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

Question

Can someone show me how:

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

where we assume that $A$ is invertible. thanks

Answer 1

Let $B = A^{-1/2}$ so that $B^2 A = A B^2 = I$

Assuming such a $B$ exists. Then

$1 = \det I = \det( B^2 A) = (\det B)^2 \det A$

so

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

Answer 2

You have $\det C^{-1} = {1 \over \det C}$. Now let $C=A^{1\over 2}$.

Answer 3

Umm. Perhaps you mean to show that if $B^2=A$ then $(det B)^2=det A$? Also, has it previously been established in your study that for every invertible $A$ there is a $B$ such that $B^2=A$?