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

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

Answers (1)

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

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