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  • provide solution for RD Sharma maths class 12 chapter 5 Determinants exercise Fill in the blanks question 34

provide solution for RD Sharma maths class 12 chapter 5 Determinants exercise  Fill in the blanks question 34

 

Answers (1)

Answer: 0

Hint: Here, we use basic concept of determinant of matrix

Given: D=\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: Let’s use separation property of determinant

                \begin{aligned} &{\left[\begin{array}{ccc} \sqrt{23} & \sqrt{5} & \sqrt{5} \\ \sqrt{15} & 5 & \sqrt{10} \\ 3 & \sqrt{15} & 5 \end{array}\right]+\left[\begin{array}{ccc} \sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15} & 5 & \sqrt{10} \\ 3 & \sqrt{15} & 5 \end{array}\right]} \\ &{\left[\begin{array}{lll} \sqrt{23} & \sqrt{5} & \sqrt{5} \\ \sqrt{2 \times 23} & \sqrt{5 \times 5} & \sqrt{2 \times 5} \\ \sqrt{5 \times 23} & \sqrt{5 \times 3} & \sqrt{5 \times 5} \end{array}\right]+\left[\begin{array}{ccc} \sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{3 \times 5} & \sqrt{5 \times 5} & \sqrt{2 \times 5} \\ \sqrt{3 \times 3} & \sqrt{5 \times 3} & \sqrt{5 \times 5} \end{array}\right]} \end{aligned}

                \begin{aligned} C_{1} & \rightarrow \frac{C_{1}}{\sqrt{3}} \\ C_{2} & \rightarrow \frac{C_{2}}{\sqrt{5}} \text { and } C_{3} \rightarrow \frac{C_{3}}{\sqrt{5}} \end{aligned}

                \sqrt{23} \sqrt{5} \sqrt{5}\left[\begin{array}{ccc} 1 & 1 & 1 \\ \sqrt{2} & \sqrt{5} & \sqrt{2} \\ \sqrt{5} & \sqrt{3} & \sqrt{5} \end{array}\right]+\sqrt{3} \sqrt{5} \sqrt{5}\left[\begin{array}{ccc} 1 & 1 & 1 \\ \sqrt{5} & \sqrt{5} & \sqrt{2} \\ \sqrt{3} & \sqrt{3} & \sqrt{5} \end{array}\right]

                C_{1} \rightarrow C_{1}-C_{3} \text { and } C_{1} \rightarrow C_{1}-C_{2}

                5 \sqrt{23}\left[\begin{array}{ccc} 0 & 1 & 1 \\ 0 & \sqrt{5} & \sqrt{2} \\ 0 & \sqrt{3} & \sqrt{5} \end{array}\right]+5 \sqrt{23}\left[\begin{array}{ccc} 0 & 1 & 1 \\ 0 & \sqrt{5} & \sqrt{2} \\ 0 & \sqrt{3} & \sqrt{5} \end{array}\right]

Here, both column has 0 value of each column

So, determinant value is zero.

Posted by

infoexpert26

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