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Question
Prove the following identities:
\[\begin{vmatrix}x + \lambda & 2x & 2x \\ 2x & x + \lambda & 2x \\ 2x & 2x & x + \lambda\end{vmatrix} = \left( 5x + \lambda \right) \left( \lambda - x \right)^2\]
Solution
\[LHS: \]
\[\begin{vmatrix}x + \lambda & 2x & 2x \\ 2x & x + \lambda & 2x \\ 2x & 2x & x + \lambda\end{vmatrix}\]
\[ = \begin{vmatrix}x + \lambda & 2x & 2x \\ 2x - x - \lambda & x + \lambda - 2x & 0 \\ 2x - x - \lambda & 0 & x + \lambda - 2x\end{vmatrix} \left[\text{ Applying }R_2 \text{ to }R_2 - R_1 \text{ and }R_3 \text{ to }R_3 - R_1 \right]\]
\[ = \begin{vmatrix}x + \lambda & 2x & 2x \\ - (\lambda - x) & \lambda - x & 0 \\ - (\lambda - x) & 0 & \lambda - x\end{vmatrix}\]
\[ = \left( \lambda - x \right)^2 \begin{vmatrix}x + \lambda & 2x & 2x \\ - 1 & 1 & 0 \\ - 1 & 0 & 1\end{vmatrix} \left[\text{ Taking }\left( \lambda - x \right)\text{ common from }R_2\text{ and }\left( \lambda - x \right)\text{ common from }R_3 \right]\]
\[ = \left( \lambda - x \right)^2 \left[ - 1\left( - 2x \right) + 1\left( x + \lambda + 2x \right) \right] \left[\text{ Expanding along last row }\right]\]
\[ = \left( \lambda - x \right)^2 \left( \lambda + 5x \right)\]
\[ = RHS\]
\[ \therefore \begin{vmatrix}x + \lambda & 2x & 2x \\ 2x & x + \lambda & 2x \\ 2x & 2x & x + \lambda\end{vmatrix} = \left( \lambda - x \right)^2 \left( \lambda + 5x \right)\]
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