Prove that:

Solution:

Let LHS $=\Delta=\mid \begin{array}{lll}1 & 1+p & 1+p+q\end{array}$

$\begin{array}{lll}2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q \mid\end{array}$

$=\mid \begin{array}{lll}1 & 1 & 1+\mathrm{p}\end{array}$

$\begin{array}{llll}3 & 6 & 10+6 p|+| 1 & p & q \\ 2 & 2 p & 2 q & \\ 3 & 3 p & 3 q \mid & \end{array}$

$\begin{array}{lllll}3 & 6 & 10+6 p & |+| 1 & p & q \\ 2 & 2 p & 2 q & & & \\ 3 & 3 p & 3 q \mid & & \end{array}$

$=\mid \begin{array}{lll}1 & 1 & 1\end{array}$

$\begin{array}{lll}2 & 3 & 4\end{array}$

$\begin{array}{lll}3 & 6 & 10\end{array}|+| \begin{array}{lll}1 & 1 & \mathrm{p}\end{array}$

$\begin{array}{lll}2 & 3 & 3 p\end{array}$

$\begin{array}{lll}3 & 6 & 6 \mathrm{p}|+(\mathrm{pq})| \begin{array}{lll}1 & 1 & 1\end{array}\end{array}$'

$\begin{array}{lll}2 & 2 & 2 \\ 3 & 3 & 3\end{array}$            [Taking out $p q$ common from last determinant]

$=\mid \begin{array}{lll}1 & 1 & 1\end{array}$

$2 \quad 3 \quad 4$

$\begin{array}{lll}3 & 6 & 10\end{array}|+(\mathrm{p})| \begin{array}{lll}1 & 1 & 1\end{array}$

$\begin{array}{lll}2 & 3 & 3 \\ 3 & 6 & 6 \mid+0\end{array}$ [Taking out $p$ common from second determinant ]

$=\mid \begin{array}{lll}1 & 1 & 1\end{array}$

$\begin{array}{lll}2 & 3 & 4\end{array}$

$\begin{array}{lll}3 & 6 & 10 \mid+0\end{array} \quad[\because$ Value of determinant with two identical columns is zero]

$=\left|\begin{array}{lll}1 & 0 & 0\end{array}\right|$

$\begin{array}{ccc}2 & 1 & 2 \\ 3 & 3 & 7\end{array}$ $\left[\right.$                 Applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}$ and $\left.\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C} 1\right]$

$=\left\{1 \times\left|\begin{array}{ll}1 & 2 \\ 3 & 7\end{array}\right|\right\} \quad\left[\right.$ Expanding along $\left.\mathrm{R}_{1}\right]$

$=7-6$

$=1$

$=\mathrm{RHS}$