Question

Prove that :

\[\begin{vmatrix}\left( a + 1 \right) \left( a + 2 \right) & a + 2 & 1 \\ \left( a + 2 \right) \left( a + 3 \right) & a + 3 & 1 \\ \left( a + 3 \right) \left( a + 4 \right) & a + 4 & 1\end{vmatrix} = - 2\]

 

Answer 1

\[\text{ Let LHS }= \Delta = \begin{vmatrix} \left( a + 1 \right)\left( a + 2 \right) & a + 2 & 1\\\left( a + 2 \right)\left( a + 3 \right) & a + 3 & 1 \\\left( a + 3 \right)\left( a + 4 \right) & a + 4 & 1 \end{vmatrix}\] 
\[ = \begin{vmatrix} \left( a + 1 \right)\left( a + 2 \right) - \left( a + 2 \right)\left( a + 3 \right) & \left( a + 2 \right) - \left( a + 3 \right) & 0\\\left( a + 2 \right)\left( a + 3 \right) - \left( a + 3 \right)\left( a + 4 \right) & \left( a + 3 \right) - \left( a + 4 \right) & 0 \\\left( a + 3 \right)\left( a + 4 \right) & \left( a + 4 \right) & 1 \end{vmatrix} \left[\text{ Applying }C_1 \to \hspace{0.167em} C_1 - C_2\text{ and }C_2 \to C_2 - C_3 \right]\] 
\[ = \begin{vmatrix} - 2\left( a + 2 \right) & - 1 & 0\\ - 2\left( a + 3 \right) & - 1 & 0\\\left( a + 3 \right)\left( a + 4 \right) & \left( a + 4 \right) & 1 \end{vmatrix}\] 
\[ = \left\{ 1 \times \begin{vmatrix} - 2\left( a + 2 \right) & - 1 \\- 2\left( a + 3 \right) & - 1 \end{vmatrix} \right\} \left[\text{ Expanding along }C_3 \right]\] 
\[ = 4 + 2a - 2a - 6\] 
\[ = - 2\] 
\[ = RHS\]

Hence proved.