The permutation \({p_1}\) of the set {1, 2, 3, 4} is defined by

\[{p_1} = \left( {\begin{array}{*{20}{c}}
  1&2&3&4 \\
  2&4&1&3
\end{array}} \right)\]

(a)     (i)     State the inverse of \({p_1}\).

  (ii)     Find the order of \({p_1}\).

(b)     Another permutation \({p_2}\) is defined by

\[{p_2} = \left( {\begin{array}{*{20}{c}}
  1&2&3&4 \\
  3&2&4&1
\end{array}} \right)\]

  (i)     Determine whether or not the composition of \({p_1}\) and \({p_2}\) is commutative.

  (ii)     Find the permutation \({p_3}\) which satisfies

\[{p_1}{p_3}{p_2} = \left( {\begin{array}{*{20}{c}}
  1&2&3&4 \\
  1&2&3&4
\end{array}} \right){\text{.}}\]

(a)     (i)     the inverse is

\(\left( {\begin{array}{*{20}{c}}
  1&2&3&4 \\
  3&1&4&2
\end{array}} \right)\)     A1

 

(ii)     EITHER

\(1 \to 2 \to 4 \to 3 \to 1\) (is a cycle of length 4)     R3

so \({p_1}\) is of order 4     A1     N2

OR

consider

\(p_1^2 = \left( {\begin{array}{*{20}{c}}
  1&2&3&4 \\
  4&3&1&2
\end{array}} \right)\)     M1A1

it is now clear that

\(p_1^4 = \left( {\begin{array}{*{20}{c}}
  1&2&3&4 \\
  1&2&3&4
\end{array}} \right)\)     A1

so \({p_1}\) is of order 4     A1     N2

[5 marks]

 

(b)     (i)     consider

\({p_1}{p_2} = \left( {\begin{array}{*{20}{c}}
  1&2&3&4 \\
  2&4&1&3
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&2&3&4 \\
  3&2&4&1
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1&2&3&4 \\
  1&4&3&2
\end{array}} \right)\)     M1A1

\({p_2}{p_1} = \left( {\begin{array}{*{20}{c}}
  1&2&3&4 \\
  3&2&4&1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&2&3&4 \\
  2&4&1&3
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1&2&3&4 \\
  2&1&3&4
\end{array}} \right)\)     A1

composition is not commutative     A1

Note: In this part do not penalize candidates who incorrectly reverse the order both times.

 

(ii)     EITHER

pre and postmultiply by \(p_1^{ - 1}\), \(p_2^{ - 1}\)to give

\({p_3} = p_1^{ - 1}p_2^{ - 1}\)     (M1)(A1)

\( = \left( {\begin{array}{*{20}{c}}
  1&2&3&4 \\
  3&1&4&2
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&2&3&4 \\
  4&2&1&3
\end{array}} \right)\)     A1

\( = \left( {\begin{array}{*{20}{c}}
  1&2&3&4 \\
  2&1&3&4
\end{array}} \right)\)     A1

OR

starting from

\(\left( {\begin{array}{*{20}{c}}
  1&2&3&4 \\
  2&4&1&3
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&2&3&4 \\
  {}&{}&{}&{}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&2&3&4 \\
  3&2&4&1
\end{array}} \right)\)     M1

successively deducing each missing number, to get

\(\left( {\begin{array}{*{20}{c}}
  1&2&3&4 \\
  2&4&1&3
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&2&3&4 \\
  2&1&3&4
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&2&3&4 \\
  3&2&4&1
\end{array}} \right)\)     A3

[8 marks]

Total [13 marks]

Many candidates scored well on this question although some gave the impression of not having studied this topic. The most common error in (b) was to believe incorrectly that \({p_1}{p_2}\) means \({p_1}\) followed by \({p_2}\). This was condoned in (i) but penalised in (ii). The Guide makes it quite clear that this is the notation to be used.