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#### Provide Solution For  R. D. Sharma Maths Class 12 Chapter determinants Exercise 5.2 Question 2 Sub Question 8 Maths Textbook Solution.

Answer: $\left|\begin{array}{ccc} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{array}\right|=0$

Hint: We will try to do any two column or row equal

Given: $\left|\begin{array}{ccc} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{array}\right|$

Solution: $\left|\begin{array}{ccc} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{array}\right|$

\begin{aligned} &\text { On multiplying } C_{1}, C_{2} \text { and } C_{3} \text { by } z, y \text { and } x \text { respectively }\\ &=\frac{1}{x y z}\left|\begin{array}{ccc} 0 & x y & y x \\ -x z & 0 & z x \\ -y z & -z y & 0 \end{array}\right| \end{aligned}

\begin{aligned} &\text { Take } y, x \text { and } z \text { common from } \mathrm{R}_{1}, R_{2} \text { and } \mathrm{R}_{3} \text { respectively }\\ &=\frac{1}{x y z}(y z x)\left|\begin{array}{ccc} 0 & x & x \\ -z & 0 & z \\ -y & -y & 0 \end{array}\right|\\ &=\left|\begin{array}{ccc} 0 & x & x \\ -z & 0 & z \\ -y & -y & 0 \end{array}\right|\\ &\text { On applying } C_{2} \rightarrow C_{2}-C_{3}\\ &=\left|\begin{array}{ccc} 0 & 0 & x \\ -z & -z & z \\ -y & -y & 0 \end{array}\right| \end{aligned}

If any two rows or columns of a determinant are identical.

The value of the determinant is zero

$=0 \quad\left(\because C_{1}=C_{2}\right)$

Hence $\left|\begin{array}{ccc} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{array}\right|=0$

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