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26 May 2008 18:25:36 IST

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Simple Challenge in Matrices - From Feynmann

None

$This \;is \;a\;really\;simple\;problem\;if \;a\;little\;bit\;of$ $imagination\;is\;employed.$

$Here \;goes\;the\;problem$ :

$Provide\;the\;inverse\;of\;the\;following\;matrix\;$ ::

$\begin{bmatrix} {c}_{0}&{c}_{1} &{c}_{2} & {c}_{3}\\ {c}_{2} & {c}_{3} &{c}_{0} &{c}_{1} \\ {c}_{3}&{-c}_{2} &{c}_{1} &{-c}_{0} \\ {c}_{1}&{-c}_{0} &{c}_{3} &{-c}_{2} \end{bmatrix}$

$where\;{c}_{0}=\frac{1+\sqrt3}{4\sqrt2}$ ${c}_{1}=\frac{3+\sqrt3}{4\sqrt2}$ ${c}_{2}=\frac{3-\sqrt3}{4\sqrt2}\;\;{c}_{3}=\frac{1-\sqrt3}{4\sqrt2}$

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Comments (6)

abhishek sinha

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26 May 2008 21:20:46 IST

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Come up Guys!!

It is not as difficult as it seems .

You virtually have to do no numerical calculation at all for matrix inversion !!

djdylan2000. bits goa eee

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26 May 2008 21:39:13 IST

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How do u find determinant of 4th order. Think its out of syllabus.

SUNDEEP ALLAMRAJU

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26 May 2008 22:12:37 IST

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Put c₁=rt3c₀ and c₂=-rt3c₃ and write the matrix and it's transpose.

Now,find A.A^T,it will be equal to (4c₀²+4c₃²)I where

4(c₀²+c₃²)=1 on substituting the values and Hence A^T=A^-1

Hurrah..I am the first to solve it.

SUNDEEP ALLAMRAJU

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26 May 2008 22:24:16 IST

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Thus,the answer is A^-1=A^T which is

Matrix c₀ c₂ c₃ c₁

c₁ c₃ -c₂ -c₀

c₂ c₀ c₁ c₃

c₃ c₁ -c₀ -c₂

Mukund

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27 May 2008 17:21:44 IST

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I'd solved this yesterday, but net went down, and I've only just been able to log on..

Anyways congratulations to allamraju..

@feynmann

By the way was that observation you were talking about this

c0^2 + c1^2 + c2^2 + c3^2 = 1

And c0 . c2 + c1. c3 = 0

Anyway, that's how I got it..

abhishek sinha

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27 May 2008 17:54:04 IST

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Excellent !!

Simple Problem Simple Answer !!

But Allamraju I should mention that in addition to AA' =I , we also have A'A=I(check this out !! )

That is why the inverse is A'.

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Comments (6)

You virtually have to do no numerical calculation at all for matrix inversion !!

Put c1=rt3c0 and c2=-rt3c3 and write the matrix and it's transpose.

Now,find A.AT,it will be equal to (4c02+4c32)I where

4(c02+c32)=1 on substituting the values and Hence AT=A-1

Hurrah..I am the first to solve it.

Thus,the answer is A-1=AT which is

Matrix c0 c2 c3 c1

c1 c3 -c2 -c0

c2 c0 c1 c3

c3 c1 -c0 -c2

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Put c₁=rt3c₀ and c₂=-rt3c₃ and write the matrix and it's transpose.

Now,find A.A^T,it will be equal to (4c₀²+4c₃²)I where

4(c₀²+c₃²)=1 on substituting the values and Hence A^T=A^-1

Thus,the answer is A^-1=A^T which is

Matrix c₀ c₂ c₃ c₁

c₁ c₃ -c₂ -c₀

c₂ c₀ c₁ c₃

c₃ c₁ -c₀ -c₂