Converted from a comment:

Knowing the successive compound matrices of $A$ without knowing their eigenvalues one can recover the elementary symmetric polynomials in the eigenvalues of $A$. Example 5.6 in Prells, U.; Friswell, M. I.; Garvey, S. D.: Use of geometric algebra: compound matrices and the determinant of the sum of two matrices, Proceedings of the Royal Society of London A: Mathematical, Physical and Engineering Sciences 459 (2003): 273-285 (over $\mathbb{C}$, says that $${\rm det}(A-\lambda I) = (-1)^n \lambda^n + \sum_{j=1}^{(n-1)^j} \lambda^j {\rm tr}C_{n-j}(A) + {\rm det}A.$$

Converted from a comment by another user:

Knowing the successive compound matrices of $A$ without knowing their eigenvalues one can recover the elementary symmetric polynomials in the eigenvalues of $A$. Example 5.6 in Prells, U.; Friswell, M. I.; Garvey, S. D.: Use of geometric algebra: compound matrices and the determinant of the sum of two matrices, Proceedings of the Royal Society of London A: Mathematical, Physical and Engineering Sciences 459 (2003): 273-285 (over $\mathbb{C}$, says that $${\rm det}(A-\lambda I) = (-1)^n \lambda^n + \sum_{j=1}^{(n-1)^j} \lambda^j {\rm tr}C_{n-j}(A) + {\rm det}A.$$