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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)$\frac{11}{20}$

in Linear Algebra by (441 points)
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+1

what is your doubt in this question ??

0

@Mk Utkarsh

can you edit the question,so everyone see easily?

0

@Lakshman Patel RJIT

Didn't you authorize to edit ?

0

a simple formula is A = $\begin{bmatrix} a & b\\ c & d \end{bmatrix}$ then A-1 = $\frac{1}{|A|}$$\begin{bmatrix} d & -b\\ -c & a \end{bmatrix}$

just substitute the values...

0

@Shaik Masthan 

I don't have the power to edit the question which is asked by anyone, otherwise, I will edit the question.

0
ok
0

we can do it also using A A-1 = I where A is an invertible matrix.

0

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$:a+b=\frac{1}{60}+\frac{1}{3}$

                         $\Rightarrow a+b=\frac{20+1}{60}$

                         $\Rightarrow a+b=\frac{21}{60}$

                          $\Rightarrow a+b=\frac{7}{20}$

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