If then show that A satisfies the following equation. |
\[{{A}^{3}}-4{{A}^{2}}+11I-3A=O\] |
OR |
If \[A+B+C=\pi ,\]show that |
\[=-\sin (A\,-B)sin(B\,-C)sin(C\,-A).\] |
Answer:
We have, \[\therefore \] Now, ?(i) \[\therefore \] ?(ii) Now, ?(iii) \[\therefore \]\[{{A}^{3}}-4{{A}^{2}}+11/3A=0\] [from Eqs. (i), (ii) and (iii)] Hence proved. OR The given determinant \[[{{C}_{1}}\to {{C}_{1}}+{{C}_{3}}]\] \[[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}},\,\,{{R}_{3}}-{{R}_{1}}]\] [taking \[2\sin (A-B)\]and\[2\sin (A-C)\]common from\[{{R}_{2}}\]and \[{{R}_{3}}\] respectively] \[[\because A+B+C=\pi ]\] \[=\sin \,(A-B)\sin \,(A-C)[\sin B\cos C-\cos B\sin C]\] \[=\sin \,(A-B)\sin \,(A-C)\sin \,(B-C)\] \[=-\sin \,(A-B)\sin \,(B-C)\sin \,(C-A)\] Hence proved.
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