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Question
Using properties of determinants prove that
\[\begin{vmatrix}x + 4 & 2x & 2x \\ 2x & x + 4 & 2x \\ 2x & 2x & x + 4\end{vmatrix} = \left( 5x + 4 \right) \left( 4 - x \right)^2\]
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Solution
\[∆ = \begin{vmatrix}x + 4 & 2x & 2x \\ 2x & x + 4 & 2x \\ 2x & 2x & x + 4\end{vmatrix}\]
\[ = \begin{vmatrix}5x + 4 & 5x + 4 & 5x + 4 \\ 2x & x + 4 & 2x \\ 2x & 2x & x + 4\end{vmatrix} \left[\text{ Applying }R_1 \text{ to }R_1 + R_2 + R_3 \right]\]
\[ = 5x + 4\begin{vmatrix}1 & 1 & 1 \\ 2x & x + 4 & 2x \\ 2x & 2x & x + 4\end{vmatrix} \left[\text{ Take out }5x + 4\text{ common from }R_1 \right]\]
\[ = 5x + 4\begin{vmatrix}1 & 0 & 0 \\ 2x & 4 - x & 0 \\ 2x & 0 & 4 - x\end{vmatrix} \left[\text{ Applying }C_2 \text{ to }C_2 - C_1\text{ and }C_3 \text{ to }C_3 - C_1 \right]\]
\[ = 5x + 4(4 - x )^2 \left[\text{ Expanding along }R_1 \right]\]
Hence proved.
