Eigenvalues of a 3x3 matrix

$$\begin{array}{rlr}& \mathbf{Mv}=\lambda \mathbf{v}& \text{(by definition of eigenvalue/eigenvector)}\\ & ⟹\mathbf{Mv}=\lambda \mathbf{Iv}& \\ & ⟹\left(\mathbf{M}-\lambda \mathbf{I}\right)\mathbf{v}=0& \\ & ⟹\det (\mathbf{M}-\lambda \mathbf{I})=0& \text{(as M maps v to the origin)}\end{array}$$

Hence

$$\det \left(\begin{array}{ccc}a-\lambda & b& c\\ d& e-\lambda & f\\ g& h& i-\lambda \end{array}\right)=0$$

We can - e.g. by doing $\left(\begin{array}{c}a-\lambda \\ d\\ g\end{array}\right)\times \left(\begin{array}{c}b\\ e-\lambda \\ h\end{array}\right)\cdot \left(\begin{array}{c}c\\ f\\ i-\lambda \end{array}\right)$ - produce a polynomial equation in $\lambda$, the solutions of which will be the eigenvalues of $\mathbf{M}$. This polynomial is called the characteristic polynomial of $\mathbf{M}$.

The Cayley-Hamilton theorem states that $\mathbf{M}$ always satisfies its own characteristic polynomial.