A) 0
B) -1
C) 1
D) None of these
Correct Answer: A
Solution :
[a] As \[{{A}^{2}}=0,{{A}^{k}}=0\forall \,k\ge 2\]. Thus, \[{{(A+I)}^{50}}=I+50A\Rightarrow {{(A+I)}^{50}}-50A=I\] \[\therefore a=1,\,b=0,\,c=0,\,d=1\] \[\operatorname{abc} + abd + bcd + acd = 0\]
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