(Please help!!) If AČ = I, prove det A = ±1


by chapone
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chapone
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#1
May23-08, 04:54 PM
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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!
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Hurkyl
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#2
May23-08, 04:57 PM
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Well, in this case, doing matrix arithmetic is going to be a lot easier than doing scalar arithmetic on the entries of the matrix.
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May23-08, 05:19 PM
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Quote Quote by chapone View Post
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!
Hi chapone! Welcome to PF!

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))?

asdfggfdsa
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#4
May26-08, 05:57 PM
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(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.)
john the gree
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#5
Oct14-10, 09:35 AM
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aČ = 1
aČ - 1 = 0
(a - 1) (a + 1) = 0
a=1 and a = -1
then a = ±1
Mark44
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#6
Oct14-10, 10:39 AM
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Quote Quote by john the gree View Post
aČ = 1
aČ - 1 = 0
(a - 1) (a + 1) = 0
a=1 and a = -1
then a = ±1
This is all well and good for a real number a, but the OP is working with a matrix A, not a scalar. As such, A [itex]\neq[/itex] 1.
Fredrik
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#7
Oct14-10, 12:41 PM
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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]
vigvig
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#8
Oct23-10, 10:17 AM
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If A^2=1, then A is a matrix of order 2, which means that A is invertible.
D H
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#9
Oct23-10, 10:43 PM
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Quote Quote by vigvig View Post
If A^2=1, then A is a matrix of order 2
There are multiple meanings of the term "order", and your meaning here is not the meaning usually meant for matrices.
vigvig
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#10
Oct24-10, 11:51 AM
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Quote Quote by D H View Post
There are multiple meanings of the term "order", and your meaning here is not the meaning usually meant for matrices.
What do you mean? Order in the sense of "group order", meaning A generates a group of order 2. Implying A must be invertible


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