Question
Let \[A = \left[ {\begin{array}{*{20}{c}}
1&2\\
3&4
\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{c}}
a&0\\
0&b
\end{array}} \right]\] where $$a, b$$ are natural numbers, then which one of the following is correct ?
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 $$B’s$$ such that $$AB = BA$$
D.
There cannot exist any $$B$$ such that $$AB = BA$$
Answer :
There exist infinitely many $$B’s$$ such that $$AB = BA$$
Solution :
\[\begin{array}{l}
AB = \left[ {\begin{array}{*{20}{c}}
1&2\\
3&4
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
a&0\\
0&b
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
a&{2b}\\
{3a}&{4b}
\end{array}} \right]\\
{\rm{and, }}\,\,BA = \left[ {\begin{array}{*{20}{c}}
a&0\\
0&b
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&2\\
3&4
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
a&{2a}\\
{3b}&{4b}
\end{array}} \right]\\
{\rm{If, }}\,\,AB = BA\\
\Rightarrow \left[ {\begin{array}{*{20}{c}}
a&{2b}\\
{3a}&{4b}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
a&{2a}\\
{3b}&{4b}
\end{array}} \right]\\
\Rightarrow a = b
\end{array}\]
From the above it is clear that there exist infinitely many $$B's$$ such that $$AB = BA.$$