By using the properties of determinants show that $(ii) \begin{vmatrix} x+y+2z&x&y \\ z&y+z+2x&y \\ z&x&z+x+2y \end{vmatrix} = 2(x+y+z)^3$

Note: This is part 2 of a 2 part question, split as 2 separate questions here.

Toolbox:
• If each element of a row (or column) of a determinant is multiplied by a constant k ,then its value gets multiplied by k.
• By this property we can take out any common factor from any one row or any one column of the given determinant.
• Elementary transformations can be done by
• 1. Interchanging any two rows or columns. rows.
• 2. Mutiplication of the elements of any row or column by a non-zero number
• 3. The addition of any row or column , the corresponding elemnets of any other row or column multiplied by any non zero number.
Let $\bigtriangleup=\begin{vmatrix}x+y+2z& x & y\\z & y+z+2x & y\\z &x & z+x+2y\end{vmatrix}$

Let us add $C_1,C_2$ and $C_3$,hence we apply $C_1\rightarrow C_1+C_2+C_3$

$\begin{vmatrix}2x+2y+2z& x & y\\2x+2y+2z & y+z+2x & y\\2x+2y+2z &x & z+x+2y\end{vmatrix}$

Take 2(x+y+z) as a common factor from $C_1$

$\bigtriangleup=2(x+y+z)\begin{vmatrix}1& x & y\\1 & y+z+2x & y\\z &x & z+x+2y\end{vmatrix}$

Now applying $R_1\rightarrow R_1-R_2$ and $R_2\rightarrow R_2-R_3$

$\bigtriangleup=2(x+y+z)\begin{vmatrix}0& -(x+y+z) & 0\\0 & x+y+z & -(x+y+z)\\1 &x & z+x+2y\end{vmatrix}$

By applying $R_1\rightarrow R_1+R_2$

$\bigtriangleup=2(x+y+z)\begin{vmatrix}0& 0 & -(x+y+z)\\0 & x+y+z & -(x+y+z)\\1 &x & z+x+2y\end{vmatrix}$

Now expanding along $R_1$ we get,

$\bigtriangleup=2(x+y+z)(-(x+y+z))[o\times x-1\times(x+y+z)]$

$\quad=2(x+y+z)(x+y+z)^2$

$\quad=2(x+y+z)^3$

Hence $\begin{vmatrix}x+y+2z & x& y\\z & y+z+2x & y\\z & x & z+x+2y\end{vmatrix}=2(x+y+z)^3$

Hence proved.