#### Explain Solution R.D. Sharma Class 12 Chapter  deteminants Exercise 5.2 Question 21 maths Textbook Solution.

Answer:$\left|\begin{array}{lll} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right|=-2$

Hint: $\text { First we will make two elements of } \mathrm{C}_{3} \text { zero. then we will expand it }$

Given:$\left|\begin{array}{lll} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right|=-2$

Solution:

$\text { L.H.S }\left|\begin{array}{lll} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right|$

\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \text { and } \mathrm{R}_{2} \rightarrow R_{2}-R_{3} \\ &=\left|\begin{array}{ccc} (a+1)(a+2)-(a+2)(a+3) & (a+2)-(a+3) & 0 \\ (a+2)(a+3)-(a+3)(a+4) & (a+3)-(a+4) & 0 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right| \\ &=\left|\begin{array}{ccc} (a+2)(a+1-a-3) & a+2-a-3 & 0 \\ (a+3)(a+2-a-4) & a+3-a-4 & 0 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right| \\ &=\left|\begin{array}{ccc} (a+2)(-2) & -1 & 0 \\ (a+3)(-2) & -1 & 0 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right| \end{aligned}

\begin{aligned} &\text { On expanding w.r.t } \mathrm{C}_{3}\\ &=0\left|\begin{array}{cc} (a+3)(-2) & -1 \\ (a+3)(a+4) & a+4 \end{array}\right|-0\left|\begin{array}{cc} (a+2)(-2) & -1 \\ (a+3)(a+4) & a+4 \end{array}\right|+1 \mid\left(\begin{array}{ll} a+2)(-2) & -1 \\ (a+3)(-2) & -1 \end{array} \mid\right.\\ &=0-0+1\{(a+2)(-2)(-1)-(-1)(-2)(a+3)\}\\ &=1\{2(a+2)-2(a+3)\}\\ &=2 a+4-2 a-6\\ &=-2\\ &=R \cdot H \cdot S \end{aligned}

Hence it is proved that

$\left|\begin{array}{lll} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right|=-2$