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explain solution RD Sharma class 12 chapter 5 Determinants exercise Fill in the blanks question 28 maths

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Answer: 675

Hint: Here, we use basic concept of determinant of matrix |K A B|=K^{n}|A||B|

Given: A is 3\times 3 matrix. So, n = 3

            |A|=5,|B|=5

Solution: |3 A B|

                Here 3 is constant K. So,

                \begin{aligned} &|K A B|=K^{n}|A||B| \\ &|3 A B|=3^{n}|A||B| \quad(K=3) \\ &|3 A B|=3^{3} \times 5 \times 5[n=3,|A|=5,|B|=5] \\ &=27 \times 25 \\ &|3 A B|=675 \end{aligned}

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