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# By using properties of determinants, show that: determinant x+y+2z x y z y+z+2x y z x z+x+2y =2(x+y+z)^3

Q : 11        By using properties of determinants, show that:

$\dpi{100} (ii) \begin{vmatrix} x+y+2z &x &y \\ z & y+z+2x & y\\ z & x &z+x+2y \end{vmatrix}=2(x+y+z)^3$

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Given determinant

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

Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$   we get;

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

Taking 2(x+y+z) factor out, we get;

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

Now, applying row transformations, $R_{1} \rightarrow R_{1} -R_{2}$  and then $R_{2} \rightarrow R_{2} -R_{3}$.

we get;

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

$=2(x+y+z)^3\begin{vmatrix} 0 &-1 &0 \\ 0 & 1 & -1\\ 1 & x &z+x+2y \end{vmatrix}$

$=2(x+y+z)^3\begin{vmatrix} -1 &0 \\ 1& -1 \end{vmatrix} = 2(x+y+z)^3$

Hence proved.

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