#### Explain Solution R D Sharma Class 12 Chapter deteminants Exercise 5.2  Question 2 Sub Question 16 maths Textbook Solution.

Answer:$\left|\begin{array}{ccc} \sqrt{23}+\sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15}+\sqrt{46} & 5 & \sqrt{10} \\ 3+\sqrt{115} & \sqrt{15} & 5 \end{array}\right|=0$

Hint: We will try to make all the elements of two row or  column zero

Given: $\left|\begin{array}{ccc} \sqrt{23}+\sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15}+\sqrt{46} & 5 & \sqrt{10} \\ 3+\sqrt{115} & \sqrt{15} & 5 \end{array}\right|$

Solution:$\left|\begin{array}{ccc} \sqrt{23}+\sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15}+\sqrt{46} & 5 & \sqrt{10} \\ 3+\sqrt{115} & \sqrt{15} & 5 \end{array}\right|$

$\text { On taking } \sqrt{5} \text { common from } \mathrm{C}_{2} \text { and } \mathrm{C}_{3}$

$=\sqrt{5} \times \sqrt{5}\left|\begin{array}{lll} \sqrt{23}+\sqrt{3} & 1 & 1 \\ \sqrt{15}+\sqrt{46} & \sqrt{5} & \sqrt{2} \\ 3+\sqrt{115} & \sqrt{3} & \sqrt{5} \end{array}\right|$

\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}-\sqrt{3} C_{2}-\sqrt{23} C_{3} \\ &=5\left|\begin{array}{ccc} \sqrt{23}+\sqrt{3}-\sqrt{3}-\sqrt{23} & 1 & 1 \\ \sqrt{15}+\sqrt{46}-\sqrt{15}-\sqrt{46} & \sqrt{5} & \sqrt{2} \\ 3+\sqrt{115}-3-\sqrt{115} & \sqrt{3} & \sqrt{5} \end{array}\right| \\ &=5\left|\begin{array}{ccc} 0 & 1 & 1 \\ 0 & \sqrt{5} & \sqrt{2} \\ 0 & \sqrt{3} & \sqrt{5} \end{array}\right| \end{aligned}

If all the elements of a row or column are zero

The value of the determinant is zero

$=5\times 0$

$=0$

Hence $\left|\begin{array}{ccc} \sqrt{23}+\sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15}+\sqrt{46} & 5 & \sqrt{10} \\ 3+\sqrt{115} & \sqrt{15} & 5 \end{array}\right|=0$