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

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, here$8\left | A \right |$ is our answer.