(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]