Undergrad linear algebra 1 ~
I'm aware of the following three rules; each for an elementary operation: *Note || denotes determinant; I find it kind of confusing with abs but that's what the book says.
E1: if the matrix B is A but with rows swapped, $|B|=-|A|$
E2: if the matrix B is A but a row is a multiple $k$ of another, $|B|=k|A|$
E3: if the matrix B is A but a multiple of another row is added/subtracted to another, $|B| = |A|$
However, I came across this question to find the determinant of $C$ given $det A = 3$ and $det B = -2$: $$ A = \begin{bmatrix} a&b\\ c&d \end{bmatrix} $$ $$ B = \begin{bmatrix} e&f\\ c&d \end{bmatrix} $$ $$ C = \begin{bmatrix} 2a-e&2b-f\\ 3c&3d \end{bmatrix} $$
I was able to solve it by the "usual way" of just multiplying $(2a-e)(3d)-(3c)(2b-f)$ because it's a simple 2 by 2, but I was wondering if there would be a way to use the logic of the elementary ops to "remove row 1 of matrix B from the matrix C" such that then matrix C has rows that are just scalar multiples of A, so then I can easily apply the E2 rule.
Now write the determinant as $$\det{C}=6(ad-bc)+3(cf-ed)=12$$ (from your expression for the determinant of $C$).