| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Rank and null space basis |
| Difficulty | Challenging +1.2 This is a structured linear algebra question requiring row reduction to find α, then determining bases for range and null spaces. While it involves multiple steps and 4×4 matrices, the techniques are standard Further Maths procedures with clear methodology—harder than typical A-level due to the Further Maths content and computational demands, but not requiring novel insight. |
| Spec | 4.03r Solve simultaneous equations: using inverse matrix4.03s Consistent/inconsistent: systems of equations |
| Answer | Marks | Guidance |
|---|---|---|
| Row reduction: \(\begin{pmatrix}1&2&0&4\\5&2&1&-3\\4&0&1&-7\\-2&4&-1&\alpha\end{pmatrix} \to \cdots \to \begin{pmatrix}1&2&0&4\\0&-8&1&-23\\0&0&0&0\\0&0&0&\alpha-15\end{pmatrix}\) | M1 A1 | Performs row operations correctly |
| \(r(\mathbf{M}) = 2 \Rightarrow \alpha = 15\) | A1 | CWO |
| Basis for range space is \(\left\{\begin{pmatrix}1\\5\\4\\-2\end{pmatrix}, \begin{pmatrix}2\\2\\0\\4\end{pmatrix}\right\}\) | B1 | Accept any two independent column vectors from \(\mathbf{M}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x + 2y + 4t = 0\); \(-8y + z - 23t = 0\) | M1 | Forms system of equations |
| \(t = \lambda\), \(z = \mu\), \(y = \frac{1}{8}\mu - \frac{23}{8}\lambda\), \(x = -\frac{1}{4}\mu + \frac{7}{4}\lambda\) | M1 | Uses two parameters |
| Basis for null space is \(\left\{\begin{pmatrix}-2\\1\\8\\0\end{pmatrix}, \begin{pmatrix}14\\-23\\0\\8\end{pmatrix}\right\}\) | A1 A1 | AEF; many alternatives possible |
## Question 5(i):
| Row reduction: $\begin{pmatrix}1&2&0&4\\5&2&1&-3\\4&0&1&-7\\-2&4&-1&\alpha\end{pmatrix} \to \cdots \to \begin{pmatrix}1&2&0&4\\0&-8&1&-23\\0&0&0&0\\0&0&0&\alpha-15\end{pmatrix}$ | M1 A1 | Performs row operations correctly |
| $r(\mathbf{M}) = 2 \Rightarrow \alpha = 15$ | A1 | CWO |
| Basis for range space is $\left\{\begin{pmatrix}1\\5\\4\\-2\end{pmatrix}, \begin{pmatrix}2\\2\\0\\4\end{pmatrix}\right\}$ | B1 | Accept any two independent column vectors from $\mathbf{M}$ |
## Question 5(ii):
| $x + 2y + 4t = 0$; $-8y + z - 23t = 0$ | M1 | Forms system of equations |
| $t = \lambda$, $z = \mu$, $y = \frac{1}{8}\mu - \frac{23}{8}\lambda$, $x = -\frac{1}{4}\mu + \frac{7}{4}\lambda$ | M1 | Uses two parameters |
| Basis for null space is $\left\{\begin{pmatrix}-2\\1\\8\\0\end{pmatrix}, \begin{pmatrix}14\\-23\\0\\8\end{pmatrix}\right\}$ | A1 A1 | AEF; many alternatives possible |
---5 The linear transformation $\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }$ is represented by the matrix $\mathbf { M }$, where
$$\mathbf { M } = \left( \begin{array} { r r r r }
1 & 2 & 0 & 4 \\
5 & 2 & 1 & - 3 \\
4 & 0 & 1 & - 7 \\
- 2 & 4 & - 1 & \alpha
\end{array} \right)$$
It is given that the rank of $\mathbf { M }$ is 2 .\\
(i) Find the value of $\alpha$ and state a basis for the range space of T .\\
(ii) Obtain a basis for the null space of T .\\
\hfill \mbox{\textit{CAIE FP1 2019 Q5 [8]}}