A) \[-2\]
B) 1
C) 0
D) \[-1\]
Correct Answer: C
Solution :
Subtracting \[{{P}^{3}}-{{P}^{2}}Q={{Q}^{3}}-{{Q}^{2}}P\] \[{{P}^{2}}(P-Q)+{{Q}^{2}}(P-Q)=0\] \[({{P}^{2}}+{{Q}^{2}})(P-Q)=0\] If \[\left| {{P}^{2}}+{{Q}^{2}} \right|\ne 0\] then \[{{P}^{2}}+{{Q}^{2}}\] is invertible \[\Rightarrow P-Q=0\] contradiction Hence \[\left| {{P}^{2}}+{{Q}^{2}} \right|=0\]You need to login to perform this action.
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