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Explain Solution R.D. Sharma Class 12 Chapter  deteminants Exercise 5.2 Question 52 Sub Question 7 maths Textbook Solution.

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Answer: x=4

Hint: Use determinant formula

Given: \left|\begin{array}{ccc} 15-2 x & 11-3 x & 7-x \\ 11 & 17 & 14 \\ 10 & 16 & 13 \end{array}\right|=0

Solution:

\begin{aligned} &\text { Apply } R_{3} \rightarrow R_{3}-R_{2} \\ &\begin{array}{ccc} 15-2 x & 11-3 x & 7-x \\ 11 & 17 & 14 \\ 1 & 1 & 1 \end{array} \mid=0 \\ &\text { Now } \mathrm{C}_{1} \rightarrow C_{1}-C_{2} \& C_{2} \rightarrow C_{2}-C_{3} \\ &\left|\begin{array}{ccc} 4+x & 4-2 x & 7-x \\ -6 & 3 & 14 \\ 0 & 0 & 1 \end{array}\right|=0 \end{aligned}

\begin{aligned} &\text { Expand from } \mathrm{R}_{3}\\ &0\left|\begin{array}{cc} 4+x & 4-2 x \\ -6 & 3 \end{array}\right|-0\left|\begin{array}{cc} 4+x & 7-x \\ -6 & 14 \end{array}\right|+1\left|\begin{array}{cc} 4+x & 4-2 x \\ -6 & 3 \end{array}\right|=0\\ &1(3(4+x)-(-6)(4-2 x))=0\\ &12+3 x+24-12 x=0\\ &36=9 x\\ &x=4 \end{aligned}

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