| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Rank and null space basis |
| Difficulty | Challenging +1.2 This is a standard Further Maths linear algebra question requiring row reduction to find rank and null space, then using a particular solution to find the general solution. While it involves 4×4 matrices and multiple steps, the techniques are routine for FP1 students with no novel insights required—just systematic application of Gaussian elimination and the relationship between particular and homogeneous solutions. |
| Spec | 4.03a Matrix language: terminology and notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}1&-1&-2&3\\5&-3&-4&25\\6&-4&-6&28\\7&-5&-8&31\end{pmatrix} \rightarrow\cdots\rightarrow \begin{pmatrix}1&-1&-2&3\\0&1&3&5\\0&0&0&0\\0&0&0&0\end{pmatrix}\) | M1A1 | |
| \(r(\mathbf{A}) = 4 - 2 = 2\) | A1 | |
| \(x - y - 2z + 3t = 0\) and \(y + 3z + 5t = 0\) | B1 | |
| \(z = \lambda,\ t = \mu \Rightarrow x = -\lambda - 8\mu,\ y = -3\lambda - 5\mu\) | M1 | |
| Basis for null space is \(\left\{\lambda\begin{pmatrix}-1\\-3\\1\\0\end{pmatrix}, \mu\begin{pmatrix}-8\\-5\\0\\1\end{pmatrix}\right\}\) | A1 A1 | OE |
| 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{A}\begin{pmatrix}-1\\1\\-1\\1\end{pmatrix} = \begin{pmatrix}3\\21\\24\\27\end{pmatrix}\) | B1 | |
| \(\mathbf{x} = \begin{pmatrix}-1\\1\\-1\\1\end{pmatrix} + \lambda\begin{pmatrix}-1\\-3\\1\\0\end{pmatrix} + \mu\begin{pmatrix}-8\\-5\\0\\1\end{pmatrix}\) | M1A1FT | OE |
| 3 |
## Question 7:
**7(i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}1&-1&-2&3\\5&-3&-4&25\\6&-4&-6&28\\7&-5&-8&31\end{pmatrix} \rightarrow\cdots\rightarrow \begin{pmatrix}1&-1&-2&3\\0&1&3&5\\0&0&0&0\\0&0&0&0\end{pmatrix}$ | M1A1 | |
| $r(\mathbf{A}) = 4 - 2 = 2$ | A1 | |
| $x - y - 2z + 3t = 0$ and $y + 3z + 5t = 0$ | B1 | |
| $z = \lambda,\ t = \mu \Rightarrow x = -\lambda - 8\mu,\ y = -3\lambda - 5\mu$ | M1 | |
| Basis for null space is $\left\{\lambda\begin{pmatrix}-1\\-3\\1\\0\end{pmatrix}, \mu\begin{pmatrix}-8\\-5\\0\\1\end{pmatrix}\right\}$ | A1 A1 | OE |
| | **7** | |
**7(ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{A}\begin{pmatrix}-1\\1\\-1\\1\end{pmatrix} = \begin{pmatrix}3\\21\\24\\27\end{pmatrix}$ | B1 | |
| $\mathbf{x} = \begin{pmatrix}-1\\1\\-1\\1\end{pmatrix} + \lambda\begin{pmatrix}-1\\-3\\1\\0\end{pmatrix} + \mu\begin{pmatrix}-8\\-5\\0\\1\end{pmatrix}$ | M1A1FT | OE |
| | **3** | |
---7 The linear transformation $\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }$ is represented by the matrix $\mathbf { A }$, where
$$\mathbf { A } = \left( \begin{array} { r r r r }
1 & - 1 & - 2 & 3 \\
5 & - 3 & - 4 & 25 \\
6 & - 4 & - 6 & 28 \\
7 & - 5 & - 8 & 31
\end{array} \right)$$
(i) Find the rank of $\mathbf { A }$ and a basis for the null space of T .\\
(ii) Find the matrix product $\mathbf { A } \left( \begin{array} { r } - 1 \\ 1 \\ - 1 \\ 1 \end{array} \right)$ and hence find the general solution of the equation $\mathbf { A } \mathbf { x } = \left( \begin{array} { r } 3 \\ 21 \\ 24 \\ 27 \end{array} \right)$.\\
$8 \quad$ Let $I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \sec ^ { n } x \mathrm {~d} x$ for $n > 0$.\\
\hfill \mbox{\textit{CAIE FP1 2017 Q7 [10]}}