Question

Using properties of determinants, prove that
$\left| \begin{matrix}
   {{(x+y)}^{2}} & zx & zy \\
   zx & {{(z+y)}^{2}} & xy \\
   zy & xy & {{(z+x)}^{2}} \\
\end{matrix} \right|=2xyz{{(x+y+z)}^{3}}$

Answer 1

Hint:here we have to find the value of the given determinant so in order to find the value we can use the properties of determinant so as to we minimize the calculation. In a determinant if we each element of a row or column is multiplied by a constant $k$then the value of new determinant is $k$times the value of the original determinant.
Here we apply ${{R}_{1}}\to z{{R}_{1,}}\text{ }{{R}_{2}}\to x{{R}_{2}},\text{ }{{R}_{3}}\to y{{R}_{3}}$. Then try to take common terms from the elements and then solve.

Complete step by step answer:
Let us consider the given determinant as $\Delta $, so we can write
$\Delta =\left| \begin{matrix}
   {{(x+y)}^{2}} & zx & zy \\
   zx & {{(z+y)}^{2}} & xy \\
   zy & xy & {{(z+x)}^{2}} \\
\end{matrix} \right|$
Now we can do the operation of given determinant as ${{R}_{1}}\to z{{R}_{1}}$, hence we can write
$\Delta =\dfrac{1}{z}\left| \begin{matrix}
   z{{(x+y)}^{2}} & {{z}^{2}}x & {{z}^{2}}y \\
   zx & {{(z+y)}^{2}} & xy \\
   zy & xy & {{(z+x)}^{2}} \\
\end{matrix} \right|$
Also, we can do the operation of given determinant as ${{R}_{2}}\to x{{R}_{2}}$, hence we can write
$\Delta =\dfrac{1}{zx}\left| \begin{matrix}
   z{{(x+y)}^{2}} & {{z}^{2}}x & {{z}^{2}}y \\
   z{{x}^{2}} & x{{(z+y)}^{2}} & {{x}^{2}}y \\
   zy & xy & {{(z+x)}^{2}} \\
\end{matrix} \right|$
Further, we can do the operation of given determinant as ${{R}_{3}}\to y{{R}_{3}}$, hence we can write
$\Delta =\dfrac{1}{xyz}\left| \begin{matrix}
   z{{(x+y)}^{2}} & {{z}^{2}}x & {{z}^{2}}y \\
   z{{x}^{2}} & x{{(z+y)}^{2}} & {{x}^{2}}y \\
   z{{y}^{2}} & x{{y}^{2}} & y{{(z+x)}^{2}} \\
\end{matrix} \right|$-------------------------(1)
Now from the determinant numbering (1) we can take common $z$ from ${{C}_{1}}$, $x$ from ${{C}_{2}}$ and $y$ from ${{C}_{3}}$. So, we can write the above determinant as
$\Delta =\dfrac{xyz}{xyz}\left| \begin{matrix}
   {{(x+y)}^{2}} & {{z}^{2}} & {{z}^{2}} \\
   {{x}^{2}} & {{(z+y)}^{2}} & {{x}^{2}} \\
   {{y}^{2}} & {{y}^{2}} & {{(z+x)}^{2}} \\
\end{matrix} \right|$
Now we can do the operation ${{C}_{1}}\to {{C}_{1}}-{{C}_{2}},{{C}_{2}}\to {{C}_{2}}-{{C}_{3}}$, we can write
$\Delta =\left| \begin{matrix}
   {{(x+y)}^{2}}-{{z}^{2}} & {{z}^{2}}-{{z}^{2}} & {{z}^{2}} \\
   {{x}^{2}}-{{(z+y)}^{2}} & {{(z+y)}^{2}}-{{x}^{2}} & {{x}^{2}} \\
   {{y}^{2}}-{{y}^{2}} & {{y}^{2}}-{{(z+x)}^{2}} & {{(z+x)}^{2}} \\
\end{matrix} \right|$
Here we use the formula
${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$, we can write
$\Delta =\left| \begin{matrix}
   (x+y+z)(x+y-z) & 0 & {{z}^{2}} \\
   (x+y+z)(x-z-y) & (z+y+x)(z+y-x) & {{x}^{2}} \\
   0 & (y-z-x)(y+z+x) & {{(z+x)}^{2}} \\
\end{matrix} \right|$
Now we can take common $(x+y+z)$from ${{C}_{1}}$and ${{C}_{2}}$, we can write
$\Delta ={{(x+y+z)}^{2}}\left| \begin{matrix}
