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W^3
Posts:
28
Registered:
4/19/11
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Re: linear algebra with inverse of matrix.
Posted:
Mar 16, 2013 1:45 PM
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In article <010c295a-ec4c-43df-a2e5-bcf30115e590@googlegroups.com>,
mina_world@hanmail.net wrote:
> Hello teacher~
>
> nonzero 2 by 2 matrix A, B
>
> A^2 + B^2 = 0 and (A+B)^2 = 0
>
> (E : Identity matrix)
>
> (1) AB = -BA
>
> (2) (A^3)(B^3) = -(B^3)(A^3)
>
> (3) A+B+E has an inverse matrix.
>
> ---------------------------------------
> (1)
> (A+B)^2 = (A+B)(A+B)
> = A^2 + AB + BA + B^2 = 0
>
> Sinc A^2 + B^2 = 0, AB + BA = 0
> so, AB = -BA
>
> (2)
> Since AB = -BA,
> (A^3)(B^3) = AAABBB = AA(-BA)BB = AA(-B)(-BA)B
> =AA(-B)(-B)(-BA) = AA(-B^3.A)
> It means that ABBB = -BBBA
> so, AAABBB = -AABBBA = --ABBBAA = ---BBBAAA = -(B^3)(A^3)
>
> (3)
> Answer : inverse matrix ==> E-A-B
> How do you find it ?
>
> Of course, it's really true.
> (A+B+E)(E-A-B) = A-AA-AB+B-BA-BB+E-A-B = E
We know (1+z)^{-1} = 1 - z + z^2 - ... under certain circumstances.
So why not try (E+(A+B))^{-1} = E - (A+B) + (A+B)^2 - (A+B)^3 + ...
= E - (A+B) + 0 - 0 + ... = E - (A+B).
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