| Date | May 2017 | Marks available | 4 | Reference code | 17M.1.hl.TZ0.10 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Determine | Question number | 10 | Adapted from | N/A |
Question
Let \(G\) denote the set of \(2 \times 2\) matrices whose elements belong to \(\mathbb{R}\) and whose determinant is equal to 1. Let \( * \) denote matrix multiplication which may be assumed to be associative.
Let \(H\) denote the set of \(2 \times 2\) matrices whose elements belong to \(\mathbb{Z}\) and whose determinant is equal to 1.
Show that \(\{ G,{\text{ }} * \} \) is a group.
Determine whether or not \(\{ H,{\text{ }} * \} \) is a subgroup of \(\{ G,{\text{ }} * \} \).
Markscheme
closure: let A, B \( \in G\)
(because AB is a \(2 \times 2\) matrix)
and det(AB) = det(A)det(B) \( = 1 \times 1 = 1\) M1A1
identity: the \(2 \times 2\) identity matrix has determinant 1 R1
inverse: let A \( \in G\). Then A has an inverse because it is non-singular (R1)
since AA\(^{ - 1} = \) I, det(A)det(A\(^{ - 1}\)) = det(I) = 1 therefore A\(^{ - 1} \in G\) R1
associativity is assumed
the four axioms are satisfied therefore \(\{ G,{\text{ }} * \} \) is a group AG
[5 marks]
closure: let A, B \( \in H\). Then AB \( \in H\) because the arithmetic involved produces elements that are integers R1
inverse: A\(^{ - 1} \in H\) because the calculation of the inverse involves interchanging the elements and dividing by the determinant which is 1 R1
the identity (and associativity) follow as above R1
therefore \(\{ H,{\text{ }} * \} \) is a subgroup of \(\{ G,{\text{ }} * \} \) A1
Note: Award the A1 only if the first two R1 marks are awarded but not necessarily the third R1.
Note: Accept subgroup test.
[4 marks]
