A) 1
B) 16
C) 1/16
D) ¼
Correct Answer: C
Solution :
Here, \[|A{{|}^{2}}.|B|=8\] ...(i) and \[\frac{|A|}{|B|}=8\] ...(ii) Solving (i) and (ii), we get\[|A|=4\,\text{and}|B|\,=\,\frac{1}{2}\] Now, det\[(B{{A}^{-1}}{{B}^{T}})\,=\,\frac{|B{{|}^{2}}}{|A|}=\frac{1}{4}.\frac{1}{4}=\frac{1}{16}\]You need to login to perform this action.
You will be redirected in
3 sec