   (x+y-z) & 0 & {{z}^{2}} \\
   (x-z-y) & (z+y-x) & {{x}^{2}} \\
   0 & (y-z-x) & {{(z+x)}^{2}} \\
\end{matrix} \right|$
Now we can do the operation ${{R}_{3}}\to {{R}_{3}}-({{R}_{2}}+{{R}_{1}})$ we can write further
\[\begin{align}
  & \Delta ={{(x+y+z)}^{2}}\left| \begin{matrix}
   (x+y-z) & 0 & {{z}^{2}} \\
   (x-z-y) & (z+y-x) & {{x}^{2}} \\
   0 & (y-z-x) & {{(z+x)}^{2}} \\
\end{matrix} \right| \\
 & \Delta ={{(x+y+z)}^{2}}\left| \begin{matrix}
   (x+y-z) & 0 & {{z}^{2}} \\
   (x-z-y) & (z+y-x) & {{x}^{2}} \\
   0-(2x-2z) & (y-z-x)-(z+y-x) & {{(z+x)}^{2}}-{{z}^{2}}-{{x}^{2}} \\
\end{matrix} \right| \\
 & \Delta ={{(x+y+z)}^{2}}\left| \begin{matrix}
   (x+y-z) & 0 & {{z}^{2}} \\
   (x-z-y) & (z+y-x) & {{x}^{2}} \\
   2z-2x & -2z & 2zx \\
\end{matrix} \right| \\
\end{align}\]
We can further do the operation
${{C}_{1}}\to {{C}_{1}}+{{C}_{2}}$
\[\begin{align}
  & \Delta ={{(x+y+z)}^{2}}\left| \begin{matrix}
   (x+y-z) & 0 & {{z}^{2}} \\
   0 & (z+y-x) & {{x}^{2}} \\
   -2x & -2z & 2zx \\
\end{matrix} \right| \\
 & \Delta =2{{(x+y+z)}^{2}}\left| \begin{matrix}
   (x+y-z) & 0 & {{z}^{2}} \\
   0 & (z+y-x) & {{x}^{2}} \\
   -x & -z & zx \\
\end{matrix} \right| \\
\end{align}\]
Here we take 2 as common from ${{R}_{3}}$
Now we expand the determinant
\[\begin{align}
  & \Delta =2{{(x+y+z)}^{2}}\left[ (x+y-z)\left| \begin{matrix}
   z+y-x & {{x}^{2}} \\
   -z & zx \\
\end{matrix} \right|-0\left| \begin{matrix}
   0 & {{z}^{2}} \\
   -z & zx \\
\end{matrix} \right|+(-x)\left| \begin{matrix}
   0 & {{z}^{2}} \\
   z+y-x & {{x}^{2}} \\
\end{matrix} \right| \right] \\
 & \Delta =2{{(x+y+z)}^{2}}\left[ (x+y-z)\{(z+y-x)(zx)-(-z)({{x}^{2}})-0-x\{(0)({{x}^{2}})-({{z}^{2}})(z+y-x)\} \right] \\
 & \Delta =2{{(x+y+z)}^{2}}\left[ (x+y-z)\{({{z}^{2}}x+xyz-z{{x}^{2}}+z{{x}^{2}})\}-x\{0-{{z}^{3}}-{{z}^{2}}y+{{z}^{2}}x\} \right] \\
 & \Delta =2{{(x+y+z)}^{2}}\left[ (x+y-z)({{z}^{2}}x+xyz)+x{{z}^{3}}+x{{z}^{2}}y-{{z}^{2}}{{x}^{2}} \right] \\
 & \Delta =2{{(x+y+z)}^{2}}z\left[ (x+y-z)(zx+xy)+x{{z}^{2}}+xzy-z{{x}^{2}} \right] \\
 & \Delta =2{{(x+y+z)}^{2}}zx\left[ (x+y-z)(z+y)+{{z}^{2}}+zy-zx \right] \\
\end{align}\]
We take $z$and $x$as common and taken outside the bracket, we can simplify further
\[\begin{align}
  & \Delta =2{{(x+y+z)}^{2}}zx\left[ (x+y-z)(z+y)+{{z}^{2}}+zy-zx \right] \\
 & \Delta =2{{(x+y+z)}^{2}}zx\left[ xz+xy+yz+{{y}^{2}}-{{z}^{2}}-yz+{{z}^{2}}+zy-zx \right] \\
 & \Delta =2{{(x+y+z)}^{2}}zx\left[ xy+yz+{{y}^{2}} \right] \\
 & \Delta =2{{(x+y+z)}^{2}}zxy\left[ x+z+y \right] \\
\end{align}\]
In the above simplification process, we take $y$as common, so, the value of the determinant is
\[\Delta =2xyz{{(x+y+z)}^{3}}\]
Hence proved.

Note:
The main theme behind the simplification of a determinant lies in obtaining the maximum possible zeros in a row (or a column) by using the properties of determinant and then to expand the determinant by that row or column.
If we get two row or two columns of a determinant is identical then the value of determinant is zero.