A) 8
B) \[8\sqrt{2}\]
C) \[16\sqrt{2}\]
D) 32
Correct Answer: C
Solution :
[c] Given that \[A{{A}^{T}}=2I\] \[\Rightarrow \,\,\,|A{{|}^{2}}=8\] and \[{{A}^{-1}}={{A}^{T}}-A\,\,adj(2{{B}^{-1}})\] \[\Rightarrow \,\,\,\,I=A{{A}^{T}}-{{A}^{2}}adj(2{{B}^{-1}})\] \[\Rightarrow \,\,\,\,I=2I-{{A}^{2}}adj(2{{B}^{-1}})\] \[\Rightarrow \,\,\,\,{{A}^{2}}\] adj \[(2{{B}^{-1}})=I\] \[\Rightarrow \,\,\,\,\,|{{A}^{2}}||adj\,\,(2{{B}^{-1}})|=1\] \[\Rightarrow \,\,\,8|2{{B}^{-1}}{{|}^{2}}=1\] \[\Rightarrow \,\,\,\,\,\,8\times \frac{64}{|B{{|}^{2}}}=1\] \[\Rightarrow \,\,\,\,\,\,\,\,|B{{|}^{2}}=64\times 8\] \[\therefore \,\,\,\,\,\,|B|=\pm 16\sqrt{2}\]You need to login to perform this action.
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