| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2016 |
| Session | November |
| 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 systematic linear algebra question requiring row reduction to find rank, then identifying pivot columns for range space basis and solving the homogeneous system for null space basis. While it involves a 4×4 matrix and multiple parts, the techniques are standard Further Maths procedures with no conceptual surprises—harder than typical A-level due to being Further Maths content and computational length, but straightforward application of learned algorithms. |
| Spec | 4.03a Matrix language: terminology and notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Row reduction to \(\begin{pmatrix} 1 & 3 & 5 & 7 \\ 0 & 2 & -3 & -5 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}\) | M1A1 | Method for row reduction, correct echelon form |
| \(r(\mathbf{A}) = 4 - 2 = 2\) | A1✓ | Follow through on echelon form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Basis for range space is \(\left\{ \begin{pmatrix}1\\2\\3\\6\end{pmatrix}, \begin{pmatrix}3\\8\\13\\24\end{pmatrix} \right\}\) | B1 | oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x + 3y + 5z + 7t = 0\) and \(2y - 3z - 5t = 0\) | B1 | |
| Solving \(\Rightarrow\) basis for null space is \(\left\{ \begin{pmatrix}-29\\5\\0\\2\end{pmatrix}, \begin{pmatrix}-19\\3\\2\\0\end{pmatrix} \right\}\) | M1, A1A1 | oe |
## Question 5:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Row reduction to $\begin{pmatrix} 1 & 3 & 5 & 7 \\ 0 & 2 & -3 & -5 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ | M1A1 | Method for row reduction, correct echelon form |
| $r(\mathbf{A}) = 4 - 2 = 2$ | A1✓ | Follow through on echelon form |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Basis for range space is $\left\{ \begin{pmatrix}1\\2\\3\\6\end{pmatrix}, \begin{pmatrix}3\\8\\13\\24\end{pmatrix} \right\}$ | B1 | oe |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x + 3y + 5z + 7t = 0$ and $2y - 3z - 5t = 0$ | B1 | |
| Solving $\Rightarrow$ basis for null space is $\left\{ \begin{pmatrix}-29\\5\\0\\2\end{pmatrix}, \begin{pmatrix}-19\\3\\2\\0\end{pmatrix} \right\}$ | M1, A1A1 | oe |
---5 The linear transformation $\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }$ is represented by the matrix $\mathbf { A }$, where
$$\mathbf { A } = \left( \begin{array} { r r r r }
1 & 3 & 5 & 7 \\
2 & 8 & 7 & 9 \\
3 & 13 & 9 & 11 \\
6 & 24 & 21 & 27
\end{array} \right)$$
Find\\
(i) the rank of $\mathbf { A }$,\\
(ii) a basis for the range space of T ,\\
(iii) a basis for the null space of T .
\hfill \mbox{\textit{CAIE FP1 2016 Q5 [8]}}