| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Properties of matrix operations |
| Difficulty | Challenging +1.2 This is a structured group theory proof requiring matrix multiplication, solving simultaneous equations from the commutative property, and verifying group axioms. While it involves multiple steps and abstract algebra concepts, each part follows standard procedures: (i) is algebraic manipulation, (ii) applies the non-singular condition, and (iii) systematically checks closure, identity, and inverses. The question is more demanding than average A-level work due to the abstract nature and proof requirements, but the techniques are well-practiced in FP3. |
| Spec | 8.03c Group definition: recall and use, show structure is/isn't a group8.03f Subgroups: definition and tests for proper subgroups |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\mathbf{AQ} = \mathbf{QA} \Rightarrow \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) | M1 | For considering \(\mathbf{AQ} = \mathbf{QA}\) with general A |
| i.e. \(\begin{pmatrix} a & a+b \\ c & c+d \end{pmatrix} = \begin{pmatrix} a+c & b+d \\ c & d \end{pmatrix}\) | A1 | For correct simplified equation |
| Hence \(a = a + c\) and \(a + b = b + d\) i.e. \(c = 0\) and \(d = a\) | M1 | For equating corresponding entries |
| A1 | For complete proof | |
| 4 | ||
| (ii) To be non-singular, \(a \neq 0\) | B1 | For stating that \(a\) is non-zero |
| (iii) Identity is \(\begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}\) as usual, since this is in \(S\) | B1 | For justifying the identity correctly |
| Inverse of \(\begin{pmatrix} a & b \\ 0 & a \end{pmatrix}\) is \(\begin{pmatrix} 1/a & -b/a^2 \\ 0 & 1/a \end{pmatrix}\), as \(a \neq 0\) | B1 | For statement of correct inverse |
| B1 | For justification via non-zero \(a\) | |
| \(\begin{pmatrix} a_1 & b_1 \\ 0 & a_1 \end{pmatrix} \begin{pmatrix} a_2 & b_2 \\ 0 & a_2 \end{pmatrix} = \begin{pmatrix} a_1 a_2 & a_1 b_2 + b_1 a_2 \\ 0 & a_1 a_2 \end{pmatrix}\) | M1 | For considering a general product |
| This is in \(S\), since \(a_1 a_2 \neq 0\), so all necessary group properties are shown | A1 | For complete proof |
| 5 | ||
| 10 |
**(i)** $\mathbf{AQ} = \mathbf{QA} \Rightarrow \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ | M1 | For considering $\mathbf{AQ} = \mathbf{QA}$ with general A
i.e. $\begin{pmatrix} a & a+b \\ c & c+d \end{pmatrix} = \begin{pmatrix} a+c & b+d \\ c & d \end{pmatrix}$ | A1 | For correct simplified equation
Hence $a = a + c$ and $a + b = b + d$ i.e. $c = 0$ and $d = a$ | M1 | For equating corresponding entries
| A1 | For complete proof
| **4** |
**(ii)** To be non-singular, $a \neq 0$ | B1 | For stating that $a$ is non-zero
**(iii)** Identity is $\begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}$ as usual, since this is in $S$ | B1 | For justifying the identity correctly
Inverse of $\begin{pmatrix} a & b \\ 0 & a \end{pmatrix}$ is $\begin{pmatrix} 1/a & -b/a^2 \\ 0 & 1/a \end{pmatrix}$, as $a \neq 0$ | B1 | For statement of correct inverse
| B1 | For justification via non-zero $a$
$\begin{pmatrix} a_1 & b_1 \\ 0 & a_1 \end{pmatrix} \begin{pmatrix} a_2 & b_2 \\ 0 & a_2 \end{pmatrix} = \begin{pmatrix} a_1 a_2 & a_1 b_2 + b_1 a_2 \\ 0 & a_1 a_2 \end{pmatrix}$ | M1 | For considering a general product
This is in $S$, since $a_1 a_2 \neq 0$, so all necessary group properties are shown | A1 | For complete proof
| **5** |
| **10** |6 The set $S$ consists of all non-singular $2 \times 2$ real matrices $\mathbf { A }$ such that $\mathbf { A Q } = \mathbf { Q A }$, where
$$\mathbf { Q } = \left( \begin{array} { l l }
1 & 1 \\
0 & 1
\end{array} \right)$$
(i) Prove that each matrix $\mathbf { A }$ must be of the form $\left( \begin{array} { l l } a & b \\ 0 & a \end{array} \right)$.\\
(ii) State clearly the restriction on the value of $a$ such that $\left( \begin{array} { l l } a & b \\ 0 & a \end{array} \right)$ is in $S$.\\
(iii) Prove that $S$ is a group under the operation of matrix multiplication. (You may assume that matrix multiplication is associative.)
\hfill \mbox{\textit{OCR FP3 Q6 [10]}}