(i)     Find the matrices \({\boldsymbol{A}^2}\) and \({\boldsymbol{A}^3}\) , and verify that \({{\boldsymbol{A}}^3} = 2{{\boldsymbol{A}}^2} - {\boldsymbol{A}}\) .

(ii)     Deduce that \({{\boldsymbol{A}}^4} = 3{{\boldsymbol{A}}^2} - 2{\boldsymbol{A}}\) .

(i)     Suggest a similar expression for \({\boldsymbol{A}^n}\) in terms of \(\boldsymbol{A}\) and \({\boldsymbol{A}^2}\) , valid for \(n \ge 3\) .

(ii)     Use mathematical induction to prove the validity of your suggestion.

(i)     \({{\boldsymbol{A}}^2} = \left( {\begin{array}{*{20}{c}}
  2&4&1 \\
  4&7&2 \\
  { - 14}&{ - 26}&{ - 7}
\end{array}} \right)\)     A1

\({{\boldsymbol{A}}^3} = \left( {\begin{array}{*{20}{c}}
  4&7&2 \\
  6&{10}&3 \\
  { - 24}&{ - 41}&{ - 12}
\end{array}} \right)\)     A1

\(2{{\boldsymbol{A}}^2} - {\boldsymbol{A}} = 2\left( {\begin{array}{*{20}{c}}
  2&4&1 \\
  4&7&2 \\
  { - 14}&{ - 26}&{ - 7}
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
  0&1&0 \\
  2&4&1 \\
  { - 4}&{ - 11}&{ - 2}
\end{array}} \right)\)     M1

\( = \left( {\begin{array}{*{20}{c}}
  4&7&2 \\
  6&{10}&3 \\
  { - 24}&{ - 41}&{ - 12}
\end{array}} \right) = {{\boldsymbol{A}}^3}\)     AG

(ii)     \({{\boldsymbol{A}}^4} = {\boldsymbol{A}}{{\boldsymbol{A}}^3}\)     M1

\( = {\boldsymbol{A}}(2{{\boldsymbol{A}}^2} - {\boldsymbol{A}})\)     A1

\( = 2{{\boldsymbol{A}}^3} - {{\boldsymbol{A}}^2}\)

\( = 2(2{{\boldsymbol{A}}^2} - {\boldsymbol{A}}) - {{\boldsymbol{A}}^2}\)     A1

\( = 3{{\boldsymbol{A}}^2} - 2{\boldsymbol{A}}\)     AG

Note: Accept alternative solutions that include correct calculation of both sides of the expression.

[6 marks]

(i)     conjecture: \({{\boldsymbol{A}}^n} = \left( {n - 1} \right){{\boldsymbol{A}}^2} - \left( {n - 2} \right){\boldsymbol{A}}\)     A1

(ii)     first check that the result is true for \(n = 3\)

the formula gives \({{\boldsymbol{A}}^3} = 2{{\boldsymbol{A}}^2} - {\boldsymbol{A}}\) which is correct     A1

assume the result for \(n = k\) , i.e.     M1

\({{\boldsymbol{A}}^k} = (k - 1){{\boldsymbol{{\rm A}}}^2} - (k - 2){\boldsymbol{A}}\)

so

\({{\boldsymbol{A}}^{k + 1}} = {\boldsymbol{A}}\left[ {\left( {k - 1} \right){{\boldsymbol{A}}^2} - \left( {k - 2} \right){\boldsymbol{A}}} \right]\)     M1

\( = \left( {k - 1} \right){{\boldsymbol{A}}^3} - \left( {k - 2} \right){{\boldsymbol{A}}^2}\)     A1

\( = \left( {k - 1} \right)\left( {2{{\boldsymbol{A}}^2} - {\boldsymbol{A}}} \right) - \left( {k - 2} \right){{\boldsymbol{A}}^2}\)     M1

\( = k{{\boldsymbol{A}}^2} - \left( {k - 1} \right){\boldsymbol{A}}\)     A1

so true for \(n = k \Rightarrow \) true for \(n = k + 1\) and since true for \(n = 3\) ,

the result is proved by induction     R1

Note: Only award the R1 mark if a reasonable attempt at a proof by induction has been made.

[8 marks]