#### Please Solve R. D.Sharma class 12 Chapter determinants Exercise 5.2 Question 1 sub question 4 Maths textbook Solution.

Answer:$\left|\begin{array}{ccc} 1 & -3 & 2 \\ 4 & -1 & 2 \\ 3 & 5 & 2 \end{array}\right|=40$

Hint: First we will use row operation to make some element 0

Given: $\left|\begin{array}{ccc} 1 & -3 & 2 \\ 4 & -1 & 2 \\ 3 & 5 & 2 \end{array}\right|$

Solution:

\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} -3 & -2 & 0 \\ 1 & -6 & 0 \\ 3 & 5 & 2 \end{array}\right| \end{aligned}

\begin{aligned} &\text { Now we will expand it w.r.t } R_{1}\\ &=-3\left|\begin{array}{cc} -6 & 0 \\ 5 & 2 \end{array}\right|-(-2)\left|\begin{array}{cc} 1 & 0 \\ 3 & 2 \end{array}\right|+0\left|\begin{array}{cc} 1 & -6 \\ 3 & 5 \end{array}\right|\\ &=-3(-12-5 \times 0)+2(1 \times 2-3 \times 0)+0\\ &=-3(-12)+2(2)\\ &=36+4\\ &=40 \end{aligned}

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