Question

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

• A. $\begin{bmatrix} 1 & 4 \\ 6 & 7 \end{bmatrix}$
• B. $\begin{bmatrix} 1 & 4 \\ 7 & 6 \end{bmatrix}$
• C. $\begin{bmatrix} 2 & 2 \\ 2 & 4 \end{bmatrix}$
• D. $\begin{bmatrix} 3 & 4 \\ 4 & 9 \end{bmatrix}$

Answer 1

Answer: (A)

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