Prove $\Delta=\left|\begin{smallmatrix} 3x&-x+y&-x+z\\ -y+x&3y&-y+z\\-z+x&-z+y&3z\end{smallmatrix}\right|=3(x+y+z)(xy+yz+zx)$

Prove that$$\Delta=\begin{vmatrix} 3x&-x+y&-x+z\\ -y+x&3y&-y+z\\ -z+x&-z+y&3z \end{vmatrix}=3(x+y+z)(xy+yz+zx)$$ using factor theorem and polynomials.

My Attempt

$$\begin{matrix} \color{red}{3x}&\color{blue}{-x+y}&\color{red}{-x+z}&\color{blue}{3x}&\color{red}{-x+y}\\ -y+x&\color{red}{3y}&\color{blue}{-y+z}&\color{red}{-y+x}&3y\\ \color{red}{-z+x}&\color{blue}{-z+y}&\color{red}{3z}&\color{blue}{-z+x}&\color{red}{-z+y} \end{matrix}$$

$\Delta$ is a symmetric polynomial of degree $3$.

If we set $x+y+z=0$ we have $\Delta=0$, thus $(x+y+z)$ is a factor, which I can guess from the column operation $C_1\to C_1+C_2+C_3$.

But, how do I find that the remaining factor is $(xy+yz+zx)$ ?

Of course I'll get $\Delta=0$ if I substitute $(xy+yz+zx)=0$, but how do I guess the term in the first place from the fact that $x+y+z$ is one term and $\Delta$ is a symmetric polynomial ?

Similar Problem

Please check the attempted solution by @David Holden in Demonstrate using determinant properties that the determinant of A is equal to $2abc(a+b+c)^3$,

where he has made the substitution $\Delta(a,b,c)=abc(a+b+c)\left(\lambda(a^2+b^2+c^2)+\mu(ab+bc+ca)\right)$ after obtaining $abc(a+b+c)$ as a factor in order to obtain the remaining factor, which is confusing for me. Is there a better way to guess the remaining factor in my case?

As you have mentioned, $x+y+z$ is a factor. As the determinant is symmetric, the factor other than $x+y+z$ must also be symmetric. As it is quadratic, it must be in the form $a(x^2+y^2+x^2)+b(xy+yz+zx)$. (We don't have to guess. $a(x^2+y^2+x^2)+b(xy+yz+zx)$ is the general form of symmetric quadratic polynomial in $x$, $y$, $z$.)

Put $x=1$, $y=z=0$, we have $a=0$.

Put $x=y=z=1$, we have $b=3$.

• thanks. i understand the remaining term should be symmetric polynomial as there is no other way to make the whole thing symmetric. Could u pls direct me into details of : $a(x^2+y^2+z^2)+b(xy+yz+zx)$ is the general form of quadratic polynomial, what about polynomials of higher order ? – ss1729 Feb 28 '18 at 14:56
• I have to add that the polynomial is not only symmetric but also is homogeneous. If you have the term $x^2$ in the quadratic polynomial, to make it symmetric, you need $y^2$ and $z^2$ with the same coefficients. This gives $x^2+y^2+z^2$. If you have the term $xy$ in the quadratic polynomial, to make it symmetric, you need $yz$ and $zx$ with the same coefficients. This gives $xy+yz+zx$. So we have the general form. – CY Aries Feb 28 '18 at 15:04

After $$C_1\to C_1+C_2+C_3$$ take $$(x+y+z)$$ out as common and then apply the the following transformations. $$R_1\to R_1-R_2$$ and $$R_2 \to R_2-R_3$$ and find the determinant to get the required factors.

• thnx, but i am basically asking is there another way to guess the term $xy+yz+zx$ rather than the row and column operations, from the fact that $\Delta$ is a symmetric polynomial of degree 3, given we have obtained $x+y+z$ as a term which is kinda easy to see ? – ss1729 Feb 28 '18 at 13:27
• This matrix is $0$ if $x+y+z=0$ since it is a common factor. Further part can be calculated in a similar way as stated in the other answer. You'll get coefficient of square terms 0 and other product terms as 1 – Your IDE Feb 28 '18 at 13:37