| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find eigenvalues of 3×3 matrix |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question combining 3D geometry with eigenvalue theory. Part (a) requires finding plane equations (straightforward), part (b) involves line equations and algebraic manipulation (routine), part (c) requires computing a 3×3 characteristic polynomial and solving it (standard but algebraically intensive), and part (d) involves diagonalization with eigenvector calculation. While each component is a standard technique, the combination across multiple topics and the computational demands of 3×3 matrices place this above average difficulty for Further Maths, though not exceptionally challenging. |
| Spec | 4.03h Determinant 2x2: calculation4.03i Determinant: area scale factor and orientation4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1+2(6)+3(0) = d_1\) or \((3)+2(-6)+3(0) = d_2\) | M1 | Substitutes point into \(x+2y+3z\) |
| \(x+2y+3z=13\) | A1 | |
| \(x+2y+3z=-9\) | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\6\\0\end{pmatrix} + \lambda\begin{pmatrix}2\\-12\\0\end{pmatrix}\) or \(\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\6\\0\end{pmatrix} + \lambda\begin{pmatrix}1\\-6\\0\end{pmatrix}\) | M1 | Forms vector equation of line |
| \(x-1 = \frac{y-6}{-6},\ z=0 \Rightarrow \frac{1}{2}x + \frac{1}{12}y = 1,\ z=0\) | A1 | AG. CWO |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix}1-\lambda & 2 & 3\\ 1 & 2-\lambda & 3\\ \frac{1}{2} & \frac{1}{12} & -\lambda\end{vmatrix}\) | M1 | Uses \(\det(\mathbf{A}-\lambda\mathbf{I})\) (attempt an expansion) |
| \((1-\lambda)\left(-\lambda(2-\lambda)-\frac{1}{4}\right) - 2\left(-\lambda - \frac{3}{2}\right) + 3\left(\frac{1}{12} - \frac{1}{4}(2-\lambda)\right) = 0\) leading to \(-\lambda^3 + 3\lambda^2 + \frac{7}{4}\lambda = 0\) | A1 | Expands determinant, expansion of brackets shown, AG. CWO. (Must have \(=0\) here or above) |
| \(\lambda = 0,\ \frac{7}{2},\ -\frac{1}{2}\) | B1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\lambda=0\): \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & 2 & 3\\ \frac{1}{2} & \frac{1}{12} & 0\end{vmatrix} = \begin{pmatrix}-\frac{1}{4}\\ \frac{3}{2}\\ -\frac{11}{12}\end{pmatrix} \sim \begin{pmatrix}3\\-18\\11\end{pmatrix}\) | M1 A1 | Uses vector product (or equations) to find corresponding eigenvectors |
| \(\lambda=\frac{7}{2}\): eigenvector \(\sim\begin{pmatrix}6\\6\\1\end{pmatrix}\); \(\lambda=-\frac{1}{2}\): eigenvector \(\sim\begin{pmatrix}-6\\-6\\7\end{pmatrix}\) | A1 A1 | |
| \(\mathbf{P} = \begin{pmatrix}3 & 6 & -6\\-18 & 6 & -6\\11 & 1 & 7\end{pmatrix}\) | M1 A1 | Or correctly matched permutations of columns. Do not accept if a column of zeros appears in P |
| \(\mathbf{D} = \begin{pmatrix}0 & 0 & 0\\ 0 & \left(\frac{7}{2}\right)^n & 0\\ 0 & 0 & \left(-\frac{1}{2}\right)^n\end{pmatrix}\) | M1 is for matching eigenvectors correctly with their eigenvalues | |
| 6 |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1+2(6)+3(0) = d_1$ or $(3)+2(-6)+3(0) = d_2$ | **M1** | Substitutes point into $x+2y+3z$ |
| $x+2y+3z=13$ | **A1** | |
| $x+2y+3z=-9$ | **A1** | |
| | **3** | |
---
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\6\\0\end{pmatrix} + \lambda\begin{pmatrix}2\\-12\\0\end{pmatrix}$ or $\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\6\\0\end{pmatrix} + \lambda\begin{pmatrix}1\\-6\\0\end{pmatrix}$ | **M1** | Forms vector equation of line |
| $x-1 = \frac{y-6}{-6},\ z=0 \Rightarrow \frac{1}{2}x + \frac{1}{12}y = 1,\ z=0$ | **A1** | AG. CWO |
