1 votes 1 votes IF A=$\begin{bmatrix} 2& -0.1\\ 0 & 3 \end{bmatrix}$ AND A-1=$\begin{bmatrix} \frac{1}{2}& a\\ 0 & b \end{bmatrix}$ then a+b ? a)$\frac{7}{20}$ b)$\frac{3}{20}$ c)$\frac{19}{60}$ d)$\frac{11}{20}$ Linear Algebra matrix + – altamash asked Oct 27, 2018 • recategorized Oct 27, 2018 by Shaik Masthan altamash 576 views answer comment Share Follow See all 9 Comments See all 9 9 Comments reply Show 6 previous comments lakshaysaini2013 commented Oct 31, 2018 reply Follow Share we can do it also using A A-1 = I where A is an invertible matrix. 0 votes 0 votes Lakshman Bhaiya commented Oct 31, 2018 reply Follow Share 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$:a+b=\frac{1}{60}+\frac{1}{3}$ $\Rightarrow a+b=\frac{20+1}{60}$ $\Rightarrow a+b=\frac{21}{60}$ $\Rightarrow a+b=\frac{7}{20}$ 0 votes 0 votes eshita1997 commented Dec 26, 2020 i moved by Lakshman Bhaiya Dec 26, 2020 reply Follow Share a is the correct answer for this 7/20. 0 votes 0 votes Please log in or register to add a comment.