# Solve the following

Question:

$\left|\begin{array}{ccc}3 x & -x+y & -x+z \\ x-y & 3 y & z-y \\ x-z & y-z & 3 z\end{array}\right|$

Solution:

Given, $\left|\begin{array}{ccc}3 x & -x+y & -x+z \\ x-y & 3 y & z-y \\ x-z & y-z & 3 z\end{array}\right|$

[Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$ ]

$=\left|\begin{array}{ccc}x+y+z & -x+y & -x+z \\ x+y+z & 3 y & z-y \\ x+y+z & y-z & 3 z\end{array}\right|$

[Taking $(x+y+z)$ common from column $C_{1}$ ]

$=(x+y+z)\left|\begin{array}{ccc}1 & -x+y & -x+z \\ 1 & 3 y & z-y \\ 1 & y-z & 3 z\end{array}\right|$

[Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$ ]

$=(x+y+z)\left|\begin{array}{ccc}1 & -x+y & -x+z \\ 0 & 2 y+x & x-y \\ 0 & x-z & 2 z+x\end{array}\right|$

[Applying $C_{2} \rightarrow C_{2}-C_{3}$ ]

$=(x+y+z)\left|\begin{array}{ccc}1 & -x+y & -x+z \\ 0 & 3 y & x-y \\ 0 & -3 z & 2 z+x\end{array}\right|$

[Expanding along first column]

= (x + y + z) . 1[3y(3z + x) + (3z)(x – y)]

= (x + y + z)(3yz + 3yx + 3xz)

= 3(x + y + z)(xy + yz + zx)