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  • Explain Solution R. D. Sharma Class 12 Chapter deteminants Exercise 5.2 Question 42 maths Textbook Solution.

Explain Solution R. D. Sharma Class 12 Chapter  deteminants Exercise 5.2 Question 42  maths Textbook Solution.

Answers (1)

Answer: (x+y+z)^{3}

Hint Use determinant formula

Given:\left|\begin{array}{ccc} 2 y & y-z-x & 2 y \\ 2 z & 2 z & z-x-y \\ x-y-z & 2 x & 2 x \end{array}\right|=(x+y+z)^{3}

\text { L.H.S }\left|\begin{array}{ccc} 2 y & y-z-x & 2 y \\ 2 z & 2 z & z-x-y \\ x-y-z & 2 x & 2 x \end{array}\right|

Solution:

\begin{aligned} &\text { Apply } \mathrm{R}_{1} \rightarrow R_{1}+R_{2}+R_{3} \\ &=\left|\begin{array}{ccc} x+y+z & x+y+z & x+y+z \\ 2 z & 2 z & z-x-y \\ x-y-z & 2 x & 2 x \end{array}\right| \\ &(x+y+z) \text { common from } \mathrm{C}_{1} \\ &=(x+y+z)\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 z & 2 z & z-x-y \\ x-y-z & 2 x & 2 x \end{array}\right| \end{aligned}

\begin{aligned} &\text { Apply } C_{2} \rightarrow C_{2}-C_{1} \& \mathrm{C}_{3} \rightarrow C_{3}-C_{1} \\ &=(x+y+z)\left|\begin{array}{ccc} 1 & 0 & 1 \\ 2 z & 0 & -(x+y+z) \\ x-y-z & x+y+z \quad & x+y+z \end{array}\right| \\ &(x+y+z) \text { common from } \mathrm{C}_{2} \& C_{3} \\ &=(x+y+z)^{3}\left|\begin{array}{ccc} 1 & 0 & 1 \\ 2 z & 0 & -1 \\ x-y-z & 1 & 1 \end{array}\right| \end{aligned}

Expanding w.r.t C_{3}

=(x+y+z)^{3}\left[1\left|\begin{array}{cc} 0 & -1 \\ 1 & 1 \end{array}\right|-0(1+0) 1\right]

\begin{aligned} &=(x+y+z)^{3}\left|\begin{array}{cc} 0 & -1 \\ 1 & 1 \end{array}\right| \\ &=(x+y+z)^{3}(0(1)-(-1) 1) \\ &=(x+y+z)^{3} \\ &=R \cdot H . S \end{aligned}

 

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