A) A
B) 2A
C) -A
D) \[l-A\]
Correct Answer: A
Solution :
We have, \[{{A}^{2}}=A\]and \[B=I-A\] \[\therefore \] \[AB+BA+I-{{(I-A)}^{2}}\] \[=A(I-A)+(I-A)A+I-(I-A)(I-A)\] \[=A-{{A}^{2}}+A-{{A}^{2}}+I-(I-2A+{{A}^{2}})\] \[=A-A+A-A+I-(I-2A+A)\] \[[\because {{A}^{2}}=A]\] \[=A\]You need to login to perform this action.
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