The formula $\DeclareMathOperator{tr}{tr}\mathrm{adj}(A)=\tfrac{1}{2}[(\tr A)^2-\tr(A^2)]I_3-[\tr A]A+A^2$ for the adjoint of a $3\times 3$ matrix 
Let $A$ be a square matrix of order $3$. Prove that 
  $$
\operatorname{adj}(A) 
= \tfrac{1}{2} \bigl[ 
(\operatorname{tr} A)^2 - \operatorname{tr}(A^2) \bigr] I_3 
- [\operatorname{tr} A] A + A^2
$$ 
  where $\operatorname{tr}A$ is the trace of $A$.

To begin, I'm not quite sure how to start proving this. I've considered using brute force, but I suspect there should be a much more elegant way of doing it. Is there a need to prove this for both $(i,i)$ and $(i,j)$ entries? I'd appreciate an explanation that is not too complex. Thanks in advance!
 A: Hint Recall that the adjugate matrix $\DeclareMathOperator{adj}{adj} \DeclareMathOperator{tr}{tr} \adj A$ of an $n \times n$ matrix $A$ satisfies $$A \adj A = (\det A) I$$ for all $A$.
Now, if $\lambda_1, \lambda_2, \lambda_3$ are the eigenvalues of the $3 \times 3$ matrix $A$, the characteristic polynomial of $A$ is
$$p_A(t) = (t - \lambda_1)(t - \lambda_2)(t - \lambda_3) = t^3 - (\lambda_1 + \lambda_2 + \lambda_3)t^2 + (\lambda_2 \lambda_3 + \lambda_3 \lambda_1 + \lambda_1 \lambda_2) t - \lambda_1 \lambda_2 \lambda_3.$$ The $t^2$ coefficient is $-\tr A$ and the constant term is $-\det A$. We can write the $t$ coefficient as
$$\tfrac{1}{2}[(\lambda_1 + \lambda_2 + \lambda_3)^2 - (\lambda_1^2 + \lambda_2^2 + \lambda_3^2)] = \tfrac{1}{2}[(\tr A)^2 - \tr (A^2)] ,$$ and hence the characteristic polynomial is
$$p_A(t) = t^3 - (\tr A) t^2 + \tfrac{1}{2}[(\tr A)^2 - \tr (A^2)] t - \det A .$$

The Cayley-Hamilton Theorem tells us that $p_A(A) = 0$, so $$0=p_A(A)=A^3-(\tr A) A^2 + \tfrac{1}{2}[(\tr A)^2 - \tr (A^2)] A - (\det A) I .$$ The first display equation above suggests rearranging: $$A [A^2 - (\tr A) A + \tfrac{1}{2}[(\tr A)^2 - \tr (A^2)] I] = (\det A) I,$$ and we also know that $$(\det A) I = A \adj A .$$ So, if $A$ is invertible, multiplying by $A^{-1}$ gives $$\adj A = A^2 - (\tr A) A + \tfrac{1}{2}[(\tr A)^2 - \tr (A^2)] I ,$$ and hence the desired identity holds in that case. On the other hand, both sides of this equation are continuous (indeed, polynomial) in the entries of $A$, and the set of invertible matrices is dense in the set of all $3 \times 3$ matrices, so by continuity the desired identity holds for all matrices.

A: Recall that $A \cdot \text{adj}(A) = \det(A) I$. If you multiply the given equation by $A$, you get
\begin{align}
\det(A) I & = \frac 12\left[(\text{tr}(A))^2 - \text{tr}(A^2)\right]A - (\text{tr}(A))A^2 + A^3.
\end{align}
Now suppose $\lambda_1, \lambda_2, \lambda_3$ are eigenvalues of $A$. Then we can rewrite $\det(A)$, $\text{tr}(A)$ and $\text{tr}(A^2)$ in terms of $\lambda_i$'s as follows:
\begin{align}
\lambda_1\lambda_2\lambda_3 I & = \frac 12\left[(\lambda_1+\lambda_2+\lambda_3)^2 - \lambda_1^2 - \lambda_2^2 - \lambda_3^2\right]A - (\lambda_1 + \lambda_2 + \lambda_3)A^2 + A^3.
\end{align}
Rearrange to get
\begin{align}
(A - \lambda_1I)(A - \lambda_2I)(A - \lambda_3I) & = 0.
\end{align}
We know this to be true (Cayley-Hamilton), so you can derive the desired equation by writing these equations from the bottom up.
