Question

By using properties of determinants, show that:
$\left| \begin{align}
  & x\,\,\,\,\,\,{{x}^{2}}\,\,\,\,\,yz \\
 & y\,\,\,\,\,\,{{y}^{2}}\,\,\,\,\,zx \\
 & z\,\,\,\,\,\,\,{{z}^{2}}\,\,\,\,\,xy \\
\end{align} \right|=\left( x-y \right)\left( y-z \right)\left( z-x \right)\left( xy+yz+zx \right)$

Answer 1

Hint: Here, we need to use the row transformations for the first row and the second row in the first part of the answer and then use row transformation for the first row, column transformation for the second row. In the end, we need to use the basic principle of the matrix of how to find the determinant and find the required result.

Complete step-by-step answer:
Here we have, on the left-hand side $\left| \begin{align}
  & x\,\,\,\,\,\,{{x}^{2}}\,\,\,\,\,yz \\
 & y\,\,\,\,\,\,{{y}^{2}}\,\,\,\,\,zx \\
 & z\,\,\,\,\,\,\,{{z}^{2}}\,\,\,\,\,xy \\
\end{align} \right|$
We shall apply row transformations to solve further, we get
Let us use, ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$, we get
$\left| \begin{align}
  & x\,\,\,\,\,\,{{x}^{2}}\,\,\,\,\,yz \\
 & y\,\,\,\,\,\,{{y}^{2}}\,\,\,\,\,zx \\
 & z\,\,\,\,\,\,\,{{z}^{2}}\,\,\,\,\,xy \\
\end{align} \right|$
$=\left| \begin{align}
  & x-y\,\,\,\,\,\,{{x}^{2}}-{{y}^{2}}\,\,\,\,\,yz-zx \\
 & \,\,\,y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{y}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,zx \\
 & \,\,\,z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{z}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,xy \\
\end{align} \right|$
Let us take $\left( x-y \right)$ common from the first row,
$=\left| \begin{align}
  & x-y\,\,\,\,\,\,\left( x-y \right)\left( x+y \right)\,\,\,\,\,-z\left( x-y \right) \\
 & \,\,\,y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{y}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,zx \\
 & \,\,\,z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{z}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,xy \\
\end{align} \right|$
$=\left( x-y \right)$$\left| \begin{align}
  & \,1\,\,\,\,\,\,\,\,\,\,\left( x+y \right)\,\,\,\,\,-z \\
 & y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{y}^{2}}\,\,\,\,\,\,\,\,\,\,zx \\
 & z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{z}^{2}}\,\,\,\,\,\,\,\,\,\,xy \\
\end{align} \right|$
Now, let us use another row transformation for the second row, we get
Let us use, ${{R}_{2}}\to {{R}_{2}}-{{R}_{3}}$, we get
$\left( x-y \right)$$\left| \begin{align}
  & \,1\,\,\,\,\,\,\,\,\,\,\left( x+y \right)\,\,\,\,\,-z \\
 & y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{y}^{2}}\,\,\,\,\,\,\,\,\,\,zx \\
 & z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{z}^{2}}\,\,\,\,\,\,\,\,\,\,xy \\
\end{align} \right|$
$=\left( x-y \right)$$\left| \begin{align}
  & \,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( x+y \right)\,\,\,\,\,\,\,\,\,\,\,\,\,-z \\
 & y-z\,\,\,\,\,\,\,\,\,\,{{y}^{2}}-{{z}^{2}}\,\,\,\,\,\,\,\,\,\,zx-xy \\
 & \,\,\,\,z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{z}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,xy \\
