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Question
Prove the following identity:
\[\begin{vmatrix}2y & y - z - x & 2y \\ 2z & 2z & z - x - y \\ x - y - z & 2x & 2x\end{vmatrix} = \left( x + y + z \right)^3\]
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Solution
\[LHS\]
\[ = \begin{vmatrix}2y & y - z - x & 2y \\ 2z & 2z & z - x - y \\ x - y - z & 2x & 2x\end{vmatrix}\]
\[ = \begin{vmatrix}2y + 2z + x - y - z & y - z - x + 2z + 2x & 2y + z - x - y + 2x \\ 2z & 2z & z - x - y \\ x - y - z & 2x & 2x\end{vmatrix} \left[\text{ Applying }R_1 \text{ to }R_1 + R_2 + R_3 \right]\]
\[ = \begin{vmatrix}x + y + z & x + y + z & x + y + z \\ 2z & 2z & z - x - y \\ x - y - z & 2x & 2x\end{vmatrix}\]
\[ = \left( x + y + z \right)\begin{vmatrix}1 & 1 & 1 \\ 2z & 2z & z - x - y \\ x - y - z & 2x & 2x\end{vmatrix} \left[\text{ Taking }\left( x + y + z \right)\text{ common from }R_1 \right]\]
\[ = \left( x + y + z \right)\begin{vmatrix}0 & 1 & 1 \\ 0 & 2z & z - x - y \\ - x - y - z & 2x & 2x\end{vmatrix} \left[\text{ Applying }C_1 \text{ to }C_1 - C_2 \right]\]
\[ = \left( x + y + z \right)^2 \begin{vmatrix}0 & 1 & 1 \\ 0 & 2z & z - x - y \\ - 1 & 2x & 2x\end{vmatrix} \left[\text{ Taking }\left( x + y + z \right)\text{ common from }C_1 \right]\]
\[ = \left( x + y + z \right)^2 \left[ - 1\left( z - x - y - 2z \right) \right] \left[\text{ Expanding along first column }\right]\]
\[ = \left( x + y + z \right)^3 \]
\[ = RHS\]
\[ \therefore \begin{vmatrix}2y & y - z - x & 2y \\ 2z & 2z & z - x - y \\ x - y - z & 2x & 2x\end{vmatrix} = \left( x + y + z \right)^3\]
