Kvadratiske matricer
rev.20.08.09. @
<Text-field style="Heading 1" layout="Heading 1">Initalisering</Text-field>restart:with(LinearAlgebra):
<Text-field style="Heading 1" layout="Heading 1"><Font encoding="UTF-8">Find X s\303\245 AX = B</Font></Text-field>Eksempel :Der er givet de to f\303\270lgende 2\303\2272 matricer :A:=< <1,0>|<-3,2> >;B:=< <-7,6>|<-4,2> >;X:=LinearSolve(A,B);Kontrol :A.X=B;Som forventet !
<Text-field style="Heading 1" layout="Heading 1"><Font encoding="UTF-8">Find E s\303\245 AE = A </Font></Text-field>Tallet 1 opfylder at der for ethvert reelt tal a g\303\246lder at LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYoLUkjbWlHRiQ2JVEiYUYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNi1RJyZzZG90O0YnL0YzUSdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGPS8lKXN0cmV0Y2h5R0Y9LyUqc3ltbWV0cmljR0Y9LyUobGFyZ2VvcEdGPS8lLm1vdmFibGVsaW1pdHNHRj0vJSdhY2NlbnRHRj0vJSdsc3BhY2VHUSYwLjBlbUYnLyUncnNwYWNlR0ZMLUkjbW5HRiQ2JFEiMUYnRjktRjY2LVEiPUYnRjlGO0Y+RkBGQkZERkZGSC9GS1EsMC4yNzc3Nzc4ZW1GJy9GTkZXRitGOQ==. Der findes ikke andre reelle tal med denne egenskab.I m\303\246ngden af n\303\227n matricer kunne dette inspirere til sp\303\270rgsm\303\245let :Hvordan skal en "enhedsmatrix" E se ud, der for enhver matrix A opfylder AE=A ?
For at n\303\246rme os dette opstiller vi en vilk\303\245rlig 3\303\2273 matrix:A:=< <a|b|c>,<d|e|f>,<g|h|i> >;og fors\303\270ger at finde E ved at l\303\270se ligningen AX=A :E:=LinearSolve(A,A);Netop \303\251n l\303\270sning og bem\303\246rk: Nydeligt resultat!
Vi kontroller om den \303\270nskede egenskab er opfyldt :
A.E=A;Som forventet !Vi har hermed sandsynliggjort (men ikke bevist!) at der i m\303\246ngden af 3\303\2273 matricer findes netop \303\251n matrix med den \303\270nskede egenskab. Den har 1-taller i diagonalen og 0'er alle andre steder. Tilsvarende for m\303\246ngden af n\303\227n matricer for vilk\303\245rligt n > 1.
Bem\303\246rk at der ogs\303\245 g\303\246lder :E.A=A;NB. Bogen bruger symbolet I for enhedsmatricen (engelsk: the identity matrix). Da Maple har reseveret dette symbol til brug i komplekse tal, benytter vi her E .
