Does the following matrix have an inverse?

  • Thread starter frankR
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  • #1
frankR
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N= [i,1;-1,i]

I used this theorem: N N-1 = In

Thus:

[i,1;-1,i]*[a,b:c,d]=[1,1;1,1]

I then found:

ia+c=1
ib+d=1
-a+ic=1
-b+id=1

Can I conclude an inverse does not exist. If so, how?

If not, what do I do?


Thanks,

Frank
 

Answers and Replies

  • #2
Hurkyl
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I2 = [1, 0; 0, 1]


What theorems have you learned about invertible matrices? (e.g. have you learned anything about how to tell if a matrix is invertible based on its determinant)


Or, you could apply the algorithm for computing inverses and see if you get an answer or if its impossible.
 
  • #3
frankR
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Originally posted by Hurkyl


Or, you could apply the algorithm for computing inverses and see if you get an answer or if its impossible.

I just found this:

N-1 exists only if:

det(NN-1 != 0

I'm a little rusty on my linear algebra, plus I got a concusion yesterday.
 
  • #4
arcnets
508
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frankR,
Hurkyl has told you what I2 is, because you got that wrong. Just redo your calculation using Hurkyl's hint and you should be able to answer this easily.
 

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