Question

Find the reciprocal of $\dfrac{{{\left( \dfrac{-3}{2} \right)}^{-1}}\times {{\left( \dfrac{2}{3} \right)}^{2}}}{{{\left( \dfrac{2}{3} \right)}^{-1}}\div {{\left( \dfrac{3}{2} \right)}^{3}}}$ .

Answer 1

Hint: Reciprocal of a number or an expression is nothing but the multiplicative inverse of it. In basic terms , the number we should multiply to another number to make it equal to 1. For example we have to $2$ . To find it’s reciprocal we simply have to find what number we should multiply to it to make it equal to one.$2\times \dfrac{1}{2}=1.$ So $\dfrac{1}{2}$ is the reciprocal or multiplicative inverse of it. Before finding the reciprocal or the multiplicative inverse, let us try simplifying it.

Complete step-by-step solution:
Now let us solve the expression.
We have $\dfrac{{{\left( \dfrac{-3}{2} \right)}^{-1}}\times {{\left( \dfrac{2}{3} \right)}^{2}}}{{{\left( \dfrac{2}{3} \right)}^{-1}}\div {{\left( \dfrac{3}{2} \right)}^{3}}}$.
We know that $a\div b=\dfrac{a}{b.}$.
We know that ${{\left( x \right)}^{-1}}=\dfrac{1}{x}$ .
Upon solving we get the following :
\[\begin{align}
  & \Rightarrow \dfrac{{{\left( \dfrac{-3}{2} \right)}^{-1}}\times {{\left( \dfrac{2}{3} \right)}^{2}}}{{{\left( \dfrac{2}{3} \right)}^{-1}}\div {{\left( \dfrac{3}{2} \right)}^{3}}} \\
 & \Rightarrow \dfrac{\dfrac{1}{{{\left( \dfrac{-3}{2} \right)}^{1}}}\times {{\left( \dfrac{2}{3} \right)}^{2}}}{{{\left( \dfrac{2}{3} \right)}^{-1}}\div {{\left( \dfrac{3}{2} \right)}^{3}}} \\
 & \Rightarrow \dfrac{\dfrac{1}{{{\left( \dfrac{-3}{2} \right)}^{1}}}\times {{\left( \dfrac{2}{3} \right)}^{2}}}{\dfrac{\left( \dfrac{3}{2} \right)}{{{\left( \dfrac{3}{2} \right)}^{3}}}} \\
\end{align}\]
Let us solve further.
Upon solving , we get :
\[\begin{align}
  & \Rightarrow \dfrac{{{\left( \dfrac{-3}{2} \right)}^{-1}}\times {{\left( \dfrac{2}{3} \right)}^{2}}}{{{\left( \dfrac{2}{3} \right)}^{-1}}\div {{\left( \dfrac{3}{2} \right)}^{3}}} \\
 & \Rightarrow \dfrac{\dfrac{1}{{{\left( \dfrac{-3}{2} \right)}^{1}}}\times {{\left( \dfrac{2}{3} \right)}^{2}}}{{{\left( \dfrac{2}{3} \right)}^{-1}}\div {{\left( \dfrac{3}{2} \right)}^{3}}} \\
 & \Rightarrow \dfrac{\dfrac{1}{{{\left( \dfrac{-3}{2} \right)}^{1}}}\times {{\left( \dfrac{2}{3} \right)}^{2}}}{\dfrac{1}{\left( \dfrac{2}{3} \right)}\div {{\left( \dfrac{3}{2} \right)}^{3}}} \\
 & \Rightarrow \dfrac{\left( \dfrac{2}{-3} \right)\times {{\left( \dfrac{2}{3} \right)}^{2}}}{\left( \dfrac{3}{2} \right)\div {{\left( \dfrac{3}{2} \right)}^{3}}} \\
 & \Rightarrow \dfrac{\left( \dfrac{2}{-3} \right)\times {{\left( \dfrac{2}{3} \right)}^{2}}}{\dfrac{\left( \dfrac{3}{2} \right)}{{{\left( \dfrac{3}{2} \right)}^{3}}}} \\
\end{align}\]
On solving further, we get
\[\begin{align}
  & \Rightarrow \dfrac{\left( \dfrac{2}{-3} \right)\times {{\left( \dfrac{2}{3} \right)}^{2}}}{\dfrac{\left( \dfrac{3}{2} \right)}{{{\left( \dfrac{3}{2} \right)}^{3}}}} \\
 & \Rightarrow \dfrac{\left( \dfrac{2}{-3} \right)\times {{\left( \dfrac{2}{3} \right)}^{2}}\times {{\left( \dfrac{3}{2} \right)}^{3}}}{\left( \dfrac{3}{2} \right)} \\
 & \Rightarrow \left( \dfrac{2}{-3} \right)\times {{\left( \dfrac{2}{3} \right)}^{2}}\times {{\left( \dfrac{3}{2} \right)}^{2}} \\
 & \Rightarrow \left( \dfrac{2}{-3} \right) \\
\end{align}\]
After solving the entire expression $\dfrac{{{\left( \dfrac{-3}{2} \right)}^{-1}}\times {{\left( \dfrac{2}{3} \right)}^{2}}}{{{\left( \dfrac{2}{3} \right)}^{-1}}\div {{\left( \dfrac{3}{2} \right)}^{3}}}$we get $\left( \dfrac{-2}{3} \right)$ .
Now let us find the reciprocal. Let us assume that number that we have to multiply to $\left( \dfrac{-2}{3} \right)$ to make it equal to $1$ is $x.$
So $\left( \dfrac{-2}{3} \right)\times x=1$ .
Let us solve this.
$\Rightarrow \left( \dfrac{-2}{3} \right)\times x=1$ .
Let us take $\left( \dfrac{-2}{3} \right)$ to the right hand side of the equation.
Upon doing so, we get the following :
$\begin{align}
  & \Rightarrow \left( \dfrac{-2}{3} \right)\times x=1 \\
 & \Rightarrow x=\left( \dfrac{3}{-2} \right) \\
\end{align}$
So the reciprocal is $\left( \dfrac{3}{-2} \right)$ .
$\therefore $ Hence, the reciprocal of $\dfrac{{{\left( \dfrac{-3}{2} \right)}^{-1}}\times {{\left( \dfrac{2}{3} \right)}^{2}}}{{{\left( \dfrac{2}{3} \right)}^{-1}}\div {{\left( \dfrac{3}{2} \right)}^{3}}}$ is $\left( \dfrac{3}{-2} \right)$.

Note: We have to be careful when we get a huge expression to find the reciprocal. We have to be careful to simplify it and then find its reciprocal. Or else we can first find the reciprocal of the expression we are given and then simplify it .We can do it both the ways as would give us the same answer. If the expression is not huge , then it is easy. Just be careful while reading the question. Likewise, we also have additive inverse. In multiplicative inverse or inverse we are finding the other number which will be the number equal to $1$ after multiplying. But in the additive inverse we find another number which makes our number equal to $0$.