Matrix Equations 
I have worked it out and determined that both equations hold. I am wondering why this is the case. Is there a reason why the equation holds for these two types of matrices?
 A: This is an application of the Cayley-Hamilton Theorem. 
For any matrix $A$, we define its characteristic polynomial as $p(\lambda) = \det(\lambda I - A)$. Then, $A$ satisfies $p(A) = 0$. 
Example: The characteristic polynomial of the first matrix, $A$, is 
$p(\lambda) = \det(\lambda I-A) = \left|\begin{matrix}\lambda-a&-x&-y\\0&\lambda-b&-z\\0&0&\lambda-c\end{matrix}\right| = (\lambda-a)(\lambda-b)(\lambda-c)$ $= \lambda^3 - (a+b+c)\lambda^2+(ab+bc+ca)\lambda - abc$. 
So, by the Cayley-Hamilton Theorem, $A$ satisfies the equation 
$A^3 - (a+b+c)A^2+(ab+bc+ca)A - abcI = 0$.
A: Indeed it's a really more general case which involves Exterior Algebras. In fact
$$\Delta(\lambda)=\lambda^{n}+tr(A)\lambda^{n-1}+...+tr(\Lambda^{n-1}A)\lambda+detA$$
where $tr(\Lambda^{k}A)$ can be practically thought as the sum of the principal determinants of order k. 
For example:
$$
det\left(\begin{array}{ccc}
a_{1}^{1}-\lambda & a_{2}^{1} & a_{3}^{1}\\
a_{1}^{2} & a_{2}^{2}-\lambda & a_{3}^{2}\\
a_{1}^{3} & \cdots & a_{3}^{3}-\lambda
\end{array}\right)
$$
Has the following 
$$\Delta(\lambda)=\lambda^{3}(-1)+\lambda^{2}(a_{3}^{3}+a_{2}^{2}+a_{1}^{1})+\lambda(a_{2}^{1}a_{1}^{2}-a_{2}^{2}a_{1}^{1}+a_{3}^{1}a_{1}^{3}+a_{3}^{2}a_{2}^{3}-a_{3}^{3}a_{1}^{1}-a_{3}^{3}a_{2}^{2})+detA
$$
To really get an further insight on the topic we should watch for Exterior Algebra... I'm not sure about which reference to give you since the book I have where it's easily treated is in italian... but I'm sure it's plenty of english references on the subject.
