Determinant/Multiplication theorem/Fact/Proof

We fix the matrix ${\displaystyle {}B}$.

Suppose first that ${\displaystyle {}\det B=0}$. Then, due to fact the matrix ${\displaystyle {}B}$ is not invertible and therefore, also ${\displaystyle {}A\circ B}$ is not invertible. Hence, ${\displaystyle {}\det A\circ B=0}$.

Suppose now that ${\displaystyle {}B}$ is invertible. In this case, we consider the well-defined mapping

${\displaystyle \delta \colon \operatorname {Mat} _{n}(K)\longrightarrow K,A\longmapsto (\det A\circ B)(\det B)^{-1}.}$

We want to show that this mapping equals the mapping ${\displaystyle {}A\mapsto \det A}$, by showing that it fulfills all the properties which, according to fact, characterize the determinant. If ${\displaystyle {}z_{1},\ldots ,z_{n}}$ denote the rows of ${\displaystyle {}A}$, then ${\displaystyle {}\delta (A)}$ is computed by applying the determinant to the rows ${\displaystyle {}z_{1}B,\ldots ,z_{n}B}$, and then by multiplying with ${\displaystyle {}(\det B)^{-1}}$. Hence the multilinearity and the alternating property follows from exercise. If we start with ${\displaystyle {}A=E_{n}}$, then ${\displaystyle {}A\circ B=B}$ and thus

${\displaystyle {}\delta (E_{n})=(\det B)\cdot (\det B)^{-1}=1\,.}$