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]