2008 Paper 2 Question 16

(a)

(i)

As an intermediate result

$A^2 = \left( \begin{matrix}0 & 0 & 1\\ 1& 0 &0\\ 0& 1& 0\end{matrix} \right)$

(ii)

$A^{3n+2} = (A^3)^n A^2 = (I)^n A^2 = A^2$

and $A^2$ is given explicitly in (i).

(iii)

Use

$M^{-1} = \frac{1}{\text{det}M }\left[ \text{cof }M \right]^T$

Now

$\text{det}A^2 = 1$

and the cofactor matrix is

$\text{cof }A^2 = \left( \begin{matrix}0 & 0 & 1\\ 1& 0 &0\\ 0& 1& 0\end{matrix} \right)$

Thus

$(A^2)^{-1}=\left( \begin{matrix}0 & 1 & 0\\ 0& 0 &1\\ 1& 0& 0\end{matrix} \right)$

This can easily be verified by multiplying by $A^2$ .

(iv)

$\begin{aligned} A^{3n+1} &= (A^3)^n A = (I)^n A = A \\ \therefore (A^{3n+1})^{-1} &= A^{-1}\end{aligned}$

We know $A^3=I$ , so

$\begin{aligned} A^2A &= I\\ A^2 = A^{-1} \end{aligned}$

so finally

$(A^{3n+1} )^{-1} = A^2$

NST First Year Maths