A) \[{{A}^{2}}=A\]
B) \[AB=0\]
C) \[BA=0\]
D) All of the above
Correct Answer: D
Solution :
It is given that \[B\] is an idempotent matrix. \[\therefore \] \[{{B}^{2}}=B\] Also, \[A=I-B\] On squaring both sides, we get \[{{A}^{2}}={{(I-B)}^{2}}\] \[{{A}^{2}}=(I-B)(I-B)\] \[=I-IB-BI+{{B}^{2}}\] \[=I-2B+{{B}^{2}}\] \[=I-2B+B\] \[=I-B\] \[=A\] \[\Rightarrow \] \[{{A}^{2}}=A\] Again, \[AB=(I-B)B\] \[=IB-{{B}^{2}}\] \[=IB-B\] \[=B-B\] \[(\because IB=B\]and\[{{B}^{2}}=B)\] \[=0\] and \[BA=B(I-B)\] \[=BI-{{B}^{2}}\] \[=B-B\] \[(\because BI=B\]and\[{{B}^{2}}=B)\] \[=0\]
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