Question

Let $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$ and $B=\left[\begin{array}{cc}a& 0\\ 0& b\end{array}\right],a,b\in N.$ Then

• A. There does not exist any $B$ such that $AB=BA$
• B. There exists more than one but finite number of $B^{\prime }s$ such that $AB=BA$
• C. There exists exactly one $B$ such that $AB=BA.$
• D. There exists infinitely many $B^{\prime }s$ such that $AB=BA$

Answer 1

The correct option is D There exists infinitely many $B^{\prime }s$ such that $AB=BA$

$A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$ and $B=\left[\begin{array}{cc}a& 0\\ 0& b\end{array}\right]\therefore AB=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]\left[\begin{array}{cc}a& 0\\ 0& b\end{array}\right]=\left[\begin{array}{cc}a& 2b\\ 3a& 4b\end{array}\right]\text{and}$
$BA=\left[\begin{array}{cc}a& 0\\ 0& b\end{array}\right]\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]=\left[\begin{array}{cc}a& 2a\\ 3b& 4b\end{array}\right]$
$\therefore AB=BA$ only when $a=b.$
Hence there exists infinitely many $B^{\prime }s$ such that $AB=BA$