| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 6 |
| 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 of a 4×4 matrix to find rank, then solving a homogeneous system to find the null space basis. While it involves multiple steps and careful arithmetic with a larger matrix, the techniques are standard Further Maths procedures with no novel insight required. The computational demand and potential for arithmetic errors elevate it slightly above average difficulty. |
| Spec | 4.03l Singular/non-singular matrices4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\begin{pmatrix}1&3&-2&4\\5&15&-9&19\\-2&-6&3&-7\\3&9&-5&11\end{pmatrix} \rightarrow \begin{pmatrix}1&3&-2&4\\0&0&1&-1\\0&0&0&0\\0&0&0&0\end{pmatrix}\) | M1A1 | Reduces matrix to echelon form |
| \(r(\mathbf{A}) = 4 - 2 = 2\) | A1 | Uses rank = dimension − nullity; Part mark: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x + 3y - 2z + 4t = 0\) and \(z - t = 0\) | B1 | Obtains a set of equations |
| Basis is \(\left\{\begin{pmatrix}-2\\0\\1\\1\end{pmatrix}, \begin{pmatrix}-3\\1\\0\\0\end{pmatrix}\right\}\) (OE) | M1A1 | Finds basis vectors; Part mark: 3 |
## Question 3(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix}1&3&-2&4\\5&15&-9&19\\-2&-6&3&-7\\3&9&-5&11\end{pmatrix} \rightarrow \begin{pmatrix}1&3&-2&4\\0&0&1&-1\\0&0&0&0\\0&0&0&0\end{pmatrix}$ | M1A1 | Reduces matrix to echelon form |
| $r(\mathbf{A}) = 4 - 2 = 2$ | A1 | Uses rank = dimension − nullity; Part mark: 3 |
## Question 3(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x + 3y - 2z + 4t = 0$ and $z - t = 0$ | B1 | Obtains a set of equations |
| Basis is $\left\{\begin{pmatrix}-2\\0\\1\\1\end{pmatrix}, \begin{pmatrix}-3\\1\\0\\0\end{pmatrix}\right\}$ (OE) | M1A1 | Finds basis vectors; Part mark: 3 |
---3 The linear transformation $\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }$ is represented by the matrix $\mathbf { M } = \left( \begin{array} { r r r r } 1 & 3 & - 2 & 4 \\ 5 & 15 & - 9 & 19 \\ - 2 & - 6 & 3 & - 7 \\ 3 & 9 & - 5 & 11 \end{array} \right)$.\\
(i) Find the rank of $\mathbf { M }$.\\
(ii) Obtain a basis for the null space of T .
\hfill \mbox{\textit{CAIE FP1 2011 Q3 [6]}}