#### Please Solve R.D. Sharma class 12 Chapter determinants Exercise 5.2 Question 1 Sub Question 2 Maths textbook Solution.

Answer: $\left|\begin{array}{lll} 67 & 19 & 21 \\ 39 & 13 & 14 \\ 81 & 24 & 26 \end{array}\right|=-43$

Hint: First we will use column operation to get a simple determinant.

Given: $\left|\begin{array}{lll} 67 & 19 & 21 \\ 39 & 13 & 14 \\ 81 & 24 & 26 \end{array}\right|$

Solution: $\left|\begin{array}{lll} 67 & 19 & 21 \\ 39 & 13 & 14 \\ 81 & 24 & 26 \end{array}\right|$

\begin{aligned} &\text { Applying } \mathrm{C}_{1} \rightarrow C_{1}-4 C_{3} \\ &=\left|\begin{array}{ccc} -17 & 19 & 21 \\ -17 & 13 & 14 \\ -23 & 24 & 26 \end{array}\right| \\ &\text { Applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \\ &=\left|\begin{array}{ccc} 0 & 6 & 7 \\ -17 & 13 & 14 \\ -23 & 24 & 26 \end{array}\right| \\ &\text { Applying } \mathrm{R}_{2} \rightarrow R_{2}-2 R_{1} \\ &\left|\begin{array}{ccc} 0 & 6 & 7 \\ -17 & 1 & 0 \\ -23 & 24 & 26 \end{array}\right| \end{aligned}

on expanding w.r.t row

\begin{aligned} &=0\left|\begin{array}{cc} 1 & 0 \\ 24 & 26 \end{array}\right|-6\left|\begin{array}{cc} -17 & 0 \\ -23 & 26 \end{array}\right|+7\left|\begin{array}{cc} -17 & 1 \\ -23 & 24 \end{array}\right| \\ &=0-6(-17 \times 26-0 \times-23)+7(-17 \times 24-1 \times-23) \\ &=-6(-442)+7(-408+23) \\ &=2652+7(-385) \\ &=2652-2695 \\ &=-43 \end{aligned}