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.

  • 2
    $\begingroup$ This is the multilinearity of the determinant as a function of its rows. $\endgroup$ – Lord Shark the Unknown Dec 5 '19 at 3:53
  • $\begingroup$ I have a feeling that's something I haven't covered; I'm going to search it up but could you give a very brief description? (I'm in the first/intro linear algebra class of undergrad) $\endgroup$ – Five9 Dec 5 '19 at 3:59
  • $\begingroup$ I think you do need to do it in the usual way and then maybe see if you can write this determinant equation using those for $A$ and $B$. Although $C=MA+NB$ for some matrices $M,N$ there is usually no way of 'splitting' this sum. So it is a good question to think about $\endgroup$ – AnyAD Dec 5 '19 at 4:08
  • $\begingroup$ The appearance of $-2,3$ for determinants and then when row operations are performed is likely what makes the answer bellow to work. So the answer to your question is 'not in general' $\endgroup$ – AnyAD Dec 5 '19 at 4:14

Now write the determinant as $$\det{C}=6(ad-bc)+3(cf-ed)=12$$ (from your expression for the determinant of $C$).


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