Yes, we can use $AA^{-1}=I,$-----$>(1)$ where $I$ is Identity Matrix.
$A=\begin{bmatrix} 2 &-0.1 \\ 0 &3 \end{bmatrix}_{2\times2},A^{-1}=\begin{bmatrix} \frac{1}{2} &a \\ 0 &b \end{bmatrix}_{2\times2}$ and $I=\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}_{2\times2}$
Now ,we can put the value in equation $(1),$
$\begin{bmatrix} 2 &-0.1 \\ 0 &3 \end{bmatrix}_{2\times2}.\begin{bmatrix} \frac{1}{2} &a \\ 0 &b \end{bmatrix}_{2\times2}=\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}_{2\times2}$
$\begin{bmatrix} 1 &2a-0.1b \\ 0 &3b \end{bmatrix}_{2\times2}=\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}_{2\times2}$
Now$,3b=1$
$\Rightarrow b=\frac{1}{3}$
and $2a-0.1b=0$
$\Rightarrow 2a=0.1b$
$\Rightarrow 2a=0.1(\frac{1}{3})$
$\Rightarrow a=0.1(\frac{1}{6})$
$\Rightarrow a=(\frac{1}{60})$ $[0.1=\frac{1}{10}$]
Now,we can find$:a+b=\frac{1}{60}+\frac{1}{3}$
$\Rightarrow a+b=\frac{20+1}{60}$
$\Rightarrow a+b=\frac{21}{60}$
$\Rightarrow a+b=\frac{7}{20}$