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need solution for RD Sharma maths class 12 chapter 5 Determinants exercise Fill in the blanks question 11

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Answer: 8\left | A \right |

Hint: Here, we use basic concept of determinant of matrix \left | KA \right |=K^{n}\left | A \right |

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

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

                \left | -2A \right |=-2A is constant. So,K=(-2)

                Let’s put value of K in below, Formula,

                \begin{aligned} &|K A|=K^{n}|A| \\ &|-2 A|=(-2)^{n}|A| \end{aligned} \quad[K=-2]

                \begin{aligned} &|-2 A|=(-2)^{3}|A| \quad[n=3] \\ &|-2 A|=-8|A| \end{aligned}          

So, here8\left | A \right | is our answer.

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