Determinant $\left|\begin{smallmatrix} y+z &x &y \\ z+x &z &x \\ x+y &y &z \\ \end{smallmatrix}\right|= (x+y+z)(z-x)^2$

Question

Tomorrow I have to attend a math exam. So I have to prove a problem on determinant. The Problem: $$\left|\begin{matrix} y+z &x &y \cr z+x &z &x \cr x+y &y &z \cr \end{matrix}\right|=(x+y+z)(x-z)^2$$

This is I have to prove.

Condition: No direct Expanding, Only Properties,no Factor Method should be used.

I tried many methods I can't able to prove it. Please someone Help to solve this.

Answer 1

This can be done using row operations. Since you want to have $x+y+z$ in the result, adding all rows together as the first step practically suggests itself. $\begin{vmatrix} y+z &x &y \\ z+x &z &x \\ x+y &y &z \\ \end{vmatrix}= \begin{vmatrix} y+z &x &y \\ z+x &z &x \\ 2(x+y+z)&x+y+z&x+y+z\\ \end{vmatrix}= (x+y+z)\begin{vmatrix} y+z &x &y \\ z+x &z &x \\ 2 &1 &1 \\ \end{vmatrix}= (x+y+z)\begin{vmatrix} y+z &x &y \\ z-x &z-x&0 \\ 2 &1 &1 \\ \end{vmatrix}= (x+y+z)(z-x)\begin{vmatrix} y+z &x &y \\ 1 &1 &0 \\ 2 &1 &1 \\ \end{vmatrix}= (x+y+z)(z-x)\begin{vmatrix} z-y &x-y&0 \\ 1 &1 &0 \\ 2 &1 &1 \\ \end{vmatrix}= (x+y+z)(z-x)\begin{vmatrix} z-x &0 &0 \\ 1 &1 &0 \\ 2 &1 &1 \\ \end{vmatrix}= (x+y+z)(z-x)^2$

Answer 2

Not sure what the "Factor Method" is. Maybe it's this, but here is an approach all the same.

The determinant is going to be a homogeneous cubic polynomial in $x,y,z$ since each entry is homogeneous linear.

Adding the three rows, it's clear that if $x+y+z=0$, then the determinant would be $0$ because it would make the three rows sum to a zero row. So $(x+y+z)$ is a factor of the determinant.

The determinant is $0$ if $x=z$ since the first column would then be the sum of the second and third columns. So $(x-z)$ is a factor.

Thus far the determinant is $$(x+y+z)(x-z)(ax+by+cz)$$

The coefficient of $x^3$ must be $1$ because there is only one product that produces $x^3$ (the $(1,2),(2,3),(3,1)$ product, which is $1x^3$). A similar observation can be made for $z$. From these observations, you can conclude $a=1$ and $c=-1$. Considering $y$ though, the coefficient of $y^3$ is $0$ since $y$ is missing from the middle row. And also the coefficient of $y^2$ is $0$, once you consider the three products that contribute to the coefficient of $y^2$. [The $(1,1),(2,3),(3,2)$ product gives $y^2$ a coefficient of $-x$; the $(1,3),(2,2),(3,1)$ product gives it a coefficient of $-z$; the $(1,3),(2,1),(3,2)$ gives it a coefficient of $(x+z)$.] So $b=0$, and the determinant is $$(x+y+z)(x-z)(x-z)$$