Suppose $A$ is an invertible matrix. Prove that $det(A^{-2}) = 1/(det(A))^2$

Question

Suppose $A$ is an invertible matrix. Prove that $det(A^{-2}) = 1/(det(A))^2$

I want to just say

$$det(A^{-2}) = 1/(det(A))^2$$

$$\Rightarrow (det(A^2))^{-1} = 1/(det(A))^2$$

$$ \Rightarrow 1/det(A^2) = 1/(det(A))^2$$

$$ \Rightarrow 1/(det(A))^2 = 1/(det(A))^2$$

but it feels like I'm just restating properties and not actually proving it, how would I prove each of these properties of determinants?

Answer 1

nevermind i found it in my notes:

A$^{-1}A = I$

$\det(A^{-1}A) = \det I$

then, $det(A^{-1}) = \frac{1}{\det A}$

let $A = A^2$

then $A^{-1} = A^{-2}$

then $\det(A^{-2}) = \frac{1}{\det(A^2)}$

thus $\det(A^-2) = \frac{1}{\det(A)^2}$

Answer 2

If A inversible, then $det (A^{-1})=1/det A$

Thus, $det (A^{-2})=det(A^{-1} \cdot A^{-1})=det(A^{-1}) \cdot(det A^{-1})=\frac{1}{det A}\cdot \frac{1}{det A}=\frac{1}{(det A)^2}$