Prove the next equality. $A^3-\text{tr}(A)A^2+\text{tr}(\text{adj}(A))A-\det(A)I_3=0$ Given $A \in \mathbb{R}^{3 \times 3}$, prove the next equality:
$$A^3-\text{tr}(A)A^2+\text{tr}(\text{adj}(A))A-\det(A)I_3=0.$$
We know that:
$$A^3-\text{tr}(A)A^2+\text{tr}(\text{adj}(A))A-\det(A)I_3=\\A^3-\sum_{i=1}^n a_{ii}A^2+\sum_{i=1}^n b_{ii}A-\det(A)I_3\implies\\A^3-\sum_{i=1}^n a_{ii}A^2+\sum_{i=1}^n b_{ii}A=\det(A)I_3\implies \\A^3-\left(\sum_{i=1}^n a_{ii}A+\sum_{i=1}^n b_{ii}\right)A=\det(A)I_3.$$
Here I'm too stuck. Can someone help me?
 A: First, assume that $A$ is invertible, i.e. $\det(A)\neq 0$ and all eigenvalues $\lambda_{1}, \lambda_{2}, \lambda_{3}$ are nonzero. As comments say, we will prove that 
$$\phi_{A}(t) = t^{3} - \mathrm{tr}(A)t^{2} + \mathrm{tr}(\mathrm{adj}(A))t - \det(A)$$
where $\phi_{A}(t)$ is a characteristic polynomial of $A$. Since $\mathrm{tr}(A)=\sum_{i}\lambda_{i}$ and $\det(A)=\prod_{i}\lambda_{i}$, we only have to show that 
$$
\mathrm{tr}(\mathrm{adj}(A))=\lambda_{1}\lambda_{2}+\lambda_{2}\lambda_{3}+\lambda_{3}\lambda_{1}.
$$
Let $J$ be a Jordan canonical form of $A$, i.e. $A=SJS^{-1}$. Then 
$$
\mathrm{adj}(A) = {\det(A)}A^{-1}={\det(A)}SJ^{-1}S\Rightarrow\mathrm{tr}(\mathrm{adj}(A))={\lambda_{1}\lambda_{2}\lambda_{3}}(\lambda_{1}^{-1}+\lambda_{2}^{-1}+\lambda_{3}^{-1})=\lambda_{1}\lambda_{2}+\lambda_{2}\lambda_{3}+\lambda_{3}\lambda_{1}.
$$
If $A$ is not invertible, we will consider $A_{s} = A+sI$ for $s\in \mathbb{R}$. Since $\det(A_{s})$ is a polynomial in $s$, it has only finitely many zero including $s=0$, so $A_{s}$ is invertible for sufficiently small $s\neq 0$. Then we can apply the above result for $A_{s}$, then take the limit $s\to 0$.
A: HINT.-Let $A$ be the matrix$$ \begin{bmatrix} 
      a_{11} & a_{12} & a_{13} \\ 
      a_{21} & a_{22} & a_{23}  \\ 
      a_{31} & a_{32} & a_{33}  \\     
   \end{bmatrix}$$
The characteristic polynomial is
$$x^3-(tr(A))x^2+(a_{11}(a_{22}+a_{33})+a_{31}(a_{23}-a_{13})+a_{22}a_{33}-a_{12}a_{21})x-\det(A)$$ By Hamilton-Cayley one has
$$A^3-(tr(A))A^2+(a_{11}(a_{22}+a_{33})+a_{31}(a_{23}-a_{13})+a_{22}a_{33}-a_{12}a_{21})A-\det(A)I_3=0$$
Consequently what remains to be proved is the equality
$$tr(adj(A)).A=\left[a_{11}(a_{22}+a_{33})+a_{31}(a_{23}-a_{13})+a_{22}a_{33}-a_{12}a_{21}\right]\cdot A$$
That $$tr(adj(A))=a_{11}(a_{22}+a_{33})+a_{31}(a_{23}-a_{13})+a_{22}a_{33}-a_{12}a_{21}$$ is  left as a simple exercise of verification for the OP.
