Show that any matrix of this form is its own inverse.

Show that $S$ forms an Abelian group under matrix multiplication.

Giving a reason, state whether or not this group is cyclic.

$\left( {\begin{array}{*{20}{c}} a&0&0\\ 0&b&0\\ 0&0&c \end{array}} \right)\left( {\begin{array}{*{20}{c}} a&0&0\\ 0&b&0\\ 0&0&c \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{a^2}}&0&0\\ 0&{{b^2}}&0\\ 0&0&{{c^2}} \end{array}} \right)$     A1M1

$= \left( {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right)$     A1

this shows that each matrix is self-inverse

[3 marks]

closure:

$\left( {\begin{array}{*{20}{c}} {{a_1}}&0&0\\ 0&{{b_1}}&0\\ 0&0&{{c_1}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{a_2}}&0&0\\ 0&{{b_2}}&0\\ 0&0&{{c_2}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{a_1}{a_2}}&0&0\\ 0&{{b_1}{b_2}}&0\\ 0&0&{{c_1}{c_2}} \end{array}} \right)$     M1A1

$= \left( {\begin{array}{*{20}{c}} {{a_3}}&0&0\\ 0&{{b_3}}&0\\ 0&0&{{c_3}} \end{array}} \right)$

where each of ${a_3}$, ${b_3}$, ${c_3}$ can only be $\pm 1$     A1

this proves closure

identity: the identity matrix is the group identity     A1

inverse: as shown above, every element is self-inverse     A1

associativity: this follows because matrix multiplication is associative     A1

$S$ is therefore a group     AG

Abelian:

$\left( {\begin{array}{*{20}{c}} {{a_2}}&0&0\\ 0&{{b_2}}&0\\ 0&0&{{c_2}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{a_1}}&0&0\\ 0&{{b_1}}&0\\ 0&0&{{c_1}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{a_2}{a_1}}&0&0\\ 0&{{b_2}{b_1}}&0\\ 0&0&{{c_2}{c_1}} \end{array}} \right)$     A1

$\left( {\begin{array}{*{20}{c}} {{a_1}}&0&0\\ 0&{{b_1}}&0\\ 0&0&{{c_1}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{a_2}}&0&0\\ 0&{{b_2}}&0\\ 0&0&{{c_2}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{a_1}{a_2}}&0&0\\ 0&{{b_1}{b_2}}&0\\ 0&0&{{c_1}{c_2}} \end{array}} \right)$     A1

Note: Second line may have been shown whilst proving closure, however a reference to it must be made here.

we see that the same result is obtained either way which proves commutativity so that the group is Abelian      R1

[9 marks]

since all elements (except the identity) are of order $2$, the group is not cyclic (since $S$ contains $8$ elements)     R1

[1 mark]