Question

Consider $A=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]$ and $B=\left[\begin{array}{cc}1& 1\\ 2& 1\end{array}\right]$ such that $AB=BA$ , then the value of $\frac{a_{12}}{a_{21}}+\frac{a_{11}}{a_{22}}$ is

• A. $2$
• B. $4$
• C. $\frac{3}{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

Answer: (C)

Given:
$AB=BA$
$\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\left[\begin{array}{cc}1& 1\\ 2& 1\end{array}\right]=\left[\begin{array}{cc}1& 1\\ 2& 1\end{array}\right]\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]$
$\left[\begin{array}{cc}{a}_{11}+2a_{12}& a_{11}+a_{12}\\ a_{21}+2a_{22}& a_{21}+a_{22}\end{array}\right]=\left[\begin{array}{cc}{a}_{11}+a_{21}& a_{12}+a_{22}\\ 2a_{11}+a_{21}& 2a_{12}+a_{22}\end{array}\right]$
$a_{11}+2a_{12}=a_{11}+a_{21}$ $\Rightarrow 2a_{12}=a_{21}$
$a_{11}+a_{12}=a_{12}+a_{22}\Rightarrow a_{11}=a_{22}$
$a_{21}+2a_{22}=2a_{11}+a_{21}\Rightarrow a_{11}=a_{22}$
$a_{21}+a_{22}=2a_{12}+a_{22}\Rightarrow a_{21}=2a_{12}$
$\frac{a_{12}}{a_{21}}=\frac{1}{2},\frac{a_{11}}{a_{22}}=1$