Expand det(λI−A) down the first column. The remaining determinant is $$\lambda\begin{vmatrix}\lambda&-1\\3&\lambda+4\end{vmatrix}.$$
Thus the characteristic polynomial is $\lambda(\lambda^2+4\lambda+3)=\lambda(\lambda+1)(\lambda+3)$. The eigenvalues are **0,−1,−3, A**. Their sum −4 and product zero also match the trace and determinant.
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