Let, A=[2−0.103]\mathrm{A}=\left[\begin{array}{cc}2 & -0.1 \\ 0 & 3\end{array}\right]A=[20−0.13] and A−1=[12a0 b]\mathrm{A}^{-1}=\left[\begin{array}{ll}\frac{1}{2} & \mathrm{a} \\ 0 & \mathrm{~b}\end{array}\right]A−1=[210a b] Then (a+b)=(a+b)=(a+b)=Choose one optionA720\frac{7}{20}207B320\frac{3}{20}203C1960\frac{19}{60}6019D1120\frac{11}{20}2011Submit50:50Try againView answerShowHideAnswer: AWorked explanation not available yet.Free practice · Your answers on this page are not saved.