A) 0
B) \[\frac{1}{8}\]
C) 8
D) 64
Correct Answer: B
Solution :
\[|B|=|2{{A}^{-1}}|={{2}^{3}}|{{A}^{-1}}|=\frac{{{2}^{3}}}{|A|}\] \[|C|=\left| \frac{adj.A}{2} \right|=\frac{1}{{{2}^{3}}}|adj.A|=\frac{|A{{|}^{2}}}{{{2}^{3}}}\] \[|A{{B}^{2}}{{C}^{3}}|=||A|\,|{{B}^{2}}|\,|{{C}^{3}}|\] \[=1\cdot \frac{{{2}^{6}}}{|A{{|}^{2}}}\cdot \frac{|A{{|}^{6}}}{{{2}^{9}}}=\frac{1}{8}\]You need to login to perform this action.
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