2
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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 $b=3$.

$\endgroup$
  • $\begingroup$ 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 ? $\endgroup$ – ss1729 Feb 28 '18 at 14:56
  • 1
    $\begingroup$ 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. $\endgroup$ – CY Aries Feb 28 '18 at 15:04
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 ? $\endgroup$ – ss1729 Feb 28 '18 at 13:27
  • $\begingroup$ 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 $\endgroup$ – Your IDE Feb 28 '18 at 13:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.