| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Find P and D for diagonalization / matrix powers |
| Difficulty | Standard +0.3 This is a standard Further Maths diagonalization question with straightforward eigenvalue/eigenvector calculations. Two eigenvalues are given, one eigenvector is provided, and the process follows routine algorithms (solving (A-λI)v=0, forming P from eigenvectors, creating diagonal D). While it requires multiple steps and matrix manipulation, it involves no novel insight or challenging problem-solving—just systematic application of well-practiced techniques, making it slightly easier than average for FM students. |
| Spec | 4.03a Matrix language: terminology and notation4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\lambda = 1\): \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 10 & -8 & 10 \\ 7 & -5 & 7 \end{vmatrix} = \begin{pmatrix} -6 \\ 0 \\ 6 \end{pmatrix} \sim \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\) oe | 2 | M1A1 |
| \(\lambda = 3\): \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 0 & 0 \\ 10 & -10 & 10 \end{vmatrix} = \begin{pmatrix} 0 \\ 20 \\ 20 \end{pmatrix} \sim \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\) oe | 1 | A1 |
| Subtotal | 3 | |
| \(\begin{pmatrix} 1 & 0 & 0 \\ 10 & -7 & 10 \\ 7 & -5 & 8 \end{pmatrix}\begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ -4 \\ -2 \end{pmatrix} = -2\begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} \Rightarrow \lambda = -2\) | 2 | M1A1 |
| \(\mathbf{D} = \begin{pmatrix} -2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix}\), \(\mathbf{P} = \begin{pmatrix} 0 & 1 & 0 \\ 2 & 0 & 1 \\ 1 & -1 & 1 \end{pmatrix}\) (or other multiples or permutations) | 2 | B1\(\wedge\) B1\(\wedge\) |
| \(\det \mathbf{P} = -1\) (or 1 depending on permutation) | 1 | B1 |
| \(\text{Adj } \mathbf{P} = \begin{pmatrix} 1 & -1 & 1 \\ -1 & 0 & 0 \\ -2 & 1 & 2 \end{pmatrix} \Rightarrow \mathbf{P}^{-1} = \begin{pmatrix} -1 & 1 & -1 \\ 1 & 0 & 0 \\ 2 & -1 & 2 \end{pmatrix}\) (or other permutations) | 2 | M1A1 |
| Subtotal | 5 |
## Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\lambda = 1$: $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 10 & -8 & 10 \\ 7 & -5 & 7 \end{vmatrix} = \begin{pmatrix} -6 \\ 0 \\ 6 \end{pmatrix} \sim \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$ oe | 2 | M1A1 |
| $\lambda = 3$: $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 0 & 0 \\ 10 & -10 & 10 \end{vmatrix} = \begin{pmatrix} 0 \\ 20 \\ 20 \end{pmatrix} \sim \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$ oe | 1 | A1 |
| **Subtotal** | **3** | |
| $\begin{pmatrix} 1 & 0 & 0 \\ 10 & -7 & 10 \\ 7 & -5 & 8 \end{pmatrix}\begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ -4 \\ -2 \end{pmatrix} = -2\begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} \Rightarrow \lambda = -2$ | 2 | M1A1 |
| $\mathbf{D} = \begin{pmatrix} -2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix}$, $\mathbf{P} = \begin{pmatrix} 0 & 1 & 0 \\ 2 & 0 & 1 \\ 1 & -1 & 1 \end{pmatrix}$ (or other multiples or permutations) | 2 | B1$\wedge$ B1$\wedge$ |
| $\det \mathbf{P} = -1$ (or 1 depending on permutation) | 1 | B1 |
| $\text{Adj } \mathbf{P} = \begin{pmatrix} 1 & -1 & 1 \\ -1 & 0 & 0 \\ -2 & 1 & 2 \end{pmatrix} \Rightarrow \mathbf{P}^{-1} = \begin{pmatrix} -1 & 1 & -1 \\ 1 & 0 & 0 \\ 2 & -1 & 2 \end{pmatrix}$ (or other permutations) | 2 | M1A1 |
| **Subtotal** | **5** | |
---6 The matrix A, where
$$\mathbf { A } = \left( \begin{array} { r r r }
1 & 0 & 0 \\
10 & - 7 & 10 \\
7 & - 5 & 8
\end{array} \right)$$
has eigenvalues 1 and 3 .\\
(i) Find corresponding eigenvectors.\\
It is given that $\left( \begin{array} { l } 0 \\ 2 \\ 1 \end{array} \right)$ is an eigenvector of $\mathbf { A }$.\\
(ii) Find the corresponding eigenvalue.\\
(iii) Find a diagonal matrix $\mathbf { D }$ and matrices $\mathbf { P }$ and $\mathbf { P } ^ { - 1 }$ such that $\mathbf { P } ^ { - 1 } \mathbf { A P } = \mathbf { D }$.\\
\hfill \mbox{\textit{CAIE FP1 2017 Q6 [10]}}