Dont know how to simply solve this determinant.

Asked Dec 14 '20 at 19:33

Active Dec 14 '20 at 19:36

Viewed 57 times

$$ \det\begin{pmatrix} -2a & a + b & a + c\\ b+a & -2b & b+c\\ c+a & c+b & -2c \end{pmatrix} $$

I solved this determinant and gave me $4(a+b)(b+c)(c+a)$ but I've done it by adding $((b+c)/2a)*\text{line 1}$ to $\text{line 2}$ and $((c+a)/2a)*\text{line 1}$ to $\text{line 3}$. I still don't see another way of getting 2 zeros on a line or column that can be done by simple additions and subtractions instead of multiplying by variables like I did.

Thank you!

edited Dec 14 '20 at 19:36

DMcMor

9,407

asked Dec 14 '20 at 19:33

Mario 04

There's nothing wrong with what you did (unless $a=0$). – Ted Shifrin Dec 14 '20 at 19:40
Ik but it seems kinda complicated – Mario 04 Dec 14 '20 at 19:40
For expressions it doesnt really matter if i divide (by a) but still it feels knid of forced to multiply like that – Mario 04 Dec 14 '20 at 19:41
You could use the rule of Sarrus and then factorize. This is not so complicated as it might appear. – Dietrich Burde Dec 14 '20 at 19:42
Yes but still there will be many calculations. I am asking if there is a neat solution to this. – Mario 04 Dec 14 '20 at 19:43
1

Check these: https://math.stackexchange.com/q/2782450/42969, https://math.stackexchange.com/q/2788320/42969. – Martin R Dec 14 '20 at 19:48
Thanks, so it turns out this determinant has in fact some calculations... – Mario 04 Dec 14 '20 at 19:53

Dont know how to simply solve this determinant.

0 Answers0