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From Wikipedia's article on adjugate matrix, Cayley–Hamilton theorem allows the adjugate of $A$ to be represented in terms of traces and powers of $A$.

For the $3 \times 3$ case:

$$\operatorname{adj}(A)=\frac12\left((\operatorname{tr}A)^2-\operatorname{tr}\left(A^2\right)\right)I_3-(\operatorname{tr}A)A+A^2$$

I have tried using the theorem that $$A \; \text{adj} A = \det (A) \; I$$ but I am unable to continue proving.

Is it possible to prove this $3 \times 3$ case without use of eigenvalues?

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Well, you can just compare coefficients. If the matrix has entries (by rows) $a,b,c; d,e,f; g,h,i$, then for instance the $(2,3)$ entry of the adjugate matrix is $-\left|{a\atop d}~{c\atop f}\right|$, while the RHS gives (nothing from $I_3$, and) $-(a+e+i)f+dc+ef+fi$, which is the same value. I'll let you figure out the other coefficients yourself. It is not very inspiring, but I guess that is what you were asking for.

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  • $\begingroup$ Indeed it's not very inspiring. Pretty long way of proving but I guess this the way to go. Thanks! $\endgroup$ – Hui Min Feb 14 '16 at 3:42

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