| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find P and D for A = PDP⁻¹ |
| Difficulty | Standard +0.8 This is a substantial Further Maths question requiring multiple techniques: finding when a system has non-unique solutions (determinant = 0), verifying an eigenvector, finding the full diagonalization P and D, and using the Cayley-Hamilton theorem to find the inverse. While each part uses standard methods, the combination of techniques and the computational complexity (3×3 matrices throughout) places this above average difficulty for Further Maths. |
| Spec | 4.03n Inverse 2x2 matrix4.03r Solve simultaneous equations: using inverse matrix4.03t Plane intersection: geometric interpretation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} 13 & 18 & -28 \\ -4 & -a & 8 \\ 2 & 6 & -5 \end{vmatrix} = 9a - 24\) | M1 | Finds determinant |
| \(a = \frac{8}{3}\) | A1 | Accept \(\frac{24}{9}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix} 13 & 18 & -28 \\ -4 & -1 & 8 \\ 2 & 6 & -5 \end{pmatrix}\begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 \\ 0 \\ -1 \end{pmatrix} \Rightarrow \lambda = -1\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} 13-\lambda & 18 & -28 \\ -4 & -1-\lambda & 8 \\ 2 & 6 & -5-\lambda \end{vmatrix} = 0\) | B1 | Sets \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\) |
| \(-\lambda^3 + 7\lambda^2 - 7\lambda - 15 = 0 \Rightarrow (\lambda+1)(\lambda-3)(\lambda-5) = 0\) | M1 | Expands determinant and factorises |
| \(\lambda = -1,\quad \lambda = 3,\quad \lambda = 5\) | A1 | |
| \(\lambda=3\): eigenvector \(\sim \begin{pmatrix}1\\1\\1\end{pmatrix}\) | M1 | Uses vector product (or equations) to find corresponding eigenvectors |
| \(\lambda=5\): eigenvector \(\sim \begin{pmatrix}-1\\2\\1\end{pmatrix}\) | A1 A1 | A1 for each correct eigenvector |
| \(\mathbf{P} = \begin{pmatrix}2&1&-1\\0&1&2\\1&1&1\end{pmatrix}\) and \(\mathbf{D} = \begin{pmatrix}-1&0&0\\0&3&0\\0&0&5\end{pmatrix}\) | M1 A1 | Or correctly matched permutations of columns. Accept scalar multiples of eigenvectors. At least two (non-zero) eigenvectors for M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(-\mathbf{A}^3 + 7\mathbf{A}^2 - 7\mathbf{A} - 15\mathbf{I} = \mathbf{0}\) | M1 | Substitutes \(\mathbf{A}\) into characteristic equation and multiplies through by \(\mathbf{A}^{-1}\). Allow missing \(\mathbf{I}\) |
| \(\mathbf{A}^{-1} = \frac{1}{15}(-\mathbf{A}^2 + 7\mathbf{A} - 7\mathbf{I})\) | A1 | Accept with \(\mathbf{A} = \begin{pmatrix}13&18&-28\\-4&-1&8\\2&6&-5\end{pmatrix}\); gives \(\mathbf{A}^{-1} = \frac{1}{15}\begin{pmatrix}43&78&-116\\4&9&-8\\22&42&-59\end{pmatrix}\) |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} 13 & 18 & -28 \\ -4 & -a & 8 \\ 2 & 6 & -5 \end{vmatrix} = 9a - 24$ | M1 | Finds determinant |
| $a = \frac{8}{3}$ | A1 | Accept $\frac{24}{9}$ |
**Total: 2 marks**
---
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix} 13 & 18 & -28 \\ -4 & -1 & 8 \\ 2 & 6 & -5 \end{pmatrix}\begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 \\ 0 \\ -1 \end{pmatrix} \Rightarrow \lambda = -1$ | B1 | |
**Total: 1 mark**
---
## Question 8(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} 13-\lambda & 18 & -28 \\ -4 & -1-\lambda & 8 \\ 2 & 6 & -5-\lambda \end{vmatrix} = 0$ | B1 | Sets $\det(\mathbf{A} - \lambda\mathbf{I}) = 0$ |
| $-\lambda^3 + 7\lambda^2 - 7\lambda - 15 = 0 \Rightarrow (\lambda+1)(\lambda-3)(\lambda-5) = 0$ | M1 | Expands determinant and factorises |
| $\lambda = -1,\quad \lambda = 3,\quad \lambda = 5$ | A1 | |
| $\lambda=3$: eigenvector $\sim \begin{pmatrix}1\\1\\1\end{pmatrix}$ | M1 | Uses vector product (or equations) to find corresponding eigenvectors |
| $\lambda=5$: eigenvector $\sim \begin{pmatrix}-1\\2\\1\end{pmatrix}$ | A1 A1 | A1 for each correct eigenvector |
| $\mathbf{P} = \begin{pmatrix}2&1&-1\\0&1&2\\1&1&1\end{pmatrix}$ and $\mathbf{D} = \begin{pmatrix}-1&0&0\\0&3&0\\0&0&5\end{pmatrix}$ | M1 A1 | Or correctly matched permutations of columns. Accept scalar multiples of eigenvectors. At least two (non-zero) eigenvectors for M1 |
**Total: 8 marks**
---
## Question 8(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-\mathbf{A}^3 + 7\mathbf{A}^2 - 7\mathbf{A} - 15\mathbf{I} = \mathbf{0}$ | M1 | Substitutes $\mathbf{A}$ into characteristic equation and multiplies through by $\mathbf{A}^{-1}$. Allow missing $\mathbf{I}$ |
| $\mathbf{A}^{-1} = \frac{1}{15}(-\mathbf{A}^2 + 7\mathbf{A} - 7\mathbf{I})$ | A1 | Accept with $\mathbf{A} = \begin{pmatrix}13&18&-28\\-4&-1&8\\2&6&-5\end{pmatrix}$; gives $\mathbf{A}^{-1} = \frac{1}{15}\begin{pmatrix}43&78&-116\\4&9&-8\\22&42&-59\end{pmatrix}$ |
**Total: 2 marks**8
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$ for which the system of equations
$$\begin{array} { r }
13 x + 18 y - 28 z = 0 \\
- 4 x - a y + 8 z = 0 \\
2 x + 6 y - 5 z = 0
\end{array}$$
does not have a unique solution.\\
The matrix $\mathbf { A }$ is given by
$$\mathbf { A } = \left( \begin{array} { r r r }
13 & 18 & - 28 \\
- 4 & - 1 & 8 \\
2 & 6 & - 5
\end{array} \right)$$
\item Find the eigenvalue of $\mathbf { A }$ corresponding to the eigenvector $\left( \begin{array} { l } 2 \\ 0 \\ 1 \end{array} \right)$.
\item Find a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { A } = \mathbf { P D P } ^ { - 1 }$.
\item Use the characteristic equation of $\mathbf { A }$ to find $\mathbf { A } ^ { - 1 }$ in terms of $\mathbf { A }$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q8 [13]}}