Question

Using properties of determinants prove the following –
$\left| \begin{matrix}
   3x & -x+y & -x+z \\
   x-y & 3y & z-y \\
   x-z & y-z & 3z \\
\end{matrix} \right|$ $=3\left( x+y+z \right)(xy+yz+xz)$

Answer 1

Hint: Try to make whole column or a row with elements $(x+y+z)$ or $(xy+yz+xz)$ and take it common from the determinant , so that you can easily reach Right Hand Side of the equality that we have to proof , then solve the determinant and get the desired equality.

Complete step by step answer:
Consider the Left Hand Side of the equality that we have to proof ,
$\left| \begin{matrix}
   3x & -x+y & -x+z \\
   x-y & 3y & z-y \\
   x-z & y-z & 3z \\
\end{matrix} \right|$Now, we will try to make a whole row or a column $(x+y+z)$ ,
So, for this we will use column operation ,
Since, we know the property of determinants that row or column operations don’t affect the determinant of a matrix , so we can use the column operation without any problem.
Notice when we add the columns 2 and 3 to column 1 , the whole column 1 gets the same element ,
$(x+y+z)$
${{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}$
By performing this operation, we get,
$\left| \begin{matrix}
   x+y+z & -x+y & -x+z \\
   x+y+z & 3y & z-y \\
   x+y+z & y-z & 3z \\
\end{matrix} \right|$
Now, take common $(x+y+z)$ from first column, we get
$=(x+y+z)\left| \begin{matrix}
   1 & -x+y & -x+z \\
   1 & 3y & z-y \\
   1 & y-z & 3z \\
\end{matrix} \right|$
Now, we will perform the row operations , so that in next step , we can easily evaluate the determinant,
$\begin{align}
  & {{R}_{2}}\to {{R}_{2}}-{{R}_{1}} \\
 & {{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \\
\end{align}$
By performing these 2 operations we get 1 element of row 2 and 3 as 0 , and so it will ease the process of evaluating the determinant. So, by performing these operations , we get
$=(x+y+z)\left| \begin{matrix}
   1 & x+2y & -x+z \\
   0 & 2y+x & -y+x \\
   0 & -z+x & 2z+x \\
\end{matrix} \right|$
$\begin{align}
  & =(x+y+z)((2z+x)(2y+x)-(-y+x)(-z+x) \\
 & =(x+y+z)(4yz+2xz+2xy+{{x}^{2}}-(yz-xy-zx+{{x}^{2}})) \\
 & =(x+y+z)(3yz+3xy+3xz) \\
 & =3(x+y+z)(xy+yz+xz)
\end{align}$

Hence, $\left| \begin{matrix}
   3x & -x+y & -x+z \\
   x-y & 3y & z-y \\
   x-z & y-z & 3z \\
\end{matrix} \right|$ $=3\left( x+y+z \right)(xy+yz+xz)$

Note: The possibility of mistake here is that when you perform row or column operations , you don’t need anything to compensate like in the operation ${{R}_{2}}\to {{R}_{2}}-{{R}_{1}}$ you don’t need to divide by -1 . Row or column operations don’t affect the determinants.