| | **2** | |
---
## Question 8(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix}1-\lambda & 2 & 3\\ 1 & 2-\lambda & 3\\ \frac{1}{2} & \frac{1}{12} & -\lambda\end{vmatrix}$ | **M1** | Uses $\det(\mathbf{A}-\lambda\mathbf{I})$ (attempt an expansion) |
| $(1-\lambda)\left(-\lambda(2-\lambda)-\frac{1}{4}\right) - 2\left(-\lambda - \frac{3}{2}\right) + 3\left(\frac{1}{12} - \frac{1}{4}(2-\lambda)\right) = 0$ leading to $-\lambda^3 + 3\lambda^2 + \frac{7}{4}\lambda = 0$ | **A1** | Expands determinant, expansion of brackets shown, AG. CWO. (Must have $=0$ here or above) |
| $\lambda = 0,\ \frac{7}{2},\ -\frac{1}{2}$ | **B1** | |
| | **3** | |
---
## Question 8(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\lambda=0$: $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & 2 & 3\\ \frac{1}{2} & \frac{1}{12} & 0\end{vmatrix} = \begin{pmatrix}-\frac{1}{4}\\ \frac{3}{2}\\ -\frac{11}{12}\end{pmatrix} \sim \begin{pmatrix}3\\-18\\11\end{pmatrix}$ | **M1 A1** | Uses vector product (or equations) to find corresponding eigenvectors |
| $\lambda=\frac{7}{2}$: eigenvector $\sim\begin{pmatrix}6\\6\\1\end{pmatrix}$; $\lambda=-\frac{1}{2}$: eigenvector $\sim\begin{pmatrix}-6\\-6\\7\end{pmatrix}$ | **A1 A1** | |
| $\mathbf{P} = \begin{pmatrix}3 & 6 & -6\\-18 & 6 & -6\\11 & 1 & 7\end{pmatrix}$ | **M1 A1** | Or correctly matched permutations of columns. Do not accept if a column of zeros appears in **P** |
| $\mathbf{D} = \begin{pmatrix}0 & 0 & 0\\ 0 & \left(\frac{7}{2}\right)^n & 0\\ 0 & 0 & \left(-\frac{1}{2}\right)^n\end{pmatrix}$ | | M1 is for matching eigenvectors correctly with their eigenvalues |
| | **6** | |8 The planes $\Pi _ { 1 }$ and $\Pi _ { 2 }$ do not intersect and are both perpendicular to $\mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k }$. The line $l$ intersects $\Pi _ { 1 }$ at the point $( 1,6,0 )$ and intersects $\Pi _ { 2 }$ at the point $( 3 , - 6,0 )$.
\begin{enumerate}[label=(\alph*)]
\item Find Cartesian equations of $\Pi _ { 1 }$ and $\Pi _ { 2 }$.
\item Express the vector equation of $l$ in the form $\left( \begin{array} { l } x \\ y \\ z \end{array} \right) = \mathbf { a } + \lambda \mathbf { b }$, where $\mathbf { a }$ and $\mathbf { b }$ are vectors to be determined, and hence show that for points on $l , \frac { 1 } { 2 } x + \frac { 1 } { 12 } y = 1$ and $z = 0$.\\
\includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-16_2715_40_144_2008}
\item Show that the characteristic equation of $\mathbf { A }$ is $- \lambda ^ { 3 } + 3 \lambda ^ { 2 } + \frac { 7 } { 4 } \lambda = 0$ and hence find the eigenvalues of $\mathbf { A }$.
The matrix $\mathbf { A }$ is given by
$$\mathbf { A } = \left( \begin{array} { c c c }
1 & 2 & 3 \\
1 & 2 & 3 \\
\frac { 1 } { 2 } & \frac { 1 } { 12 } & 0
\end{array} \right)$$
\includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-17_194_1711_484_212}
\item Find a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { A } ^ { n } = \mathbf { P D P } ^ { - 1 }$, where $n$ is a positive integer. [6]\\
\includegraphics[max width=\textwidth, alt={}]{27485e4a-cd34-43e3-aa92-767820a9f6f9-18_65_1581_335_322} ........................................................................................................................................\\
\includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-18_72_1579_511_324}\\
\includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-18_2718_35_144_2012}
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q8 [14]}}