• # question_answer Let$A=\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \\ \end{matrix} \right]$and$B=\left[ \begin{matrix} a & 0 \\ 0 & b \\ \end{matrix} \right],a,b,\in N$. Then,     AIEEE  Solved  Paper-2006 A) there exist more than one but finite number of B's such that$AB=BA$ B) there exists exactly one B such that$AB=BA$ C) there exist infinitely many 8's such that$AB=BA$ D) there cannot exist any B such that$AB=BA$

Since, $A=\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \\ \end{matrix} \right]$and$B=\left[ \begin{matrix} a & 0 \\ 0 & b \\ \end{matrix} \right]$ Now, $AB=\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} a & 0 \\ 0 & b \\ \end{matrix} \right]=\left[ \begin{matrix} a & 2b \\ 3a & 4b \\ \end{matrix} \right]$ and$BA=\left[ \begin{matrix} a & 0 \\ 0 & b \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \\ \end{matrix} \right]=\left[ \begin{matrix} a & 2a \\ 3b & 4b \\ \end{matrix} \right]$ If  $AB=BA\Rightarrow a=b$ Hence, $AB=BA$ is possible for infinitely many $B's$.