Question

If $A$ is a square matrix of order $2\times 2$ and $B=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right],$ such that $AB=BA,$ then $A$ can be

• A. $\left[\begin{array}{cc}1& 4\\ 6& 7\end{array}\right]$
• B. $\left[\begin{array}{cc}1& 4\\ 7& 6\end{array}\right]$
• C. $\left[\begin{array}{cc}2& 2\\ 2& 4\end{array}\right]$
• D. $\left[\begin{array}{cc}3& 4\\ 4& 9\end{array}\right]$

Answer 1

Answer: (A)

Let, $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ then $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
$\Rightarrow a+3b=a+2c,2a+4b=b+2d$
$c+3d=3a+4c,2c+4d=3b+4d$
$\Rightarrow 3b=2c,a+c=d,$
Hence, $A=\left[\begin{array}{cc}a& b\\ \frac{3b}{2}& a+\frac{3b}{2}\end{array}\right]$ where $a,b\in R$
Now, we see only first option is correct