A) \[n\pi +{{(-1)}^{n}}\frac{\pi }{4}\]
B) \[{{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{3}\]
C) \[n\pi +\frac{\pi }{4}-\frac{\pi }{3}\]
D) \[n\pi +{{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{3}\]
Correct Answer: D
Solution :
Since matrices A and B are symmetric \[\therefore \] \[{{A}^{T}}=A\]and \[{{B}^{T}}=B\] ...(i) Now, \[{{(A+B)}^{T}}={{A}^{T}}+{{B}^{T}}\] \[\Rightarrow \] \[{{(A+B)}^{T}}=A+B\] [using Eq. (i)] \[\therefore \]\[A+B\]is symmetric. \[{{(AB-BA)}^{T}}={{(AB)}^{T}}-{{(BA)}^{T}}\] \[={{B}^{T}}{{A}^{T}}-{{A}^{T}}{{B}^{T}}\] (By reversal law) \[=BA-AB\] [using Eq. (i)] \[=-(AB-BA)\] \[\therefore \]\[AB-BA\] is skew symmetric. Now, \[{{(AB+BA)}^{T}}\] \[={{(AB)}^{T}}+{{(BA)}^{T}}\] \[={{B}^{T}}{{A}^{T}}+{{A}^{T}}{{B}^{T}}\] (By reversal law) \[=BA+AB\] [using Eq. (i)] \[=AB+BA\] \[\therefore \] \[AB+BA\] is symmetric.
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