question_answer

Prove that the determinant  is independent of \[\theta .\]

Answer:

Let  Expand along \[{{R}_{1}},\]we get \[|A|\,\,=x\,(-\,{{x}^{2}}-1)-\sin \theta \,(-\,x\sin \theta -\cos \theta )\]\[+\,\cos \theta \,(-\sin \theta +x\cos \theta )\] \[=-\,{{x}^{3}}-x+x{{\sin }^{2}}\theta +\sin \theta \cos \theta \] \[-\sin \theta \cos \theta +x{{\cos }^{2}}\theta \] \[=-\,{{x}^{3}}-x+x{{\sin }^{2}}\theta +x{{\cos }^{2}}\theta \] \[=-\,{{x}^{3}}-x+x\,({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )\] \[=-\,{{x}^{3}}-x+x\]  \[[\because {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1]\] \[|A|\,\,=-\,{{x}^{3}}\]                                      ?(i) Hence, from Eq. (i), we conclude A is independent of\[\theta \]. Hence proved.