A) 1
B) 2
C) 3
D) 50
Correct Answer: A
Solution :
\[\det {{[(I+B)]}^{50}}-50B]\] \[=\det {{[}^{50}}{{C}_{0}}I{{+}^{50}}{{C}_{1}}B{{+}^{50}}{{C}_{2}}{{B}^{2}}{{+}^{50}}{{C}_{3}}{{B}^{3}}+...\] \[{{+}^{50}}{{C}_{50}}{{B}^{50}}{{B}^{50}}-50B]\] {All terms having \[{{B}^{n}},2\le n\le 50\]will be zero because given that B2 = 0} \[=\det [I+50B-50B]=det[I]=1\]
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