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(Please help ) If AČ = I, prove det A = ±1by chapone
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#1
May2308, 04:54 PM

P: 3

If AČ = I, show that det A = ±1
This book is very unclear, but I am assuming by "I" they mean the identity matrix with a size of 2x2. I have tried putting in A for row 1 column 1  B for 1,2  C for 2,1 and D for 2,2 multiplying and setting the results equal to the values of the identity matrix. I thought I was close, but now am doubting that I am going about this the right way. Any help is MUCH appreciated! Thank you! 


#2
May2308, 04:57 PM

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PF Gold
P: 16,091

Well, in this case, doing matrix arithmetic is going to be a lot easier than doing scalar arithmetic on the entries of the matrix.



#3
May2308, 05:19 PM

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Thanks
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What makes you think they mean 2x2? This theorem is true for any n x n matrix. Do you know any formulas for determinants (for example, for det (AB))? 


#4
May2608, 05:57 PM

P: 27

(Please help ) If AČ = I, prove det A = ±1
Take the determinant of both sides of A[tex]^{2}[/tex] = I.
(Any don't make any assumptions beyond what the book gives you.) 


#5
Oct1410, 09:35 AM

P: 1

aČ = 1
aČ  1 = 0 (a  1) (a + 1) = 0 a=1 and a = 1 then a = ±1 


#6
Oct1410, 10:39 AM

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#7
Oct1410, 12:41 PM

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PF Gold
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chapone got 3 good answers already, so I'll just add to what Mark44 said (also a good post, but not an answer for chapone) by providing a counterexample that shows that John's argument gets the wrong result for matrices:
[tex]\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}=\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}[/tex] 


#8
Oct2310, 10:17 AM

P: 21

If A^2=1, then A is a matrix of order 2, which means that A is invertible.



#9
Oct2310, 10:43 PM

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#10
Oct2410, 11:51 AM

P: 21




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