Problem related with characteristic polynomial I have been trying to solve the following problem:
The coefficient of $\lambda^{3}$ in the characteristic polynomial $f(\lambda)$ of \begin{pmatrix}
2 &3  &0  &1 \\ 
4&-1  &0  &0 \\ 
 0&3  &-4  &8 \\ 
 2&1  &-4  &2 
\end{pmatrix}
is

(a) $2$  (b) $3$  (c) $0$  (d) $1$. 

My question is: Is there any other way to find the coefficient of $\lambda^{3}$,apart from finding the characteristic polynomial of the given matrix? I have solved it by finding the characteristic polynomial which is lengthy. Any other alternative suggestions to tackle the problem in a better way will be appreciated. Thanks in advance for your time.
 A: Hint:
Lemma: For any square matrix $\,A\,$ of order $\,n\times n\,$ , the coefficient of $\,x^{n-1}\,$ in the characteristic polynomial of $\,A\,$ is the trace of $\,A\,\,\,,\,tr.(A)\,$,  multipied by $\,(-1)^{n-1}$
A: Finding the characteristic polynomial is probably the most direct way, though it can be admittedly tedious.
If you happened to know that the eigenvalues of the matrix are $\lambda_1,...,\lambda_4$ (not necessarily distinct, but with appropriate algebraic multiplicities), then the coefficient of $\lambda^3$ in the characteristic polynomial would be $-(\lambda_1+...+\lambda_4)$. (Why?) This, though, is simply the opposite of the trace of the matrix! (Why?) Certainly, that's the simplest way to find the desired coefficient, though you need more tools in your arsenal to be able to justify it, than if you just find the characteristic polynomial.
If you happen to know that this matrix is similar to some other, nicer matrix (diagonal, or upper/lower triangular, say) then you need only calculate the characteristic polynomial of the nicer matrix, since it will be the same as that of the matrix you're looking at. (Why?)
