Question

Prove that :

\[\begin{vmatrix}1 & 1 + p & 1 + p + q \\ 2 & 3 + 2p & 4 + 3p + 2q \\ 3 & 6 + 3p & 10 + 6p + 3q\end{vmatrix} = 1\]

 

Answer 1

\[\text{ Let LHS }= \Delta = \begin{vmatrix} 1 & 1 + p & 1 + p + q\\2 & 3 + 2p & 4 + 3p + 2q\\3 & 6 + 3p & 10 + 6p + 3q \end{vmatrix}\] 
\[ = \begin{vmatrix} 1 & 1 & 1 + p\\2 & 3 & 4 + 3p\\3 & 6 & 10 + 6p \end{vmatrix} + \begin{vmatrix} 1 & p & q\\2 & 2p & 2q\\3 & 3p & 3q \end{vmatrix}\] 
\[ = \begin{vmatrix} 1 & 1 & 1\\2 & 3 & 4\\3 & 6 & 10 \end{vmatrix} + \begin{vmatrix} 1 & 1 & p\\2 & 3 & 3p\\3 & 6 & 6p \end{vmatrix} + \left( pq \right) \begin{vmatrix} 1 & 1 & 1\\2 & 2 & 2\\3 & 3 & 3 \end{vmatrix} \left[\text{ Taking out pq common from last determinant }\right]\] 
\[ = \begin{vmatrix} 1 & 1 & 1\\2 & 3 & 4\\3 & 6 & 10 \end{vmatrix} + \left( p \right)\begin{vmatrix} 1 & 1 & 1\\2 & 3 & 3\\3 & 6 & 6 \end{vmatrix} + 0 \left[\text{ Taking out p common from second determinant }\right]\] 
\[ = \begin{vmatrix} 1 & 1 & 1\\2 & 3 & 4\\3 & 6 & 10 \end{vmatrix} + 0 \left[ \because\text{ Value of determinant with two identical columns is zero }\right]\] 
\[ = \begin{vmatrix} 1 & 0 & 0\\2 & 1 & 2\\3 & 3 & 7 \end{vmatrix} \left[\text{ Applying }C_2 \to C_2 - C_1\text{ and }C_3 \to C_3 - C1 \right]\] 
\[ = \left\{ 1 \times \begin{vmatrix}1 & 2 \\ 3 & 7\end{vmatrix} \right\} \left[\text{ Expanding along }R_1 \right]\] 
\[ = 7 - 6\] 
\[ = 1 \] 
\[ = RHS\]