Question

Let $A=\left[\begin{array}{cc}2& -0.1\\ 0& 3\end{array}\right]$ and $A^{-1}=\left[\begin{array}{cc}\frac{1}{2}& a\\ 0& b\end{array}\right]$. Then $(a+b)=$

• A. $\frac{7}{20}$
• B. $\frac{3}{20}$
• C. $\frac{19}{60}$
• D. $\frac{11}{20}$

Answer 1

The correct option is A $\frac{7}{20}$

$A=\left[\begin{array}{cc}2& -0.1\\ 0& 3\end{array}\right]$,

$A^{-1}=\left[\begin{array}{cc}\frac{1}{2}& a\\ 0& b\end{array}\right]$

$A{A}^{-1}=I$

$\Rightarrow$ $\left[\begin{array}{cc}2& -0.1\\ 0& 3\end{array}\right]$ $\left[\begin{array}{cc}\frac{1}{2}& a\\ 0& b\end{array}\right]$ = $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}1& 2a-\left(0.1\right)b\\ 0& 3b\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow 2a-\left(0.1\right)b=0$ & $3b=1$

$\text{So},b=\frac{1}{3}$ & $a=\frac{1}{60}$

$\text{Hence},a+b=\frac{1}{60}+\frac{1}{3}=\frac{21}{60}=\frac{7}{20}$