The matrix A is given by A = $\left( {\begin{array}{*{20}{c}}1&2&1\\1&1&2\\2&3&1\end{array}} \right)$.

(a)     Given that A$^3$ can be expressed in the form A$^3 = a$A$^2 = b$A $+ c$I, determine the values of the constants $a$, $b$, $c$.

(b)     (i)     Hence express A$^{ - 1}$ in the form A$^{ - 1} = d$A$^2 = e$A $+ f$I where $d,{\text{ }}e,{\text{ }}f \in \mathbb{Q}$.

(ii)     Use this result to determine A$^{ - 1}$.

(a)     successive powers of A are given by

A$^2 =$ $\left( {\begin{array}{*{20}{c}}5&7&6\\6&9&5\\7&{10}&9\end{array}} \right)$     (M1)A1

A$^3 =$ $\left( {\begin{array}{*{20}{c}}{24}&{35}&{25}\\{25}&{36}&{29}\\{35}&{51}&{36}\end{array}} \right)$     A1

it follows, considering elements in the first rows, that

$5a + b + c = 24$

$7a + 2b = 35$

$6a + b = 25$     M1A1

solving,     (M1)

$(a,{\text{ }}b,{\text{ }}c) = (3,{\text{ }}7,{\text{ }}2)$     A1

 

Note: Accept any other three correct equations.

 

Note: Accept the use of the Cayley–Hamilton Theorem.

 

[7 marks]

 

(b)     (i)     it has been shown that

A$^3 = 3$A$^2 + 7$A$+ 2$I

multiplying by A$^{ - 1}$,     M1

A$^2 = 3$A$+ 7$I$+ 2$A$^{ - 1}$     A1

whence

A$^{ - 1} = 0.5$A$^2 - 1.5$A $- 3.5$I     A1

(ii)     substituting powers of A,

A$^{ - 1} = 0.5$$\left( {\begin{array}{*{20}{c}}5&7&6\\6&9&5\\7&{10}&9\end{array}} \right) - 1.5\left( {\begin{array}{*{20}{c}}1&2&1\\1&1&2\\2&3&1\end{array}} \right) - 3.5\left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)$     M1

=$\left( {\begin{array}{*{20}{c}}{ - 2.5}&{0.5}&{1.5}\\{1.5}&{ - 0.5}&{ - 0.5}\\{0.5}&{0.5}&{ - 0.5}\end{array}} \right)$    A1

 

Note: Follow through their equation in (b)(i).

 

Note: Line (ii) of (ii) must be seen.

 

[5 marks]

[N/A]