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The question is #653 from Golan's Linear Algebra Every Graduate Student Should Know and while it doesn't explicitly say not to just expand and factor it, I think that's the spirit of the question.

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

Since the answer is $$ 4(a+b)(b+c)(a+c) $$

I am inclined to think it has something to do with wisely dividing out (a+b), etc. from particular rows or some other linear combination tricks but I can't seem to quite figure it out. Any ideas?

Luka Djevenica
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3 Answers3

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Consider $F(a,b,c) = \begin{vmatrix} -2a & a+b & a+c\\ a+b & -2b & b+c \\ c+a & c+b & -2c \\ \end{vmatrix}$

With some work we can show that $F(a,b,c)$ is a cyclic symmetric polynomial of degree $3$.

Now, we can use the properties of such polynomials to evaluate the determinant. When $a=-b$ the determinant becomes $0$. Hence $(a+b)$ is a factor. Similarly $(b+c)$, $(c+a)$ are factors.

Hence $F(a,b,c) = k(a+b)(b+c)(c+a)$. Now, to determine $k$ set $a=1, b=1, c=0$. We get $k=4$.

Hence $F(a,b,c) = 4(a+b)(b+c)(c+a)$

Hari Shankar
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  • +1. I like the idea of symmetry hunting. However, it seems to me that here we do not need to check that $F$ is a cyclic polynomial. Since $F(a, -a, c)=0$, we can already conclude that $(a+b)$ is a factor of $F$. Am I wrong? – Giuseppe Negro Dec 10 '17 at 19:20
  • @Giuseppe Negro: that is an important clarification that I missed. The symmetry of course is used when we say that if $(a+b)$ is a factor, so are $(b+c)$ and $(c+a)$. In this case, since the other two cases are easily checked, I guess we can do away with looking for symmetry. – Hari Shankar Dec 11 '17 at 08:16
  • The symmetry check is much cooler than directly verifying that $F(a, -a, c)=F(a, b, -b)=F(-c, b, c)=0$. Moreover, checking the symmetry is easy if we use Sarrus rule. The summands in Sarrus rule are all cyclic polynomials: $8abc, (a+b)(b+c)(c+a), $etc... This is quickly established by inspection. I would leave the symmetry check in place. – Giuseppe Negro Dec 11 '17 at 10:12
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I don't see any obvious tricks, but if you put $(x,y,z)=(a+b,\,b+c,\,c+a)$ and rewrite the matrix as $$ A=\pmatrix{ y-x-z &x &z\\ x &z-x-y &y\\ z &y &x-y-z}, $$ then $$ PAP=B=\pmatrix{ 0&2x&2z\\ 2x&0&2y\\ 2z&2y&0},\ \text{ where }P=\pmatrix{ 0&1&1\\ 1&0&1\\ 1&1&0}. $$ Hence $\det(A)=\det(B)/\det(P)^2$. You don't need to expand $\det(P)$ to calculate its value. In fact, since $P=ee^T-I$ (here $e$ denotes the all-one vector), we immediately get $\det(P)=e^Te-1=2$. However, to calculate $\det(B)$, I cannot think of a better method than Sarrus' rule.

user1551
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  • Impressive. Do you mind adding some details on how to discover that $PAP=B?$ Specifically, where did $P$ come out? Thank you. I would also add that $B$ reminds me of the matrix $$\begin{bmatrix} 0 & x & z \ -x & 0 & y \ -z & - y & 0 \end{bmatrix}, $$ which represents the operator $\boldsymbol v\in\mathbb R^3 \mapsto (x, y, z)\times \boldsymbol v\in\mathbb R^3, $ but I don't know if this observation leads somewhere, probably not. – Giuseppe Negro Dec 10 '17 at 18:35
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    @GiuseppeNegro It's just a lucky educated guess. While the OP doesn't want to expand the determinant, what makes expansion difficult is the presence of many high-order terms. So, my first idea was to at least get rid of the cubic powers, even though we don't want to apply Sarrus' rule. E.g. along the main diagonal, we have an $x^3$ term. To eliminate it, we may add the first row to the second row. To retain some sort of symmetry in the matrix's structure, we do the similar to other rows and columns. It turns out that this does give rise to a pretty nice structure. – user1551 Dec 10 '17 at 22:24
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Ah, I only now realized the question was asked before. To atone for my repost sin, I'll give an answer that I hadn't seen around and seems to me to be likely what the problem aims at (inspired by the substitution proposed in the other answers), please do check that it is correct:

Using the $(a+b)=x,(b+c)=y,(a+c)=z$ substitution, we have

$$ \begin{vmatrix} y-x-z & x & z\\ x & z-x-y & y \\ z & y & x-y-z \\ \end{vmatrix} $$

Now $R1=R1+R2+R3$:

$$ \begin{vmatrix} y & z & x\\ x & z-x-y & y \\ z & y & x-y-z \\ \end{vmatrix} $$

Now you multiply column 1 by $xz$, column 2 by $xy$ and column 3 by $yz$, in order to make $xyz$ across the row. To compensate you put $\frac{1}{(xyz)^2}$ in front:

$$\frac{1}{(xyz)^2} \begin{vmatrix} xyz & xyz & xyz\\ x^2z & xy(z-x-y) & y^2z \\ z^2x & xy^2 & yz(x-y-z) \\ \end{vmatrix}$$

Now you take out $xyz$ from the first row and do $C2=C2-C1$, $C3=C3-C1$, and you get something like:

$$\frac{1}{(xyz)} \begin{vmatrix} 1 & 0 & 0\\ ... & ... & ... \\ ... & ... & ... \\ \end{vmatrix}$$

Where $...$ stands in for polynomials I don't really want to write out, but you can see only one 2x2 determinant survives, which when expanded should yield the desired $4x^2y^2z^2$, which then divides by $xyz$ to get the result $4xyz$.

Luka Djevenica
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  • I think that, first of all, one should say something about the possible vanishing of $x, y$ or $z$. Apart from that, I don't understand what you do after "you take out $xyz$ etc...". There is a simpler way to do that, starting from $$\begin{bmatrix} y& z & x \ x & z-x-y & y\ z & y & x-y-z\end{bmatrix} $$ replace $R2$ with $yR2-xR1$ (determinant gets multiplied by $y$) and $R3$ with $yR3-zR1$ (determinant gets multiplied by $z$). You obtain $$\begin{bmatrix} y & \ast & \ast \ 0 & \ast & \ast \ 0& \ast & \ast\end{bmatrix}.$$ – Giuseppe Negro Dec 11 '17 at 10:30
  • Yes, I believe that is an equivalent way - I just chose to get a $1$ instead of the $y$. – Luka Djevenica Dec 11 '17 at 21:40
  • Agreed, but I am afraid that there is something wrong in your computations. I do not understand how operating on R2 and R3 you can annihilate the entries in R1. – Giuseppe Negro Dec 12 '17 at 15:32
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    Absolutely, I was referring to the columns - edited appropriately. Thank you! – Luka Djevenica Dec 13 '17 at 21:12