Answer
$$A^{-1}=\left[ \begin {array}{cccccc} 0&- 2& 0.8\\ - 10& 4.0& 4.4 \\ 10&- 2&- 3.2\end {array} \right] .$$
Work Step by Step
To find $A^{-1}$, we have $$\left[ A \ \ I \right]= \left[ \begin {array}{cccccc} 0.1& 0.2& 0.3&1&0&0 \\ - 0.3& 0.2& 0.2&0&1&0\\ 0.5& 0.5& 0.5&0&0&1\end {array} \right] . $$ Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows $$\left[I \ \ A^{-1} \right]= \left[ \begin {array}{cccccc} 1& 0& 0& 0&- 2& 0.8\\ 0& 1& 0&- 10& 4.0& 4.4 \\ 0& 0& 1& 10&- 2&- 3.2\end {array} \right] . $$ Then $A^{-1}$ is given by $$A^{-1}=\left[ \begin {array}{cccccc} 0&- 2& 0.8\\ - 10& 4.0& 4.4 \\ 10&- 2&- 3.2\end {array} \right] .$$