\end{align} \right|$
Let us take $\left( y-z \right)$ common from the second row, we get
$=\left( x-y \right)\left( y-z \right)$$\left| \begin{align}
  & \,\,1\,\,\,\,\,\,\,\,\,\,\left( x+y \right)\,\,\,\,\,\,\,\,-z \\
 & \,\,1\,\,\,\,\,\,\,\,\,\,\left( y+z \right)\,\,\,\,\,\,\,\,-x \\
 & \,\,z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{z}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,xy \\
\end{align} \right|$
To further solve, let us use the row transformation for the first row.
Let us use ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$, we get
$\left( x-y \right)\left( y-z \right)$$\left| \begin{align}
  & \,\,1\,\,\,\,\,\,\,\,\,\,\left( x+y \right)\,\,\,\,\,\,\,\,-z \\
 & \,\,1\,\,\,\,\,\,\,\,\,\,\left( y+z \right)\,\,\,\,\,\,\,\,-x \\
 & \,\,z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{z}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,xy \\
\end{align} \right|$
$=\left( x-y \right)\left( y-z \right)$\[\left| \begin{align}
  & \,\,1\,\,\,\,\,\,\,\,\,\,x+y-y-z\,\,\,\,\,\,\,-z+x \\
 & \,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( y+z \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-x \\
 & \,\,z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{z}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,xy \\
\end{align} \right|\]
$=\left( x-y \right)\left( y-z \right)$\[\left| \begin{align}
  & \,\,0\,\,\,\,\,\,\,\,\,\,x-z\,\,\,\,\,\,\,x-z \\
 & \,\,1\,\,\,\,\,\,\,\,\,\left( y+z \right)\,\,\,\,\,\,\,-x \\
 & \,\,z\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{z}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,xy \\
\end{align} \right|\]
We can write the first row in this way
$\left( x-y \right)\left( y-z \right)$\[\left| \begin{align}
  & \,\,0\,\,\,\,\,\,\,\,\,\,x-z\,\,\,\,\,\,\,x-z \\
 & \,\,1\,\,\,\,\,\,\,\,\,\left( y+z \right)\,\,\,\,\,\,\,-x \\
 & \,\,z\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{z}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,xy \\
\end{align} \right|\]
$=\left( x-y \right)\left( y-z \right)$\[\left| \begin{align}
  & \,\,0\,\,\,\,\,-\left( z-x \right)\,\,\,\,\,\,\,-\left( z-x \right) \\
 & \,\,1\,\,\,\,\,\,\,\,\,\left( y+z \right)\,\,\,\,\,\,\,\,\,\,\,\,-x \\
 & \,\,z\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{z}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,xy \\
\end{align} \right|\]
Now, let us take $\left( z-x \right)$ common from the first row, we get
$\left( x-y \right)\left( y-z \right)$\[\left| \begin{align}
  & \,\,0\,\,\,\,\,-\left( z-x \right)\,\,\,\,\,\,\,-\left( z-x \right) \\
 & \,\,1\,\,\,\,\,\,\,\,\,\left( y+z \right)\,\,\,\,\,\,\,\,\,\,\,\,-x \\
 & \,\,z\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{z}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,xy \\
\end{align} \right|\]
$=\left( x-y \right)$$\left( y-z \right)$$\left( z-x \right)$\[\left| \begin{align}
  & \,\,0\,\,\,\,\,\,\,\,\,-1\,\,\,\,\,\,\,\,\,-1 \\
 & \,\,1\,\,\,\,\,\,\left( y+z \right)\,\,\,\,\,-x \\
