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Prove $$\Delta=\begin{vmatrix} (y+z)^2 & x^2 & x^2 \\ y^2 & (z+x)^2 & y^2 \\ z^2 & z^2 & (x+y)^2 \\ \end{vmatrix} = 2xyz(x+y+z)^3$$ using factor theorem.

This is solved in Demonstrate using determinant properties that the determinant of A is equal to $2abc(a+b+c)^3$ using factor theorem.

My Attempt:

$$ x=0\text{ or }y=0\text{ or }z=0\implies\Delta=0\text{ , So $x,y,z$ are factors of }\Delta.\\ (x+y+z)=0\implies \Delta=\begin{vmatrix}x^2&x^2&x^2\\y^2&y^2&y^2\\z^2&z^2&z^2\end{vmatrix}=0\text{ , So $(x+y+z)$ is a factor of $\Delta$.} $$ $\color{black}{\text{But how do i extract the remaining term $(x+y+z)^2$ to prove $\Delta=2xyz(x+y+z)^3$ }\color{red}{ ?}}$

Similar Example:

Please check answer of @user348749 in How to solve this determinant, $$ \Delta'=\begin{vmatrix} (b+c)^2&ab&ca\\ ab&(a+c)^2&bc\\ ac&bc&(a+b)^2 \end{vmatrix}=2abc(a+b+c)^3 $$ it is said that $$ (a+b+c)=0\implies\Delta'=\begin{align*} \begin{vmatrix} c^2 & ca & bc \\ ca & a^2 & ab \\ bc & ab & b^2 \\ \end{vmatrix} =abc\begin{vmatrix} c & a & b \\ c & a & b \\ c & a & b \\ \end{vmatrix} \end{align*}=0 $$ "Since all rows are identical, $(a+b+c)^2$ is a factor. The determinant is a polynomial of degree 6 and hence the remaining factor is linear and since it is symmetric, the factor must be $k(a+b+c)$."

$\color{black}{\text{How can we say this }\color{red}{ ?}}$

My Understanding:

If the problem was similar to this, answer of @Saibal in Factorise a matrix using the factor theorem, $$\Delta''= \begin{vmatrix} x&y&z\\ x^2&y^2&z^2\\ x^3&y^3&z^3\\ \end{vmatrix}$$ I could without doubt do as below: $$ x=0\text{ or }y=0\text{ or }z=0\implies\Delta''=0\\ x=y\text{ or }y=z\text{ or }z=x\implies\Delta''=0 $$ Thus, $x,y,z,(x-y),(y-z),(z-x)$ are factors of $\Delta''$. ie. $\Delta''=kxyz(x-y)(y-z)(z-x)$

Sooraj S
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1 Answers1

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Your attempt is fine. And all the further explanations you need are already given by user348749 in How to solve this determinant I could just rephrase this here: Since for $x+y+z=0$, all three columns are identical (same argument as all three rows), $(x+y+z)^2$ is a factor (see user348749's explanation). So the remaining factor is linear. Since $\Delta$ does not change under $x \leftrightarrow y$ or any other permutation, the remaining linear factor must be symmetric under these exchanges, and only $C \cdot (x+y+z)$ is such a factor. Again, not my arguments so far, but user348749's.

The thing that remains to be done is to calculate the constant. Choose e.g. $x=y=z=1$ and you get $$ \Delta=\begin{vmatrix} 4 & 1 & 1 \\ 1 & 4 & 1 \\ 1 & 1 & 4 \\ \end{vmatrix} = 4\cdot 15 - 2\cdot 3 =54 = C \cdot 3^3 = C \cdot 27 $$ which clearly gives $C =2$.

Andreas
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  • thnx. So is it like, whn i put $x=-y-z$ and taken out $x,y,z$ from $R_2,R_3,R_1$ respectively, then $R_1=R_2=R_3$. ie. When we take $x,y,z$ from $R_2,R_3,R_1$ respectively, and apply $R_1\to R_1-R_2$ and $R_3\to R_3-R_2$ and set $x=-y-z$ then $R_1=0$ and $R_3=0$, that means $(x+y+z)$ can be taken out from both $R_1-R_2$ and $R_3-R_2$ along with $xyz$, Hence $(x+y+z)^2$ can be taken out as a factor. right ? – Sooraj S Feb 22 '18 at 05:45
  • I think for people who are unfamiliar of symmetric polynomials its is kinda confusing to just mention symmetry in such problems. I had to do a lot of reading myself to even get a hint into understanding the solutions. – Sooraj S Feb 23 '18 at 19:57
  • "the determinant is invariant under $S_3$. That means that all its factors are also symmetric polynomials". I do not think that this argument is correct broadly, as "Not all symmetric polynomials factor into symmetric factors", Ex:$(x+y)(y+z)(z+x)$. – Sooraj S Feb 23 '18 at 20:25
  • So far we have obtained $xyz(x+y+z)^2$. Thus the remaining factor must be none other than $(x+y+z)$ else it'd never be symmetric. Ex: If it was $x$ or $y$ or $z$ then $x.xyz(x+y+z)^2$ is not symmetric. Two factors alone cannot sustain a permutation of the variables, as $(x+y)xyz(x+y+z)^2$ not symmetric. Thus the only possible factor which preserve the polynomial to be symmetric is $(x+y+z)$. – Sooraj S Feb 23 '18 at 20:25