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#### Provide Solution For  R D Sharma Maths Class 12 Chapter  determinants Exercise 5.2 Question 26 Maths Textbook Solution.

Answer:$\left|\begin{array}{ccc} 1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q \end{array}\right|=1$

Hint we will convert element of $C_{1}$ into zero

Given:$\left|\begin{array}{ccc} 1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q \end{array}\right|=1$

Solution:

\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} 1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q \end{array}\right|\\ &\text { On applying } \mathrm{R}_{2} \rightarrow R_{2}-2 R_{1} \text { and } \mathrm{R}_{3} \rightarrow R_{3}-3 R_{1}\\ &=\left|\begin{array}{ccc} 1 & 1+p & 1+p+q \\ 0 & 1 & 2+p \\ 0 & 3 & 7+3 p \end{array}\right| \end{aligned}

\begin{aligned} &\text { On expanding w.r.t } \mathrm{C}_{1}\\ &=1\left|\begin{array}{cc} 1 & 2+p \\ 3 & 7+3 p \end{array}\right|-0\left|\begin{array}{cc} 1+p & 1+p+q \\ 3 & 7+3 p \end{array}\right|+0\left|\begin{array}{cc} 1+p & 1+p+q \\ 1 & 2+p \end{array}\right|\\ &=1\{1(7+3 p)-3(2+p)\}-0+0\\ &=7+3 p-6-3 p\\ &=1\\ &=R \cdot H \cdot S \end{aligned}

Hence it is proved that

$\left|\begin{array}{ccc} 1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q \end{array}\right|=1$