 & \,\,z\,\,\,\,\,\,\,\,\,\,\,{{z}^{2}}\,\,\,\,\,\,\,\,\,\,xy \\
\end{align} \right|\]
To further solve, we need to use the column transformation for column 2.
Let us take ${{C}_{2}}\to {{C}_{2}}-{{C}_{3}}$, we get
$\left( x-y \right)$$\left( y-z \right)$$\left( z-x \right)$\[\left| \begin{align}
  & \,\,0\,\,\,\,\,\,\,\,\,-1\,\,\,\,\,\,\,\,\,-1 \\
 & \,\,1\,\,\,\,\,\,\left( y+z \right)\,\,\,\,\,-x \\
 & \,\,z\,\,\,\,\,\,\,\,\,\,\,{{z}^{2}}\,\,\,\,\,\,\,\,\,\,xy \\
\end{align} \right|\]
$=\left( x-y \right)$$\left( y-z \right)$$\left( z-x \right)$\[\left| \begin{align}
  & \,\,0\,\,\,\,\,\,\,\,\,-1\,-\left( -1 \right)\,\,\,\,\,\,\,\,\,\,-1 \\
 & \,\,1\,\,\,\,\,\,\left( y+z \right)-\left( -x \right)\,\,\,\,\,-x \\
 & \,\,z\,\,\,\,\,\,\,\,\,\,\,{{z}^{2}}-xy\,\,\,\,\,\,\,\,\,\,\,\,\,\,xy \\
\end{align} \right|\]
$=\left( x-y \right)$$\left( y-z \right)$$\left( z-x \right)$\[\left| \begin{align}
  & \,\,0\,\,\,\,\,\,\,\,\,-1\,+1\,\,\,\,\,\,\,\,\,\,\,-1 \\
 & \,\,1\,\,\,\,\,\,\left( y+z \right)+x\,\,\,\,\,-x \\
 & \,\,z\,\,\,\,\,\,\,\,\,\,{{z}^{2}}-xy\,\,\,\,\,\,\,\,xy \\
\end{align} \right|\]
$=\left( x-y \right)$$\left( y-z \right)$$\left( z-x \right)$\[\left| \begin{align}
  & \,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-1 \\
 & \,\,1\,\,\,\,\,\,x+y+z\,\,\,\,\,\,\,\,-x \\
 & \,\,z\,\,\,\,\,\,\,{{z}^{2}}-xy\,\,\,\,\,\,\,\,\,\,\,xy \\
\end{align} \right|\]…… (1)
Now, let us follow the basic principle of matrix of evaluating the determinant and solve the above expression, we get
\[\left| \begin{align}
  & \,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-1 \\
 & \,\,1\,\,\,\,\,\,x+y+z\,\,\,\,\,\,\,\,-x \\
 & \,\,z\,\,\,\,\,\,\,{{z}^{2}}-xy\,\,\,\,\,\,\,\,\,\,\,xy \\
\end{align} \right|\] $=0\left[ \left( x+y+z \right)\left( xy \right)-\left( {{z}^{2}}-xy \right)\left( -x \right) \right]$$-\,0\left[ \left( 1 \right)\left( xy \right)-\left( -x \right)\left( z \right) \right]$
                                             $+\left( -1 \right)\left[ \left( 1 \right)\left( {{z}^{2}}-xy \right)-\left( x+y+z \right)\left( z \right) \right]$
                                         $=0-0-1\left[ {{z}^{2}}-xy-zx-yz-{{z}^{2}} \right]$
                                         $-1\left( -xy-zx-yz \right)$
\[\left| \begin{align}
  & \,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-1 \\
 & \,\,1\,\,\,\,\,\,x+y+z\,\,\,\,\,\,\,\,-x \\
 & \,\,z\,\,\,\,\,\,\,{{z}^{2}}-xy\,\,\,\,\,\,\,\,\,\,\,xy \\
\end{align} \right|=\] $\left( xy+yz+zx \right)$…… (2)
From, equations (1) and (2), we get
$\left| \begin{align}
  & x\,\,\,\,\,\,{{x}^{2}}\,\,\,\,\,yz \\
 & y\,\,\,\,\,\,{{y}^{2}}\,\,\,\,\,zx \\
 & z\,\,\,\,\,\,\,{{z}^{2}}\,\,\,\,\,xy \\
\end{align} \right|$$=\left( x-y \right)$$\left( y-z \right)$$\left( z-x \right)$$\left( xy+yz+zx \right)$
Hence, proved.

Note: A matrix is a rectangular array or table of numbers, symbols, expressions, equations arranged in rows and columns. The plural form of a matrix is called matrices. The size of the matrix is given by the number of rows and the number of columns.