Without directly evaluting, show that the determinant of $A = 0$

Question

Without directly evaluting, show that $det \left[ \begin{array}{ccc} b + c & c + a & b + a \\ a & b & c \\ 1 & 1 & 1 \end{array} \right] =0$

I am stuck on this one. I can only do this by evuating.

Things that I know:
1) Square matrix $A$ with two proportional rows or columns as a $det(A) = 0$
2) Square matrix $A$ has $det(A)=0$ if it has a row or column of zeros.

Can anyone help? Thanks.

Hint: adding the first two rows gives a multiple of the third. — lulu, Jul 01 '17 at 16:18
Ok @lulu. So you're saying if I had $R2$ to $R1$, then I will have $(a + b +c)\times R3$. Is that correct? I like this. Thankyou. — Bucephalus, Jul 01 '17 at 16:24
I like your answer @lulu. If you can put that as an answer I can then accept it. — Bucephalus, Jul 01 '17 at 16:27

score 5 · Accepted Answer · answered Jul 01 '17 at 16:32

5

The row vectors are linearly dependent.

Specifically, denoting the row vectors as $$\vec r_1=(b+c,a+c,a+b)\quad \vec r_2=(a,b,c)\quad \vec r_3=(1,1,1)$$ then we have $$\vec r_1+\vec r_2=(a+b+c)\,\vec r_3$$

answered Jul 01 '17 at 16:32

lulu

70,402

score 0 · Answer 2 · answered Jul 01 '17 at 16:21

0

Here's one possibility: For a fixed value of $a,b$, $\det(A)$ is a degree two polynomial in $c$. It is obviously $0$ for $c=0,a,b$, and hence the polynomial is identically $0$. (Unless $a=b$, $a=0$, etc, but then you can use the same reasoning with the other variables).

answered Jul 01 '17 at 16:21

TomGrubb

12,909

Thanks for your input @ThomasGrubb. – Bucephalus Jul 01 '17 at 16:28

Without directly evaluting, show that the determinant of $A = 0$

2 Answers2