determinant of invertible matrix

Question

Calculate that, if $A$ is invertible, then $|A^{−1}| = \dfrac{1}{|A|}$.

I think they are saying that $|A^{-1}|$ and $A^{-1}$ are the same? I think this can be proven with determinant theorems but I'm not sure.

Answer 1

By definition it is $A\cdot A^{-1}=I.$ Taking determinants $$\det(A\cdot A^{-1})=\det(I)=1.$$ Since $\det(A\cdot B)=\det(A)\cdot \det(B)$ we have that

$$\det(A)\cdot\det( A^{-1})=1,$$ from where it follows that $$\det(A^{-1})=\dfrac{1}{\det(A)}.$$

Answer 2

That is not the absolute sign. It is determinant. You can take an arbitrary matrix of 2x2 and apply this property.