| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Discrete (Further AS Paper 2 Discrete) |
| Year | 2022 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Complete or analyse Cayley table |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question requiring routine matrix multiplication of 2×2 matrices with simple entries (0s and 1s), identifying the identity element from a completed table, and commenting on commutativity. While it's a Further Maths topic, the calculations are mechanical and the conceptual demands are minimal—students just need to recognize that commutativity of this specific set doesn't imply general commutativity of matrix multiplication. |
| Spec | 8.03a Binary operations: and their properties on given sets8.03c Group definition: recall and use, show structure is/isn't a group |
| A | B | C | D | |
| A | A | D | ||
| B | B | |||
| C | C | |||
| D | D |
| Answer | Marks |
|---|---|
| Completes correctly at least 2 rows or 2 columns of Cayley table | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(A\) | \(B\) | \(C\) |
| \(A\) | \(A\) | \(A\) |
| \(B\) | \(A\) | \(B\) |
| \(C\) | \(D\) | \(C\) |
| \(D\) | \(D\) | \(D\) |
| A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Row and column for \(B\) are same as row and column heading, therefore \(B\) is the identity element of \(S\) | E1 | Explains that left and right multiplication by \(B\) leaves matrix unchanged |
| Answer | Marks | Guidance |
|---|---|---|
| Cayley table is symmetrical about leading diagonal so matrix multiplication is commutative for \(S\) | M1 | |
| However matrix multiplication is not a commutative operation in general; therefore Sam's statement is not valid | A1 | Must state commutative for \(S\) but not in general, concluding statement is not valid |
## Question 6(a):
Completes correctly at least 2 rows or 2 columns of Cayley table | M1 |
Completes full Cayley table correctly:
| | $A$ | $B$ | $C$ | $D$ |
|---|---|---|---|---|
| $A$ | $A$ | $A$ | $D$ | $D$ |
| $B$ | $A$ | $B$ | $C$ | $D$ |
| $C$ | $D$ | $C$ | $C$ | $D$ |
| $D$ | $D$ | $D$ | $D$ | $D$ |
| A1 |
---
## Question 6(b):
Row and column for $B$ are same as row and column heading, therefore $B$ is the identity element of $S$ | E1 | Explains that left and right multiplication by $B$ leaves matrix unchanged
---
## Question 6(c):
Cayley table is symmetrical about leading diagonal so matrix multiplication is commutative for $S$ | M1 |
However matrix multiplication is not a commutative operation in general; therefore Sam's statement is not valid | A1 | Must state commutative for $S$ but not in general, concluding statement is not valid
---6 The set $S$ is given by $S = \{ \mathbf { A } , \mathbf { B } , \mathbf { C } , \mathbf { D } \}$ where\\
$\mathbf { A } = \left[ \begin{array} { l l } 1 & 0 \\ 0 & 0 \end{array} \right]$\\
$\mathbf { B } = \left[ \begin{array} { l l } 1 & 0 \\ 0 & 1 \end{array} \right]$\\
$\mathbf { C } = \left[ \begin{array} { l l } 0 & 0 \\ 0 & 1 \end{array} \right]$\\
$\mathbf { D } = \left[ \begin{array} { l l } 0 & 0 \\ 0 & 0 \end{array} \right]$
6
\begin{enumerate}[label=(\alph*)]
\item Complete the Cayley table for $S$ under matrix multiplication.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
& A & B & C & D \\
\hline
A & A & & D & \\
\hline
B & & B & & \\
\hline
C & & & C & \\
\hline
D & & & & D \\
\hline
\end{tabular}
\end{center}
6
\item Using the Cayley table above, explain why $\mathbf { B }$ is the identity element of $S$ under matrix multiplication.\\[0pt]
[1 mark]
6
\item Sam states that the Cayley table in part (a) shows that matrix multiplication is commutative.
Comment on the validity of Sam's statement.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Discrete 2022 Q6 [5]}}