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Linear Algebra - Identity matrices

  • #1
I am having some difficulty with identity matrices in linear algebra at the moment. I am sure it is fairly simple to solve, but I just cannot follow the logic behind this particular problem.

I need to come up with a matrix B (2x2), such that B =/= I but B2 = I

Since
Code:
I = (1 0)
    (0 1)
If I am attempting this correctly, I have to find a number that isnt 1, but if you square it, it equals 1. So I came up with this.
Code:
B = (-1 0)
    (0 -1)
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
956
Well? Have you tried it? What is
[tex]\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right][/tex]
 
  • #3
1,631
4
well, dang, i don't know how to type matirices in latex, so i'll just try to give some hints.

let the matrix [tex] B=[B_1,B_2][/tex] where [tex] B_1,B_2[/tex] are the columns of B, so

[tex] B_1=[a.... b]^T, and, B_2=[c ... d]^T[/tex] now multiply [tex] BB=[BB_1 .... BB_2][/tex] now we want the following to hold

[tex]BB_1=e_1,BB_2=e_2[/tex] where [tex] I=[e_1 .... e_2][/tex], [tex]e_1=[1 .... 0]^T, e_2=[0 .... 1]^T[/tex]

now you will end up with some eq. involving in it a, b, c ,d so you can figure out the conditions that a, b, c, and d have to satisfy in order for BB=I. THis way you can find more than one such matrix.

I hope i was of any help
 

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