About $\det(A^{-1}) = \frac{1}{\lambda_1} \frac{1}{\lambda_2} \frac{1}{\lambda_3} \frac{1}{\lambda_4}$ (Gilbert Strang's lecture on YouTube)

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I am watching Gilbert Strang's lecture on YouTube.

I think his answer to this problem is not perfect.

Is his answer ok?

Let $A$ be an invertible $4$ by $4$ matrix.
Let $\lambda_1, \lambda_2, \lambda_3, \lambda_4$ be the eigenvalues of $A$

Prove that $\det(A^{-1}) = \frac{1}{\lambda_1} \frac{1}{\lambda_2} \frac{1}{\lambda_3} \frac{1}{\lambda_4}$.

His answer is like the following:
$\frac{1}{\lambda_1}, \frac{1}{\lambda_2}, \frac{1}{\lambda_3}, \frac{1}{\lambda_4}$ are the eigenvalues of $A^{-1}$.
So, $\det(A^{-1}) = \frac{1}{\lambda_1} \frac{1}{\lambda_2} \frac{1}{\lambda_3} \frac{1}{\lambda_4}$.

I think he needs to show the multiplicity of a eigenvalue $\lambda$ of $A$ is equal to the multiplicity of the eigenvalue $\frac{1}{\lambda}$ of $A^{-1}$.
But he didn't show the fact.
So I think his answer is not perfect.
Am I correct or not?

By the way, my anser is the following:
$1 = \det(I) = \det(A A^{-1}) = \det(A) \det(A^{-1}).$
And
$\det(A) = \lambda_1 \lambda_2 \lambda_3 \lambda_4$.

So, $\det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{\lambda_1 \lambda_2 \lambda_3 \lambda_4}$.