Let’s find the characteristic polynomial of a 3x3 matrix. Let \[A=\begin{bmatrix}3&2&1\\0&1&1\\1&2&0 \end{bmatrix}\]
To get the eigenvalues, we need to solve \[ det(\begin{bmatrix}3&2&1\\0&1&1\\1&2&0\end{bmatrix}- \begin{bmatrix}\lambda&0&0\\0&\lambda&0\\0&0&\lambda \end{bmatrix})=0 \]
That is, \[det(\begin{bmatrix}3-\lambda&2&1\\0&1-\lambda&1\\1&2&0-\lambda \end{bmatrix})=0 \]
Which reduces to \[(3-\lambda)det(\begin{bmatrix}1-\lambda&1\\2&0-\lambda\end{bmatrix})+ det(\begin{bmatrix}2&1\\1-\lambda&1\end{bmatrix})=0 \]
\[(3-\lambda)[(1-\lambda)(0-\lambda)-2]+2-(1-\lambda)=0 \\ (3-\lambda)(\lambda^{2}-\lambda-2)+\lambda+1=0\\ -\lambda^{3}+4\lambda^{2}-\lambda-6+\lambda+1=0\\ -\lambda^{3}+4\lambda^{2}-5=0\]
which gives us the characteristic polynomial as \[p_{A}(x)=-\lambda^{3}+4\lambda^{2}-5\]