I am doing the following exercise in preparation for an exam:
I have solved (a) and (b)
(a) $\lambda(2\lambda^2-6\lambda+4)$
(b) $\lambda = {0,1,2}$
However I don't know the answer to (c). If I substitue $\lambda = 3$ into the matrix, I'm left with the following:
$$ \begin{matrix} -8 & -4 & 10 \\ 4 & -3 & 7 \\ 0 & 0 & -3 \\ \end{matrix} $$
Of course, I cannot calculate this to the power of 2011 to say something about the number of solutions. I'm guessing the answer is that there is only one solution since there are no zero rows and b is not a zero column?
For part c) because $\lambda$ is not one of the values 0, 1, or 2, $det(A)$ is nonzero and therefore $A$ is invertible. Therefore $A^{2011}$ is invertible and the system $A^{2011}x=b$ has a single solution, $x=A^{-2011}b$.