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Topic: linear algebra with inverse of matrix.
Replies: 3   Last Post: Mar 18, 2013 12:47 PM

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 mina_world Posts: 2,144 Registered: 12/13/04
linear algebra with inverse of matrix.
Posted: Mar 16, 2013 11:36 AM

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

Date Subject Author
3/16/13 mina_world
3/16/13 W^3
3/16/13 quasi
3/18/13 AP