Show that $R$ is an equivalence relation.

The relationship between $a$ , $b$ , $c$ and $d$ is changed to $ad - bc = n$ . State, with a reason, whether or not there are any non-zero values of $n$ , other than $1$, for which $R$ is an equivalence relation.

since $\boldsymbol{A} = \boldsymbol{AI}$ where $\boldsymbol{I}$ is the identity     A1

and $\det (\boldsymbol{I}) = 1$ ,     A1

$R$ is reflexive

$\boldsymbol{ARB} \Rightarrow \boldsymbol{A} = \boldsymbol{BX}$ where $\det (\boldsymbol{X}) = 1$     M1

it follows that $\boldsymbol{B} = \boldsymbol{A}{\boldsymbol{X}^{ - 1}}$     A1

and $\det ({\boldsymbol{X}^{ - 1}}) = \det{(\boldsymbol{X})^{ - 1}} = 1$     A1

$R$ is symmetric

$\boldsymbol{ARB}$ and $\boldsymbol{BRC} \Rightarrow \boldsymbol{A} = \boldsymbol{BX}$ and $\boldsymbol{B} = \boldsymbol{CY}$ where $\det (\boldsymbol{X}) = \det (\boldsymbol{Y}) = 1$     M1

it follows that $\boldsymbol{A} = \boldsymbol{CYX}$     A1

$\det (\boldsymbol{YX}) = \det (\boldsymbol{Y})\det (\boldsymbol{X}) = 1$     A1

$R$ is transitive

hence $R$ is an equivalence relation     AG

[8 marks]

for reflexivity, we require $\boldsymbol{ARA}$ so that $\boldsymbol{A} = \boldsymbol{AI}$ (for all $\boldsymbol{A} \in S$ )     M1

since $\det (\boldsymbol{I}) = 1$ and we require $\boldsymbol{I} \in S$ the only possibility is $n = 1$     A1

[2 marks]

This question was not well done in general, again illustrating that questions involving both matrices and equivalence relations tend to cause problems for candidates. A common error was to assume, incorrectly, that ARB and BRC $\Rightarrow A = BX$ and $B = CX$ , not realizing that a different "$x$" is required each time. In proving that $R$ is an equivalence relation, consideration of the determinant is necessary in this question although many candidates neglected to do this.

In proving that $R$ is an equivalence relation, consideration of the determinant is necessary in this question although many candidates neglected to do this.