<Text-field style="Heading 1" layout="Heading 1"><Font size="16">Find </Font><Equation executable="false" style="Heading 1" input-equation="" display="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">LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUklbXN1cEdGJDYlLUkjbWlHRiQ2JlEiQUYnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1ElYm9sZEYnLyUrZm9udHdlaWdodEdGNy1GIzYmLUkjbW9HRiQ2L1EqJnVtaW51czA7RicvJSVzaXplR1EjMTJGJy8lJWJvbGRHRjQvRjZRJ25vcm1hbEYnLyUmZmVuY2VHRjQvJSpzZXBhcmF0b3JHRjQvJSlzdHJldGNoeUdGNC8lKnN5bW1ldHJpY0dGNC8lKGxhcmdlb3BHRjQvJS5tb3ZhYmxlbGltaXRzR0Y0LyUnYWNjZW50R0Y0LyUnbHNwYWNlR1EsMC4yMjIyMjIyZW1GJy8lJ3JzcGFjZUdGVy1JI21uR0YkNiZRIjFGJ0ZARkNGRUY1RjgvJTFzdXBlcnNjcmlwdHNoaWZ0R1EiMEYnLUY9Ni9RIn5GJ0ZARkNGRUZHRklGS0ZNRk9GUUZTL0ZWUSYwLjBlbUYnL0ZZRl9vRjVGOA==</Equation><Font encoding="UTF-8" size="16">s\303\245 </Font><Equation executable="false" style="2D Math" input-equation="" display="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">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</Equation><Font bold="true" style="Text" size="18"> = E</Font></Text-field>En n\303\227n matrix A kaldes invertibel hvis der findes en n\303\227n matrix ALUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUklbXN1cEdGJDYlLUkjbWlHRiQ2I1EhRictRiM2JS1JI21vR0YkNi1RKiZ1bWludXMwO0YnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGPS8lKXN0cmV0Y2h5R0Y9LyUqc3ltbWV0cmljR0Y9LyUobGFyZ2VvcEdGPS8lLm1vdmFibGVsaW1pdHNHRj0vJSdhY2NlbnRHRj0vJSdsc3BhY2VHUSwwLjIyMjIyMjJlbUYnLyUncnNwYWNlR0ZMLUkjbW5HRiQ2JFEiMUYnRjhGOC8lMXN1cGVyc2NyaXB0c2hpZnRHUSIwRidGOA== s\303\245 der g\303\246lder at AALUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUklbXN1cEdGJDYlLUkjbWlHRiQ2I1EhRictRiM2JS1JI21vR0YkNi1RKiZ1bWludXMwO0YnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGPS8lKXN0cmV0Y2h5R0Y9LyUqc3ltbWV0cmljR0Y9LyUobGFyZ2VvcEdGPS8lLm1vdmFibGVsaW1pdHNHRj0vJSdhY2NlbnRHRj0vJSdsc3BhY2VHUSwwLjIyMjIyMjJlbUYnLyUncnNwYWNlR0ZMLUkjbW5HRiQ2JFEiMUYnRjhGOC8lMXN1cGVyc2NyaXB0c2hpZnRHUSIwRidGOA=== E og ALUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUklbXN1cEdGJDYlLUkjbWlHRiQ2I1EhRictRiM2JS1JI21vR0YkNi1RKiZ1bWludXMwO0YnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGPS8lKXN0cmV0Y2h5R0Y9LyUqc3ltbWV0cmljR0Y9LyUobGFyZ2VvcEdGPS8lLm1vdmFibGVsaW1pdHNHRj0vJSdhY2NlbnRHRj0vJSdsc3BhY2VHUSwwLjIyMjIyMjJlbUYnLyUncnNwYWNlR0ZMLUkjbW5HRiQ2JFEiMUYnRjhGOC8lMXN1cGVyc2NyaXB0c2hpZnRHUSIwRidGOA==A = E .Eksempel 1:Lad os unders\303\270ge om flg. 3\303\2273 matrix er invertibel :A:=< <1|2|3>,<4|5|6>,<7|8|9> >;Vi fors\303\270ger at l\303\270se ligningen AX=E vha. totalmatrix:E:=IdentityMatrix(3);T:=<A|E>;ReducedRowEchelonForm(T);Ser ikke lovende ud ..Rank(A);Rank(T);Der er ingen l\303\270sning ! En kvadratisk matrix der ikke har fuld rang, kaldes singul\303\246r. A er alts\303\245 en singul\303\246r matrix.Der g\303\246lder altid at en singul\303\246r