I'm rather unsure what is going on with the last part of question c any help would be greatly appreciated!
Hi there,
Your matrix is diagonalizable. In particular,
$\displaystyle
$B=PDP^{-1}$, where
$D=\begin{pmatrix}
6 & 0 & 0 \\
0 & 6 & 0 \\
0 & 0 & 3\\
\end{pmatrix}\\
$
Hence,
$\displaystyle
B^n = (PDP^{-1})^n = (PDP^{-1})(PDP^{-1}) ... (PDP^{-1})\\
= PD^nP^{-1}\\
$
Substituting this into your equation yields,
$\displaystyle
a_nPD^nP^{-1} + ... + a_1PDP^{-1} + a_0& = C\\
$
and pre and post multiplying both sides by P inv. and P (respectively) yields:
$\displaystyle
a_nD^n + ... + a_0& = P^{-1}CP\\
$
where C is that diagonal matrix you are given on the right-hand side of the equation (with entries 1, 2, 3).
However, you will notice that when you multiply P^{-1}CP out, you will get a matrix that is not diagonal on the right hand side of the last equation.
But a power of a diagonal matrix is again a diagonal matrix, a scalar multiple of a diagonal matrix is also diagonal, and the sum of diagonal matrices is again diagonal. Hence, you will never get the RHS, so it is impossible.
Don't worry about it. It's something I might not have figured out some other day.
When I am facing problems like this, and a potential way of solving it doesn't occur to me within a few minutes, I consciously remind myself of "what I have in my toolbox." This involves not only summoning up what I know about diagonalizable matrices (and perhaps even making lists and drawing arrows to indicate potential connections between ideas), but also reminding myself of the basic methods of proof: direct proof, contrapositive, contradiction, etc. When in doubt -- proof by induction!