LINEAR ALGEBRA

Question

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)$\frac{11}{20}$

Comments

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

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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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a is the correct answer for this 7/20.

Answer 1

Simply..
adjoint (A) = [ 3   0.1
                       0    2]
And |A| =  6.

So, A^(-1) = 1/6  x  [3  0.1
                                0     2]    
                  =  [1/2   1/60
                         0      1/3]

So, a+b = 1/60 + 1/3 = 21/60 = 7/20.!

So, option (a.) is correct...!!