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W^3
Posts:
29
Registered:
4/19/11


Re: linear algebra with inverse of matrix.
Posted:
Mar 16, 2013 1:45 PM


In article <010c295aec4c43dfa2e5bcf30115e590@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 ==> EAB > How do you find it ? > > Of course, it's really true. > (A+B+E)(EAB) = AAAAB+BBABB+EAB = 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).



