forum zespołu muzyki folk Portfolk
Zapraszamy do dyskusji... TO SCOTLAND!!!
FAQ
Szukaj
Użytkownicy
Grupy
Galerie
Rejestracja
Profil
Zaloguj się, by sprawdzić wiadomości
Zaloguj
Forum forum zespołu muzyki folk Portfolk Strona Główna
->
Doradź nam coś.
Napisz odpowiedź
Użytkownik
Temat
Treść wiadomości
Emotikony
Więcej Ikon
Kolor:
Domyślny
Ciemnoczerwony
Czerwony
Pomarańćzowy
Brązowy
Żółty
Zielony
Oliwkowy
Błękitny
Niebieski
Ciemnoniebieski
Purpurowy
Fioletowy
Biały
Czarny
Rozmiar:
Minimalny
Mały
Normalny
Duży
Ogromny
Zamknij Tagi
Opcje
HTML:
NIE
BBCode
:
TAK
Uśmieszki:
TAK
Wyłącz BBCode w tym poście
Wyłącz Uśmieszki w tym poście
Kod potwierdzający: *
Wszystkie czasy w strefie EET (Europa)
Skocz do:
Wybierz forum
KATEGORIE
----------------
Koncerty
Doradź nam coś.
Co o nas myślisz?
O wszystkim i o niczym
Przegląd tematu
Autor
Wiadomość
beioiog7y3l
Wysłany: Pon 15:42, 23 Wrz 2013
Temat postu: moncler sito ufficiale Logical Equivalence
Morgan is a writer who works with Uk Best Writing service, He has experience of more than ten years in acdemic writing. He provides students with Uk research papers, ssays writing service UK and Uk essay service online.
Two statements q and r are said to be logically equivalent if they have the same truth-value or logical content. Thus, q is true [url=http://www.1855sacramento.com/moncler.php]moncler sito ufficiale[/url] if and only if p is true. An n x n matrix A is invertible if it has an inverse B such that AB = I (identity matrix) where B = A-1. Alternatively, A is invertible iff det (A) ≠ 0 (null matrix) To prove that the ten statements are equivalent you need to show that 1→2→3…..9→ 10→1
If A is invertible m [url=http://www.gotprintsigns.com/monclerpascher/]moncler pas cher[/url] x n, then there exist an n x m matrix B such that AB= I (identity matrix) in this case the [url=http://www.achbanker.com/home.php]hollister[/url] identity matrix would be of dimension n x n. Thus, the number of rows in [url=http://www.1855sacramento.com/woolrich.php]woolrich outlet[/url] A is [url=http://www.mquin.com/giuseppezanotti.php]giuseppe zanotti sneakers[/url] equivalent to number of rows in I ie 1→2. If A is an n x n identity matrix, then A has n leading ones [url=http://www.sandvikfw.net/shopuk.php]hollister sale[/url] and each of these leading ones occupies a pivot position. Thus, A if [url=http://www.1855sacramento.com/peuterey.php]peuterey outlet[/url] n x n identity matrix then A has n pivot position or 2→3.
If A has n pivot positions then A has n leading ones in its low echelon form. If Dx =0 where D is the low echelon form of A, D ≠ 0. This implies that x = 0 (trivial solution). Thus if A has n pivotal position then Ax = 0 has only trivial solutions. Alternatively, 3→4. If Ax = 0 has only trivial solution then there exists a matrix b in R^n such that A (x +b) ≠ 0. This implies that Ax + Ab ≠ 0, since Ab ≠ 0 then Ax≠ 0. This indicates that there exists b in R^n such that Ax=b or 4→5.
If Ax=b has at least one solution b in R^n, then the columns of A in low echelon form a basis for R^n. Or, precisely, each of the vectors in R^n can be written as a linear combination of column [url=http://www.ilyav.com/uggpascher.php]boots ugg pas cher[/url] vectors of A. Thus, A spans R^n or 5→6. If [url=http://www.rtnagel.com/airjordan.php]jordan pas cher[/url] the column vectors of A spans R^n, then every vector (matrix) in R^n can be expressed as a linear combination of the column vectors of A and Ax is contained in R^n for all [url=http://www.thehygienerevolution.com/barbour.php]barbour paris[/url] x in [url=http://www.sandvikfw.net/shopuk.php]hollister outlet sale[/url] R^n. Thus, x-->Ax is a one to one and onto function. Therefore, x-->Ax maps R^n onto R^n or 6→7.
If the linear transformation [url=http://www.ilyav.com/uggpascher.php]ugg pas cher[/url] x-->Ax maps R^n onto R^n, then the set of all linear transformations x-->Ax such that Ax is a member of R^n form a Group. All elements in the group have inverses. Therefore, there exists C such that CA=I. In this case, C is the inverse of A. If A and C are n x n matrix such that CA =I, then C is the inverse of A and, thus, CA = I = AC, or AD=I where D is an n x n matrix, thus, 8→9. If D is an n x n matrix such that AD=I the transpose of A is invertible. This shows that the column vectors of A are linearly independent and they span R^n or A forms a basis for R^n, thus, 9→10.
If the columns of A forms a basis of R^n, then they are linearly independent. Additionally, the column vectors of A are non-zero and they span R^n. Thus, the basis of R^n forms an n x n identity matrix. This indicates that there exist an n x n matrix B such that AB= I or A is invertible, thus, 10→1
相关的主题文章:
[url=http://cgi.ebay.co.uk/350W-SHEEP-GOATS-SHEARING-CLIPPER-WOOL-SHEARS-RRP899new-/290398329820?pt=LH_DefaultDomain_0&hash=item439d16dbdc]louboutin pas cher Finding Th[/url]
[url=http://www.lzhcdc.cn/E_GuestBook.asp]hollister[/url]
[url=http://www.jnxlygg.com/viewthread.php?tid=16197&extra=]barbour paris Why Should you Hire a Social Security Disability Attorney[/url]
fora.pl
- załóż własne forum dyskusyjne za darmo
Powered by
phpBB
© 2001 phpBB Group
Chronicles phpBB2 theme by
Jakob Persson
(
http://www.eddingschronicles.com
). Stone textures by
Patty Herford
.
Regulamin