| Date | None Specimen | Marks available | 9 | Reference code | SPNone.1.hl.TZ0.14 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Show that | Question number | 14 | Adapted from | N/A |
Question
The set \(S\) contains the eight matrices of the form\[\left( {\begin{array}{*{20}{c}}
a&0&0\\
0&b&0\\
0&0&c
\end{array}} \right)\]where \(a\), \(b\), \(c\) can each take one of the values \( + 1\) or \( - 1\) .
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.
Markscheme
\(\left( {\begin{array}{*{20}{c}}
a&0&0\\
0&b&0\\
0&0&c
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
a&0&0\\
0&b&0\\
0&0&c
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{{a^2}}&0&0\\
0&{{b^2}}&0\\
0&0&{{c^2}}
\end{array}} \right)\) 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]
