array} \right)$$
\begin{enumerate}[label=(\alph*)]
\item Find the eigenvalues of $\mathbf { A }$.
\item Use the characteristic equation of $\mathbf { A }$ to show that $\mathbf { A } ^ { - 1 } = p \mathbf { A } ^ { 2 } + q \mathbf { l }$, where $p$ and $q$ are constants to be determined.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q3 [7]}}