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![[Post New]](/templates/default/images/icon_minipost_new.gif) 4 Apr 2008 17:56:37 IST
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I mean to say without calculating adjacent and determinants........?
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 4 Apr 2008 19:55:24 IST
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No, it is not possible to calculate it by trick. Upto 2by2 you can use solution of simultaneous equation but it is difficult to solve three simultaneous equations trice
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Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 4 Apr 2008 20:06:17 IST
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no there exists a std methos which works for most of the cases but unfortunately i dont seem to remember it. Im sure its given in arihant
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 4 Apr 2008 20:08:39 IST
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r u sure , sandeep
even i have no clue on this
anybody with info. on this pls post
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all the best ... |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 4 Apr 2008 20:11:05 IST
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there's some std method like we can follow this pattern for making elements 0 or 1 A11=0 A21=0 A31=1 And sth like this im not sure but one thing im sure is the username by the name einstien knows this
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 11 May 2008 17:23:55 IST
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i too want he method pls anyone???
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 11 May 2008 18:02:13 IST
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I can think of one method using characterstic eqn.. But for that you'll have to find the square of the matrix... If you find finding that easier than finding adjoint, then maybe it's of some use.. Should i tell?
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 11 May 2008 18:13:28 IST
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yea go ahead...
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"Believe in yourself."
If you think you can do it , then you will do it .
Your fate is in your hands. |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 11 May 2008 18:18:55 IST
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You find the characterstic equation of a matrix, by subtracting K from all the diagonal elements and then solving the determinant to get a cubic in K.. Replace K by A, and 1 by I and you will get an equation which the matrix always satisfies. This is called the characterstic equation of the matrix. Now suppose the equation is A^3 + pA^2 + qA + rI = 0 Now multiply both sides by A(inv).. You get A^2 + pA + qI + rA(inv) = 0.. So if you now A^2 then it becomes a simple matter of rearranging terms.. But again, this is subject to knowing A^2..
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