| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| 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 systematic Further Maths question on null spaces requiring row reduction and basis finding. While it involves multiple parts and the conceptual understanding of subspaces, the techniques are standard for FP1 linear algebra with no novel insights required. The parameter α adds mild complexity but the row reduction is straightforward. |
| Spec | 4.03a Matrix language: terminology and notation4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Reduces to row echelon form: \(\begin{pmatrix}1&2&\alpha&-1\\2&6&-3&-3\\3&10&-6&-5\end{pmatrix}\sim\begin{pmatrix}1&2&\alpha&-1\\0&2&-3-2\alpha&-1\\0&4&-6-3\alpha&-2\end{pmatrix}\) | M1 | Good attempt at REF |
| \(\sim\begin{pmatrix}1&2&\alpha&-1\\0&2&-3-2\alpha&-1\\0&0&\alpha&0\end{pmatrix}\) | A1 | |
| Solves system of equations: \(x+2y+\alpha z-t=0\) | M1 | Forms system of equations from row echelon matrix |
| \(2y-(3+2\alpha)z-t=0\); \(\alpha z=0\); Since \(\alpha\neq 0\), \(z=0\) | A1 | Three correct equations |
| \((K_1=)\left\{\begin{pmatrix}0\\1\\0\\2\end{pmatrix}\right\}\) | A1 | AEF |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| If \(\alpha=0\) then \(x+2y-t=0\); \(2y-3z-t=0\) | M1 | Forms system of 2 equations |
| \(\Rightarrow K_2=\text{any 2 of } \begin{pmatrix}-6\\3\\2\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\\2\end{pmatrix}\ or\ \begin{pmatrix}3\\0\\-1\\3\end{pmatrix}\) | A1 A1 | AE |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| A basis vector for \(K_1\) forms part of a basis for \(K_2\) or a linear combination of basis vectors from \(K_2\) form basis for \(K_1\) | M1 | Credit should be given for a correct conclusion using their bases |
| Therefore \(K_1\) is a subspace of \(K_2\) | A1 FT | |
| Total: 2 |
## Question 8(i):
| Reduces to row echelon form: $\begin{pmatrix}1&2&\alpha&-1\\2&6&-3&-3\\3&10&-6&-5\end{pmatrix}\sim\begin{pmatrix}1&2&\alpha&-1\\0&2&-3-2\alpha&-1\\0&4&-6-3\alpha&-2\end{pmatrix}$ | M1 | Good attempt at REF |
|---|---|---|
| $\sim\begin{pmatrix}1&2&\alpha&-1\\0&2&-3-2\alpha&-1\\0&0&\alpha&0\end{pmatrix}$ | A1 | |
| Solves system of equations: $x+2y+\alpha z-t=0$ | M1 | Forms system of equations from row echelon matrix |
| $2y-(3+2\alpha)z-t=0$; $\alpha z=0$; Since $\alpha\neq 0$, $z=0$ | A1 | Three correct equations |
| $(K_1=)\left\{\begin{pmatrix}0\\1\\0\\2\end{pmatrix}\right\}$ | A1 | AEF |
| **Total: 5** | | |
---
## Question 8(ii):
| If $\alpha=0$ then $x+2y-t=0$; $2y-3z-t=0$ | M1 | Forms system of 2 equations |
| $\Rightarrow K_2=\text{any 2 of } \begin{pmatrix}-6\\3\\2\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\\2\end{pmatrix}\ or\ \begin{pmatrix}3\\0\\-1\\3\end{pmatrix}$ | A1 A1 | AE |
| **Total: 3** | | |
---
## Question 8(iii):
| A basis vector for $K_1$ forms part of a basis for $K_2$ or a linear combination of basis vectors from $K_2$ form basis for $K_1$ | M1 | Credit should be given for a correct conclusion using their bases |
|---|---|---|
| Therefore $K_1$ is a subspace of $K_2$ | A1 FT | |
| **Total: 2** | | |
---8 The linear transformation $\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 3 }$ is represented by the matrix $\mathbf { M }$, where
$$\mathbf { M } = \left( \begin{array} { r r c r }
1 & 2 & \alpha & - 1 \\
2 & 6 & - 3 & - 3 \\
3 & 10 & - 6 & - 5
\end{array} \right)$$
and $\alpha$ is a constant. When $\alpha \neq 0$ the null space of T is denoted by $K _ { 1 }$.\\
(i) Find a basis for $K _ { 1 }$.\\
When $\alpha = 0$ the null space of T is denoted by $K _ { 2 }$.\\
(ii) Find a basis for $K _ { 2 }$.\\
(iii) Determine, justifying your answer, whether $K _ { 1 }$ is a subspace of $K _ { 2 }$.\\
\hfill \mbox{\textit{CAIE FP1 2018 Q8 [10]}}