| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2008 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Rank and null space basis |
| Difficulty | Challenging +1.8 This question requires systematic row reduction of a 4×4 matrix with a parameter, determining rank in two cases, and finding a null space basis. While the techniques are standard for Further Maths, the multi-step nature, parameter handling, and requirement to find two linearly independent null space vectors make this significantly harder than routine matrix operations but not exceptionally difficult for FP1 students. |
| Spec | 4.03a Matrix language: terminology and notation4.03s Consistent/inconsistent: systems of equations |
| Answer | Marks |
|---|---|
| (i) Reduction of \(\mathbf{A}\) to echelon form, e.g., \(\begin{pmatrix} 1 & -1 & -2 & -3 \\ 0 & -1 & 3 & -4 \\ 0 & 0 & 0 & \alpha - 9 \\ 0 & 0 & 0 & 0 \end{pmatrix}\) | M1A1 |
| \(\alpha = 9 \Rightarrow\) last 2 rows consist entirely of zeros \(\Rightarrow r(\mathbf{A}) = 2\) | A1 |
| A basis for the null space of \(\mathbf{A}\) is \(\left\{\begin{pmatrix}5\\3\\1\\0\end{pmatrix}, \begin{pmatrix}1\\4\\0\\-1\end{pmatrix}\right\}\) or equivalent | M1A1 |
| (ii) \(\alpha - 9 \neq 0\) | M1 |
| \(r(\mathbf{A}) = 3\) | A1 |
**(i)** Reduction of $\mathbf{A}$ to echelon form, e.g., $\begin{pmatrix} 1 & -1 & -2 & -3 \\ 0 & -1 & 3 & -4 \\ 0 & 0 & 0 & \alpha - 9 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ | M1A1 |
$\alpha = 9 \Rightarrow$ last 2 rows consist entirely of zeros $\Rightarrow r(\mathbf{A}) = 2$ | A1 |
A basis for the null space of $\mathbf{A}$ is $\left\{\begin{pmatrix}5\\3\\1\\0\end{pmatrix}, \begin{pmatrix}1\\4\\0\\-1\end{pmatrix}\right\}$ or equivalent | M1A1 |
**(ii)** $\alpha - 9 \neq 0$ | M1 |
$r(\mathbf{A}) = 3$ | A1 |6 The matrix $\mathbf { A }$ is defined by
$$\mathbf { A } = \left( \begin{array} { r r r r }
1 & - 1 & - 2 & - 3 \\
- 2 & 1 & 7 & 2 \\
- 3 & 3 & 6 & \alpha \\
7 & - 6 & - 17 & - 17
\end{array} \right) .$$
(i) Show that if $\alpha = 9$ then the rank of $\mathbf { A }$ is 2, and find a basis for the null space of $\mathbf { A }$ in this case.\\
(ii) Find the rank of $\mathbf { A }$ when $\alpha \neq 9$.
\hfill \mbox{\textit{CAIE FP1 2008 Q6 [7]}}