#### explain solution RD Sharma class 12 chapter 5 Determinants exercise Fill in the blanks question 20 maths

Hint: Here, we use basic concept of determinant

$|K A B|=K^{n}|A||B|$

Given:  $n = 3$

$\left | A \right |=-2$

$\left | B \right |=4$

Solution: A and B are $3\times 3$ order of matrix

So, n = 3

Here, $\left | 2AB \right |\; \; \; \; \; \; \; So,K=2$

So, now

$|K A B|=K^{n}|A||B|$

$|2 A B|=2^{3} \times-2 \times 4 \; \; \; \; \; \; \; \; \; \quad[n=3,|A|=-2,|B|=4]$

\begin{aligned} &|2 A B|=8 \times-2 \times 4 \\ &|2 A B|=8 \times-8 \\ &|2 A B|=-64 \end{aligned}