| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2023 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find P and D for A² = PDP⁻¹ or A⁻¹ = PDP⁻¹ |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on diagonalization with an upper triangular matrix (eigenvalues immediately visible on diagonal). Part (a) requires finding eigenvectors and constructing P and D for A⁻¹, which is routine but involves matrix inversion of eigenvalues. Part (b) uses Cayley-Hamilton theorem to find A⁻¹ algebraically. While multi-step, both parts follow standard Further Maths procedures without requiring novel insight. Slightly above average difficulty due to the two-part nature and Further Maths content, but well within expected exam techniques. |
| Spec | 4.03i Determinant: area scale factor and orientation4.03l Singular/non-singular matrices4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\lambda = -6,\ \lambda = -2,\ \lambda = 8\) | B1 | |
| \(\lambda = -6\): \(\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\0&4&5\\0&0&2\end{vmatrix} = \begin{pmatrix}8\\0\\0\end{pmatrix} \sim \begin{pmatrix}1\\0\\0\end{pmatrix}\) | M1 A1 | Uses vector product (or equations) to find corresponding eigenvectors |
| \(\lambda = -2\): \(\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-4&2&13\\0&0&5\end{vmatrix} = \begin{pmatrix}10\\20\\0\end{pmatrix} \sim \begin{pmatrix}1\\2\\0\end{pmatrix}\) | A1 | |
| \(\lambda = 8\): \(\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-14&2&13\\0&-10&5\end{vmatrix} = \begin{pmatrix}140\\70\\140\end{pmatrix} \sim \begin{pmatrix}2\\1\\2\end{pmatrix}\) | A1 | |
| \(\mathbf{P} = \begin{pmatrix}1&1&2\\0&2&1\\0&0&2\end{pmatrix}\) and \(\mathbf{D} = \begin{pmatrix}-\frac{1}{6}&0&0\\0&-\frac{1}{2}&0\\0&0&\frac{1}{8}\end{pmatrix}\) | M1 A1 | Or correctly matched permutations of columns. M1 for their (non-zero) eigenvectors matched to their eigenvalues |
| 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A^3 - 52A - 96\mathbf{I} = 0\) | M1 | Substitutes \(\mathbf{A}\) into characteristic equation |
| \(96\mathbf{A}^{-1} = \mathbf{A}^2 - 52\mathbf{I}\) | M1 | Multiplies through by \(\mathbf{A}^{-1}\) |
| \(\mathbf{A}^2 = \begin{pmatrix}36&-16&36\\0&4&30\\0&0&64\end{pmatrix}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A^{-1} = \begin{pmatrix} -\frac{1}{6} & -\frac{1}{6} & \frac{3}{8} \\ 0 & -\frac{1}{2} & \frac{5}{16} \\ 0 & 0 & \frac{1}{8} \end{pmatrix}\) | A1 | |
| 4 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\lambda = -6,\ \lambda = -2,\ \lambda = 8$ | B1 | |
| $\lambda = -6$: $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\0&4&5\\0&0&2\end{vmatrix} = \begin{pmatrix}8\\0\\0\end{pmatrix} \sim \begin{pmatrix}1\\0\\0\end{pmatrix}$ | M1 A1 | Uses vector product (or equations) to find corresponding eigenvectors |
| $\lambda = -2$: $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-4&2&13\\0&0&5\end{vmatrix} = \begin{pmatrix}10\\20\\0\end{pmatrix} \sim \begin{pmatrix}1\\2\\0\end{pmatrix}$ | A1 | |
| $\lambda = 8$: $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-14&2&13\\0&-10&5\end{vmatrix} = \begin{pmatrix}140\\70\\140\end{pmatrix} \sim \begin{pmatrix}2\\1\\2\end{pmatrix}$ | A1 | |
| $\mathbf{P} = \begin{pmatrix}1&1&2\\0&2&1\\0&0&2\end{pmatrix}$ and $\mathbf{D} = \begin{pmatrix}-\frac{1}{6}&0&0\\0&-\frac{1}{2}&0\\0&0&\frac{1}{8}\end{pmatrix}$ | M1 A1 | Or correctly matched permutations of columns. M1 for their (non-zero) eigenvectors matched to their eigenvalues |
| | **7** | |
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A^3 - 52A - 96\mathbf{I} = 0$ | M1 | Substitutes $\mathbf{A}$ into characteristic equation |
| $96\mathbf{A}^{-1} = \mathbf{A}^2 - 52\mathbf{I}$ | M1 | Multiplies through by $\mathbf{A}^{-1}$ |
| $\mathbf{A}^2 = \begin{pmatrix}36&-16&36\\0&4&30\\0&0&64\end{pmatrix}$ | B1 | |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A^{-1} = \begin{pmatrix} -\frac{1}{6} & -\frac{1}{6} & \frac{3}{8} \\ 0 & -\frac{1}{2} & \frac{5}{16} \\ 0 & 0 & \frac{1}{8} \end{pmatrix}$ | A1 | |
| | **4** | |
---7 The matrix $\mathbf { A }$ is given by
$$A = \left( \begin{array} { r r r }
- 6 & 2 & 13 \\
0 & - 2 & 5 \\
0 & 0 & 8
\end{array} \right) .$$
\begin{enumerate}[label=(\alph*)]
\item Find a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { A } ^ { - 1 } = \mathbf { P D P } ^ { - 1 }$.
\item Use the characteristic equation of $\mathbf { A }$ to find $\mathbf { A } ^ { - 1 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q7 [11]}}