| 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 | Rank and null space basis |
| Difficulty | Challenging +1.2 This question requires row reduction to find rank and null space basis, then solving a matrix equation with given constraints. While it involves multiple steps and Further Maths content (linear transformations, null spaces), the techniques are algorithmic and standard for FP1. The computational work is substantial but straightforward, making it moderately above average difficulty. |
| Spec | 4.03a Matrix language: terminology and notation4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix} 1 & -2 & -3 & 1 \\ 3 & -5 & -7 & 7 \\ 5 & -9 & -13 & 9 \\ 7 & -13 & -19 & 11 \end{pmatrix} \sim \begin{pmatrix} 1 & -2 & -3 & 1 \\ 0 & 1 & 2 & 4 \\ 0 & 1 & 2 & 4 \\ 0 & 1 & 2 & 4 \end{pmatrix} \sim \begin{pmatrix} 1 & -2 & -3 & 1 \\ 0 & 1 & 2 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}\) | 2 | M1A1 |
| \(r(\mathbf{M}) = 4 - 2 = 2\) | 1 | A1 |
| \(x - 2y - 3z + t = 0\) and \(y + 2z + 4t = 0\) | 1 | M1 |
| E.g. Set \(z = \lambda\) and \(t = \mu \Rightarrow y = -2\lambda - 4\mu\) and \(x = -\lambda - 9\mu\) | 1 | M1 |
| \(\Rightarrow\) Basis is \(\left\{ \begin{pmatrix} -1 \\ -2 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} -9 \\ -4 \\ 0 \\ 1 \end{pmatrix} \right\}\) | 1 | A1 |
| Subtotal | 6 | |
| \(\mathbf{x} = \begin{pmatrix} 1 \\ 2 \\ 3 \\ 4 \end{pmatrix} + \lambda\begin{pmatrix} -1 \\ -2 \\ 1 \\ 0 \end{pmatrix} + \mu\begin{pmatrix} -9 \\ -4 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} a \\ b \\ -1 \\ -1 \end{pmatrix}\) | 1 | M1 |
| Solving: \(\lambda = -4\) and \(\mu = -5\) \(\Rightarrow a = 50\), \(b = 30\) | 3 | M1 A1 A1 |
| Subtotal | 4 |
## Question 7:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix} 1 & -2 & -3 & 1 \\ 3 & -5 & -7 & 7 \\ 5 & -9 & -13 & 9 \\ 7 & -13 & -19 & 11 \end{pmatrix} \sim \begin{pmatrix} 1 & -2 & -3 & 1 \\ 0 & 1 & 2 & 4 \\ 0 & 1 & 2 & 4 \\ 0 & 1 & 2 & 4 \end{pmatrix} \sim \begin{pmatrix} 1 & -2 & -3 & 1 \\ 0 & 1 & 2 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ | 2 | M1A1 |
| $r(\mathbf{M}) = 4 - 2 = 2$ | 1 | A1 |
| $x - 2y - 3z + t = 0$ and $y + 2z + 4t = 0$ | 1 | M1 |
| E.g. Set $z = \lambda$ and $t = \mu \Rightarrow y = -2\lambda - 4\mu$ and $x = -\lambda - 9\mu$ | 1 | M1 |
| $\Rightarrow$ Basis is $\left\{ \begin{pmatrix} -1 \\ -2 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} -9 \\ -4 \\ 0 \\ 1 \end{pmatrix} \right\}$ | 1 | A1 |
| **Subtotal** | **6** | |
| $\mathbf{x} = \begin{pmatrix} 1 \\ 2 \\ 3 \\ 4 \end{pmatrix} + \lambda\begin{pmatrix} -1 \\ -2 \\ 1 \\ 0 \end{pmatrix} + \mu\begin{pmatrix} -9 \\ -4 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} a \\ b \\ -1 \\ -1 \end{pmatrix}$ | 1 | M1 |
| Solving: $\lambda = -4$ and $\mu = -5$ $\Rightarrow a = 50$, $b = 30$ | 3 | M1 A1 A1 |
| **Subtotal** | **4** | |
---7 The linear transformation $\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }$ is represented by the matrix $\mathbf { M }$, where
$$\mathbf { M } = \left( \begin{array} { r r r r }
1 & - 2 & - 3 & 1 \\
3 & - 5 & - 7 & 7 \\
5 & - 9 & - 13 & 9 \\
7 & - 13 & - 19 & 11
\end{array} \right)$$
(i) Find the rank of $\mathbf { M }$ and a basis for the null space of T .\\
(ii) The vector $\left( \begin{array} { l } 1 \\ 2 \\ 3 \\ 4 \end{array} \right)$ is denoted by $\mathbf { e }$. Show that there is a solution of the equation $\mathbf { M x } = \mathbf { M e }$ of the form $\mathbf { x } = \left( \begin{array} { c } a \\ b \\ - 1 \\ - 1 \end{array} \right)$, where the constants $a$ and $b$ are to be found.\\
\hfill \mbox{\textit{CAIE FP1 2017 Q7 [10]}}