How to prove $\det (A^{-1}) = \frac{1}{\det A}$ without using the property that $\det (AA^{-1})=\det (A) \det(A^{-1})$?

Question

Let A be an invertible matrix. How to prove $\det (A^{-1}) = \frac{1}{\det A}$ without using the property that $\det (AA^{-1})=\det (A) \det(A^{-1})$? Thanks. I think $\det (A^{-1}) = \frac{1}{\det A}$ holds even without this property, given A is invertible.

Answer 1

Thank you, guys. I think I figured it out. All I have to prove is that for any eigenvalue $\lambda$ of A, $\lambda^{-1}$ is an eigenvalue of $A^{-1}$. In this way, I don't need that property to prove this question.