Invert: $\sum\limits_{d|n} \mu(d) \lambda(d)=2^{\omega(n)}$ 
Inverting $\displaystyle\sum_{d|n} \mu(d) \lambda(d)=2^{\omega(n)}$ into $\displaystyle\sum_{d|n} \lambda(n/d) 2^{\omega(d)}=1$ ,where $n \geq1$, by using Mobius Inversion Formula.

I'm able to solve the latter without Inversion, and in problem too it's not necessary to use inversion, but I'm fascinated to know how to do that, because $2^{\omega(n)}$ comes to L.H.S side from R.H.S and becomes $2^{\omega(d)}$, also $\lambda(d)$ is converted into $\lambda(n/d)$  Please help. and if it's not possible then comment, I'll remove this problem from MSE.
 A: (You should specify that $\lambda$ refers to the Liouville lambda function and not the Carmichael lambda function, which is also quite common.)
One way to do this is to use the total multiplicativity of $\lambda$ (and the convenient fact that $\lambda = 1/\lambda$) to write $\lambda(n/d)$ as both the product and the quotient of $\lambda(n)$ and $\lambda(d)$.
We rewrite the first sum as $$\lambda(n) \sum_{d\mid n} \mu(d) \lambda(n/d),$$
or in functional notation $\lambda \cdot (\mu \star \lambda)$.  Here I'm using $\cdot$ to mean pointwise multiplication and $\star$ to mean Dirichlet convolution.
Since this is equal to $2^{\omega(n)}$ by the first equality, we multiply/divide both sides by $\lambda$ to get $\mu \star \lambda = \lambda \cdot 2^{\omega(n)}$, and Möbius inversion then gives $\lambda = 1 \star (\lambda \cdot 2^{\omega(n)})$.  Writing this back in summation form gives
$$\sum_{d\mid n} \lambda(d) 2^{\omega(d)} = \lambda(n),$$
and dividing/multiplying through by $\lambda(n)$ yields the desired equality.
