
#1
Oct803, 07:55 PM

P: 91

N= [i,1;1,i]
I used this theorem: N N^{1} = I_{n} 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 



#2
Oct803, 08:12 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

I_{2} = [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
Oct803, 08:44 PM

P: 91

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
Oct1003, 06:17 PM

P: 513

Does the following matrix have an inverse?
frankR,
Hurkyl has told you what I_{2} 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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