Solve the following
Question:

$\left(\frac{-3}{2}\right)^{-1}$ is equal to

(a) $\frac{2}{3}$

(b) $-\frac{2}{3}$

(c) $\frac{3}{2}$

(d) none of these

Solution:

(b) $-\frac{2}{3}$

We have:

$\left(\frac{-3}{2}\right)^{-1}=\frac{1}{(-3) / 2} \quad—>\left(a^{-1}=1 / a\right)$

$=\frac{2}{-3}$