Since det(AB)=det(A)det(B) for all square nxn matrices A,B, why does det(A^-1) = 1/(det(A)) for all invertible nxn matrices A?

Question

Please provide a detailed explanation, thank you!

 

Answer 1

If  a matrix is invertible, then its determinant is not zero.

For an invertible matrix A, the equation above reads, with B=A^-1

det(AA^-1)=detAdetA^-1.  Since AA^-1=I, the identity matrix, whose determinant is 1, we have

1=detAdetA^-1 or that detA^-1=1/detA.