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Given a Matrix $$A = \begin{vmatrix} a^2 & (s-a)^2 & (s-a)^2\\ (s-b)^2 & b^2 & (s-b)^2 \\ (s-c)^2 & (s-c)^2 & c^2\\ \end{vmatrix}$$

and $$a+b+c = 2s$$ , how can I find its determinant without expanding?

I've tried finding all factors which make two columns (/rows) similar or proportional. Only with s=0 all the rows and columns becomes 1 (after taking out the common value). With s=a $$\left|\matrix{ a^2 & 0 & 0\\ (a-b)^2=c^2 & b^2 & (a-b)^2=c^2 \\ (a-c)^2=b^2 & (a-c)^2=b^2 & c^2\\ }\right|=0$$

Here is a list of trivial factors (s-a), (s-b), (s-c), (s-0). I'm given a hint that the determinant is a degree 6 polynomial, which I don't understand cause max degree of s is 4, please do explain how degree is 6.

I think det(A) = (s-a)(s-b)(s-c)(s-0).

Are there any more factors? What are they?

Edit: Please also explain how to arrive at the solution.

dshrikant
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  • Try to evaluate the determinant at $s=a,b$ or $c$. It is likely that you will find 0 or something which will give you some hints about the zeros of the degree six determinant. A naive guess would be that the determinant is something like $(s-a)^2(s-b)^2(s-c)^2$. – Nicolas Dec 20 '20 at 13:11
  • Seing Jan Eerland post, my naive guess seems to be a bit too naive! – Nicolas Dec 20 '20 at 13:15
  • I've tried replacing $$s=a$$ and found the above matrix [second one in question], I don't understand how the degree is 6 when there are 3 variables. – dshrikant Dec 20 '20 at 13:20
  • @SirDevil Each component is of degree $2$, so $A$ is of degree $6$. – J.G. Dec 20 '20 at 13:32
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    It's of degree $6$ in $\ s\ $, which you can see from the third term, $ (s-a)^2\big((s-b)^2(s-c)^2-b^2(s-c)^2 \big)\ $, in the expansion of the determinant by the first row. – lonza leggiera Dec 20 '20 at 13:35
  • @lonzaleggiera can't upvote yet, thanx – dshrikant Dec 20 '20 at 13:37
  • https://www.toppr.com/ask/question/if-2s-a-b-c-beginvmatrixa2-s-a2-s/ . Check it –  Dec 20 '20 at 14:42

2 Answers2

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The equation $a+b+c=2s$ has a geometric interpretation: $s$ is the semiperimeter of a triangle of sides $a,\,b,\,c$. If this triangle is degenerate, say with $a=b+c$, then $s=a$ and$$A=\left|\begin{array}{ccc} \left(b+c\right)^{2} & 0 & 0\\ c^{2} & b^{2} & c^{2}\\ b^{2} & b^{2} & c^{2} \end{array}\right|=0,$$so $(s-a)(s-b)(s-c)|A$. A similar treatment of $-a=b+c\implies s=0$ can prove $s|A$. Since $A$ is of dimension $6$ and is homogeneous when all sides are rescaled, the symmetries require a result proportional to $s^3(s-a)(s-b)(s-c)$. An equilateral triangle with $a=b=c=2,\,s=3,\,A=54$ proves the coefficient is $2$, as in @JanEerland's answer.

J.G.
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    could you explain how A has 6 dimension? And in $(s-a)(s-b)(s-c)|A$ what does $|A$ means? – dshrikant Dec 20 '20 at 13:33
  • @SirDevil I explained its dimension in a comment under the OP. $|$ denotes oine polynomial dividing another. – J.G. Dec 20 '20 at 13:36
  • Is there any other way to understand (besides symmetries) why $s^2$ should be a factor and not say $(s-c_{1}) (s-c_{2})$? – dshrikant Dec 20 '20 at 13:53
  • @SirDevil We determine factors by finding when $A=0$, not by using symmetries. – J.G. Dec 20 '20 at 13:54
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Not a 'real' answer, but it was too big for a comment.

I wrote and ran some Mathematica-code:

In[1]:=Clear["Global`*"];
FullSimplify[
 Det[{{a^2, (s - a)^2, (s - a)^2}, {(s - b)^2, 
    b^2, (s - b)^2}, {(s - c)^2, (s - c)^2, c^2}}], 
 Assumptions -> a + b + c == 2 s]

Running the code gives:

Out[1]=-2 (b - s) (b + c - s) s^3 (-c + s)

If you're not only looking for the solution, it is not very hard to show that this is correct.

Jan Eerland
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