| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Rank and null space basis |
| Difficulty | Challenging +1.2 This is a systematic row-reduction problem requiring students to find rank, range space basis, and null space basis for a 4×4 matrix. While it involves multiple steps and careful arithmetic with larger matrices than typical A-level, the procedures are algorithmic (row reduction, back-substitution) without requiring novel insight. It's harder than standard A-level due to matrix size and being Further Maths content, but remains a textbook-style exercise. |
| Spec | 4.03s Consistent/inconsistent: systems of equations4.03t Plane intersection: geometric interpretation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Row reduces matrix M to echelon form: \(\begin{pmatrix}3&4&2&5\\0&-1&1&-2\\0&0&0&0\\0&0&0&0\end{pmatrix}\) | M1A1 | Reduces matrix to echelon form |
| \(\Rightarrow R(\mathbf{M}) = 2\) | A1 | States rank |
| Basis for range space: \(\left\{\begin{pmatrix}3\\6\\9\\15\end{pmatrix}, \begin{pmatrix}4\\7\\9\\16\end{pmatrix}\right\}\) (OE) | A1 | Basis for range space; Part mark: 4 |
| Alternatively: \(2\mathbf{c}_1 = \mathbf{c}_2 + \mathbf{c}_3\) and \(\mathbf{c}_4 = \mathbf{c}_1 + \mathbf{c}_2 - \mathbf{c}_3 \Rightarrow\) lin. dep. | M1 A1 | Shows linear dependence |
| But any two e.g. \(\mathbf{c}_1\), \(\mathbf{c}_2\) are lin. indep.; Hence \(R(\mathbf{M})=2\) | A1 A1 | States rank and basis |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3x + 4y + 2z + 5t = 0\) and \(-y + z - 2t = 0\) | M1 | Forms equations |
| \((t=\lambda,\ z=\mu,\ y=\mu-2\lambda,\ x=\lambda-2\mu)\) | Gives two parameter solution | |
| Basis of null space is \(\left\{\begin{pmatrix}1\\-2\\0\\1\end{pmatrix}, \begin{pmatrix}-2\\1\\1\\0\end{pmatrix}\right\}\) or \(\left\{\begin{pmatrix}-3\\0\\2\\1\end{pmatrix}, \begin{pmatrix}0\\-3\\1\\2\end{pmatrix}\right\}\) | A1A1 | States basis of null space; Part mark: 3; Total: 7 |
# Question 4(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Row reduces matrix **M** to echelon form: $\begin{pmatrix}3&4&2&5\\0&-1&1&-2\\0&0&0&0\\0&0&0&0\end{pmatrix}$ | M1A1 | Reduces matrix to echelon form |
| $\Rightarrow R(\mathbf{M}) = 2$ | A1 | States rank |
| Basis for range space: $\left\{\begin{pmatrix}3\\6\\9\\15\end{pmatrix}, \begin{pmatrix}4\\7\\9\\16\end{pmatrix}\right\}$ (OE) | A1 | Basis for range space; Part mark: 4 |
| Alternatively: $2\mathbf{c}_1 = \mathbf{c}_2 + \mathbf{c}_3$ and $\mathbf{c}_4 = \mathbf{c}_1 + \mathbf{c}_2 - \mathbf{c}_3 \Rightarrow$ lin. dep. | M1 A1 | Shows linear dependence |
| But any two e.g. $\mathbf{c}_1$, $\mathbf{c}_2$ are lin. indep.; Hence $R(\mathbf{M})=2$ | A1 A1 | States rank and basis |
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# Question 4(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3x + 4y + 2z + 5t = 0$ and $-y + z - 2t = 0$ | M1 | Forms equations |
| $(t=\lambda,\ z=\mu,\ y=\mu-2\lambda,\ x=\lambda-2\mu)$ | | Gives two parameter solution |
| Basis of null space is $\left\{\begin{pmatrix}1\\-2\\0\\1\end{pmatrix}, \begin{pmatrix}-2\\1\\1\\0\end{pmatrix}\right\}$ or $\left\{\begin{pmatrix}-3\\0\\2\\1\end{pmatrix}, \begin{pmatrix}0\\-3\\1\\2\end{pmatrix}\right\}$ | A1A1 | States basis of null space; Part mark: 3; Total: 7 |4 The linear transformation $\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }$ is represented by the matrix $\mathbf { M }$, where
$$\mathbf { M } = \left( \begin{array} { r r r r }
3 & 4 & 2 & 5 \\
6 & 7 & 5 & 8 \\
9 & 9 & 9 & 9 \\
15 & 16 & 14 & 17
\end{array} \right)$$
Find\\
(i) the rank of $\mathbf { M }$ and a basis for the range space of T ,\\
(ii) a basis for the null space of T .
\hfill \mbox{\textit{CAIE FP1 2011 Q4 [7]}}