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.
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3It's hard to speculate about whether a true statement holds "without" another true statement. – Oct 17 '20 at 07:43
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1If you interpret the determinate as ratio of volume, then the result is immediate. – Lynnx Oct 17 '20 at 07:45
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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.
Charlie Xu
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1well, you also want to show that multiplicities are the same, but it's probably easy – Peter Franek Oct 17 '20 at 08:12