Solution
= Solution
Use the multiplicativity of the <determinant> on $AA^{-1}=I$:
$$
\det A\,\det(A^{-1})=1,\qquad\boxed{\det(A^{-1})=\frac1{\det A}.}
$$
The denominator is nonzero because $A$ is an <invertible matrix>.
= Solution
Use the multiplicativity of the <determinant> on $AA^{-1}=I$:
$$
\det A\,\det(A^{-1})=1,\qquad\boxed{\det(A^{-1})=\frac1{\det A}.}
$$
The denominator is nonzero because $A$ is an <invertible matrix>.