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Question
Prove that :
Solution
\[\text{ Let LHS }= ∆ = \begin{vmatrix} a & a + b & a + 2b\\a + 2b & a & a + b\\a + b & a + 2b & a \end{vmatrix}\]
\[ \Delta = \begin{vmatrix} 3a + 3b & 3a + 3b & 3a + 3b\\ a + 2b & a & a + b\\a + b & a + 2b & a \end{vmatrix}\left[\text{ Applying }R_1 \to R_1 + R_2 + R_3 \right]\]
\[ = 3 \left( a + b \right)\begin{vmatrix} 1 & 1 & 1\\ a + 2b & a & a + b\\a + b & a + 2b & a \end{vmatrix} \left[\text{ Taking out 3 }\left( a + b \right)\text{ common from }R_1 \right]\]
\[ = 3 \left( a + b \right) \begin{vmatrix} 0 & 0 & 1\\2b & - b & a + b\\ - b & 2b & a \end{vmatrix}\left[\text{ Applying }C_1 \to C_1 - C_2 \hspace{0.167em}\text{ and }C_2 \to C_2 - C_3 \right]\]
\[ = 3 \left( a + b \right) b^2 \begin{vmatrix} 0 & 0 & 1\\ 2 & - 1 & a + b\\ - 1 & 2 & a \end{vmatrix} \left[\text{ Taking out b common from }C_1\text{ and }C_2 \right]\]
\[ = 3 \left( a + b \right) b^2 \times 3\]
\[ = 9\left( a + b \right) b^2 \]
\[ = RHS\]
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