#### Please Solve R. D. Sharma class 12 Chapter determinants Exercise 5.2 Question 1 sub question 6 Maths textbook Solution.

Answer:$\left|\begin{array}{ccc} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{array}\right|=0$

Hint: In this question we will convert some elements into 0

Given:$\left|\begin{array}{ccc} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{array}\right|$

Solution:

$\left|\begin{array}{ccc} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{array}\right|$

\begin{aligned} &\text { Applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \text { and } \mathrm{R}_{2} \rightarrow R_{2}-R_{3} \\ &=\left|\begin{array}{ccc} 4 & -2 & 0 \\ 12 & -6 & 0 \\ -10 & 5 & 2 \end{array}\right| \end{aligned}

On expanding it w.r.t $R_{1}$

\begin{aligned} &=4\left|\begin{array}{ll} -6 & 0 \\ 5 & 2 \end{array}\right|-(-2)\left|\begin{array}{cc} 12 & 0 \\ -10 & 2 \end{array}\right|+0\left|\begin{array}{cc} 12 & -6 \\ -10 & 5 \end{array}\right| \\ &=4(-6 \times 2-5 \times 0)+2(12 \times 2-(-10) \times 0)+0 \\ &=4(-12)+2(24) \\ &=-48+48 \\ &=0 \end{aligned}

Hence$\left|\begin{array}{ccc} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{array}\right|=0$