Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 17 / b

The determinant is multiplicative under matrix multiplication:This also holds when $B$ is singular, in which case both sides vanish.

Multiplying an $n\times n$ matrix by $\alpha$ scales every row, giving a factor ${\alpha }^n$ in its determinant. Also $|A^{-1}|=1/|A|$, since $A{A}^{-1}=I$. HenceOnly $A$ needs to be invertible; the result remains valid when $\alpha =0$.

Let $D_{\beta }=\mathrm{diag}({\beta }_1,\dots ,{\beta }_n)$ and $D_{\gamma }=\mathrm{diag}({\gamma }_1,\dots ,{\gamma }_n)$. The specified entries mean $C=D_{\beta }A{D}_{\gamma }$. Multiplication of determinants therefore givesThis is also the direct row- and column-scaling rule, and it includes zero scaling factors.