JEE MAIN - Mathematics (2006 - No. 8)
Let $$A = \left( {\matrix{
1 & 2 \cr
3 & 4 \cr
} } \right)$$ and $$B = \left( {\matrix{
a & 0 \cr
0 & b \cr
} } \right),a,b \in N.$$ Then
there cannot exist any $$B$$ such that $$AB=BA$$
there exist more then one but finite number of $$B'$$s such that $$AB=BA$$
there exists exactly one $$B$$ such that $$AB=BA$$
there exist infinitely many $$B'$$s such that $$AB=BA$$
Explanation
$$A = \left[ {\matrix{
1 & 2 \cr
3 & 4 \cr
} } \right]\,\,\,\,B = \left[ {\matrix{
a & 0 \cr
0 & b \cr
} } \right]$$
$$AB = \left[ {\matrix{ a & {2b} \cr {3a} & {4b} \cr } } \right]$$
$$BA = \left[ {\matrix{ a & 0 \cr 0 & b \cr } } \right]\left[ {\matrix{ 1 & 2 \cr 3 & 4 \cr } } \right] = \left[ {\matrix{ a & {2a} \cr {3b} & {4b} \cr } } \right]$$
Hence, $$AB=BA$$ only when $$a=b$$
$$\therefore$$ There can be infinitely many $$B's$$
for which $$AB=BA$$
$$AB = \left[ {\matrix{ a & {2b} \cr {3a} & {4b} \cr } } \right]$$
$$BA = \left[ {\matrix{ a & 0 \cr 0 & b \cr } } \right]\left[ {\matrix{ 1 & 2 \cr 3 & 4 \cr } } \right] = \left[ {\matrix{ a & {2a} \cr {3b} & {4b} \cr } } \right]$$
Hence, $$AB=BA$$ only when $$a=b$$
$$\therefore$$ There can be infinitely many $$B's$$
for which $$AB=BA$$
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