matrix ikke har en invers matrix. Eksempel 2:Vi fors\303\270ger nu med en regul\303\246r matrix, dvs. den har fuld rang :A:=< <0|2|3>,<4|5|6>,<7|8|9> >;Rank(A);T:=<A|E>;ReducedRowEchelonForm(T);Det ses at der er \303\251n l\303\270sning. Den tr\303\246kker vi ud s\303\245ledes:invA:=SubMatrix(%,[1..3],[4..6]);Vi kontrollerer at AALUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictSSVtc3VwR0YkNiVGKy1GIzYlLUkjbW9HRiQ2LVEoJm1pbnVzO0YnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGPS8lKXN0cmV0Y2h5R0Y9LyUqc3ltbWV0cmljR0Y9LyUobGFyZ2VvcEdGPS8lLm1vdmFibGVsaW1pdHNHRj0vJSdhY2NlbnRHRj0vJSdsc3BhY2VHUSwwLjI3Nzc3NzhlbUYnLyUncnNwYWNlR0ZMLUkjbW5HRiQ2JFEiMUYnRjhGOC8lMXN1cGVyc2NyaXB0c2hpZnRHUSIwRidGK0Y4 = E og ALUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictSSVtc3VwR0YkNiVGKy1GIzYlLUkjbW9HRiQ2LVEoJm1pbnVzO0YnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGPS8lKXN0cmV0Y2h5R0Y9LyUqc3ltbWV0cmljR0Y9LyUobGFyZ2VvcEdGPS8lLm1vdmFibGVsaW1pdHNHRj0vJSdhY2NlbnRHRj0vJSdsc3BhY2VHUSwwLjI3Nzc3NzhlbUYnLyUncnNwYWNlR0ZMLUkjbW5HRiQ2JFEiMUYnRjhGOC8lMXN1cGVyc2NyaXB0c2hpZnRHUSIwRidGK0Y4A = EA.invA=invA.A;OK !Vi kan osse finde den inverse direkte med:A^(-1);Der g\303\246lder altid at en regul\303\246r matrix har netop \303\251n invers matrix.
<Text-field style="Heading 1" layout="Heading 1"><Font encoding="UTF-8" size="16"> AX = B l\303\270st vha. </Font><Equation executable="false" style="2D Math" input-equation="" display="LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2I1EhRicvJSxtYXRodmFyaWFudEdRJ25vcm1hbEYn">LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2I1EhRic=</Equation><Font bold="true" style="Text" size="16">A</Font><Equation executable="false" style="2D Math" input-equation="" display="LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictSSVtc3VwR0YkNiVGKy1GIzYlLUkjbW9HRiQ2LlEoJm1pbnVzO0YnLyUlc2l6ZUdRIzE2RicvJSxtYXRodmFyaWFudEdRJ25vcm1hbEYnLyUmZmVuY2VHUSZmYWxzZUYnLyUqc2VwYXJhdG9yR0ZALyUpc3RyZXRjaHlHRkAvJSpzeW1tZXRyaWNHRkAvJShsYXJnZW9wR0ZALyUubW92YWJsZWxpbWl0c0dGQC8lJ2FjY2VudEdGQC8lJ2xzcGFjZUdRJjAuMGVtRicvJSdyc3BhY2VHRk8tSSNtbkdGJDYlUSIxRidGOEY7RjsvJTFzdXBlcnNjcmlwdHNoaWZ0R1EiMEYnRitGOw==">LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictSSVtc3VwR0YkNiVGKy1GIzYlLUkjbW9HRiQ2LlEoJm1pbnVzO0YnLyUlc2l6ZUdRIzE2RicvJSxtYXRodmFyaWFudEdRJ25vcm1hbEYnLyUmZmVuY2VHUSZmYWxzZUYnLyUqc2VwYXJhdG9yR0ZALyUpc3RyZXRjaHlHRkAvJSpzeW1tZXRyaWNHRkAvJShsYXJnZW9wR0ZALyUubW92YWJsZWxpbWl0c0dGQC8lJ2FjY2VudEdGQC8lJ2xzcGFjZUdRJjAuMGVtRicvJSdyc3BhY2VHRk8tSSNtbkdGJDYlUSIxRidGOEY7RjsvJTFzdXBlcnNjcmlwdHNoaWZ0R1EiMEYnRitGOw==</Equation></Text-field>Vi vender tilbage til det f\303\270rste eksempel og l\303\270ser ligningen p\303\245 en alternativ m\303\245de:A:=< <1,0>|<-3,2> >;B:=< <-7,6>|<-4,2> >;A har en invers matrix idet den klart er regul\303\246r .Vi kan derfor l\303\270se ligningen AX = B p\303\245 flg. m\303\245de (hver side ganges med ALUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYlLUklbXN1cEdGJDYlLUkjbWlHRiQ2I1EhRictRiM2JS1JI21vR0YkNi1RKiZ1bWludXMwO0YnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGPS8lKXN0cmV0Y2h5R0Y9LyUqc3ltbWV0cmljR0Y9LyUobGFyZ2VvcEdGPS8lLm1vdmFibGVsaW1pdHNHRj0vJSdhY2NlbnRHRj0vJSdsc3BhY2VHUSwwLjIyMjIyMjJlbUYnLyUncnNwYWNlR0ZMLUkjbW5HRiQ2JFEiMUYnRjhGOC8lMXN1cGVyc2NyaXB0c2hpZnRHUSIwRictRjU2LVEifkYnRjhGO0Y+RkBGQkZERkZGSC9GS1EmMC4wZW1GJy9GTkZaRjg=fra venstre) :X:=A^(-1).B;hvilket stemmer overens med det resultat vi fandt ovenfor !
<Text-field style="Heading 1" layout="Heading 1">En anvendelse af determinant</Text-field>P\303\245 Lille Dag i sidste uge indholdt ligningssystemet i Opgave 6 en ubekendt koefficient, og opgaven var ikke helt nem at l\303\270se.Vi viser nu en alternativ l\303\270sning der bygger p\303\245 vores nye kendskab til determinant.Ligningssystemet har f\303\270lgende koefficientmatrix A og h\303\270jreside b :A:=< <a|1|1>,<1|a|1>,<1|1|a> >;b:=<1,1,1>;Rank(A);Bem\303\246rk at Maple helt generelt svarer at koefficientmatricen har rang 3 - og i overenssstemmelse hermed angiver netop \303\251n l\303\270sning for et givet a :x:=<x1,x2,x3>:
x=LinearSolve(A,b);Men er rangen af A n\303\270dvendigvis 3 ??I denne opgave, hvor vi har at g\303\270re med en kvadratisk koefficient-matrix, kan vi finde alle s\303\246rtilf\303\246ldene ved en unders\303\270gelse af determinanten:detA:=Determinant(A);Vi finder r\303\270dderne i dette 3. gradspolynomium:solve(detA=0,a);L\303\270sningen ovenfor g\303\246lder atls\303\245 kun n\303\245r a \342\211\240 -2 og a \342\211\240 1.
L\303\270sningen findes nu for s\303\246rtilf\303\246ldet a = 1 :A1:=subs(a=1,A);x=LinearSolve(A1,b,free=t);P\303\245 strukturel form: For a = 1 har ligningssystemet l\303\270sningen :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\303\270sningen findes nu for s\303\246rtildet a = -2 :A2:=subs(a=-2,A);LinearSolve(A2,b);Maple h\303\246vder at systemet er inkonsistent. Hvorfor?Vi m\303\245 finde grunden:T:=<A2|b>;
ReducedRowEchelonForm(T);Her bem\303\246rkes at LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEnJiM5NjE7RicvJSdpdGFsaWNHUSZmYWxzZUYnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJ0Yy(T) = 3, mens LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEnJiM5NjE7RicvJSdpdGFsaWNHUSZmYWxzZUYnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJ0Yy(A) = 2. Ligningssystemet har derfor ingen l\303\270sning for LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2JVEiYUYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNi1RIj1GJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRj0vJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUnbHNwYWNlR1EsMC4yNzc3Nzc4ZW1GJy8lJ3JzcGFjZUdGTC1JI21uR0YkNiRRKyZ1bWludXMwOzJGJ0Y5Rjk=. Hermed har vi l\303\270st ligningssystemet for enhver v\303\246rdi af a.