A) \[24\]
B) \[3\]
C) \[\frac{1}{3}\]
D) \[\frac{1}{24}\]
Correct Answer: D
Solution :
Given, \[|A{{|}_{3\times 3}}\ne 0\] and \[|A|=3\] Then, \[|{{(2A)}^{-1}}|=\left| \frac{1}{2A} \right|=\frac{1}{|2A|}\] \[=\frac{1}{{{(2)}^{3}}}.\frac{1}{|A|}\] \[(\because \,\,|aA|={{a}^{3}}|A|)\] \[=\frac{1}{8}.\frac{1}{3}=\frac{1}{24}\]You need to login to perform this action.
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