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  • Explain Solution R.D. Sharma Class 12 Chapter deteminants Exercise 5.2 Question 48 maths Textbook Solution.

Explain Solution R.D. Sharma Class 12 Chapter  deteminants Exercise 5.2 Question 48 maths Textbook Solution.

Answers (1)

Answer: 0

Hint \alpha, \beta, \gamma \text { are in AP }

Given:\left|\begin{array}{lll} x-3 & x-4 & x-\alpha \\ x-2 & x-3 & x-\beta \\ x-1 & x-2 & x-\gamma \end{array}\right|=0 \text { where } \alpha, \beta, \gamma \text { are in } \mathrm{AP}

Solution:

$$ \begin{aligned} &\alpha, \beta, \gamma \text { are in } \mathrm{AP}\\ &a_{2}-a_{1}=a_{3}-a_{1}\\ &\beta-\gamma=\gamma-\beta\\ &\beta+\beta=\gamma+\alpha\\ &2 \beta=\gamma+\alpha\\ &\gamma+\alpha-2 \beta=0 \end{aligned}                                ...(1)

$$ \begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} x-3 & x-4 & x-\alpha \\ x-2 & x-3 & x-\beta \\ x-1 & x-2 & x-\gamma \end{array}\right| \\ &\text { Now } R_{1} \rightarrow R_{1}+R_{3}-2 R_{2} \\ &=\left|\begin{array}{ccc} 0 & 0 & 2 \beta-(\alpha+\gamma) \\ x-2 & x-3 & x-\beta \\ x-1 & x-2 & x-\gamma \end{array}\right| \end{aligned}

$$ \begin{aligned} &=\left|\begin{array}{ccc} 0 & 0 & 2 \beta-2 \beta \\ x-2 & x-3 & x-\beta \\ x-1 & x-2 & x-\gamma \end{array}\right| \ldots \ldots \ldots \text { from(1) } \\ &=\left|\begin{array}{ccc} 0 & 0 & 0 \\ x-2 & x-3 & x-\beta \\ x-1 & x-2 & x-\gamma \end{array}\right| \end{aligned}

Expanding along R_{1},

=0 ..hence proved

 

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