If \(A\) is an invertible \(n \times n\) matrix over a field, show that \(\det A \neq 0\).
Solution
Since \(A\) is invertible, there is a matrix \(A^{-1}\) with \[ AA^{-1}=I. \] By multiplicativity of the determinant, \[ 1=\det I=\det(AA^{-1})=(\det A)(\det A^{-1}). \] Thus \(\det A\) has a multiplicative inverse in the field, namely \(\det(A^{-1})\), and therefore \(\det A\ne0\).