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IX. Opgaver – 5. Januar 2005 596 Løsning. 1) Lad Så har matricen ⎡ A = ⎣ ⎡ A − bI = ⎣ 0 0 1 0 b 0 4 0 0 ⎤ ⎦. −b 0 1 0 0 0 4 0 −b en 0-række. Dermed er nulrummet ikke trivielt og b er en egenværdi, [LA] 9. 2) Den opgivne matrix er A for b = 2. Nulrummet beregnes ved elimination. Rækkereduktionen ⎡ −2 A − 2I = ⎣ 0 0 0 ⎤ ⎡ 1 0 ⎦ ∼ ⎣ ⎤ ⎦ 4 0 −2 giver det reducerede ligningssystem x1 = 1 2 x3 En vektor (x1,x2,x3) i nulrummet opskrives ⎡ ⎣ x1 x2 x3 ⎤ ⎡ ⎦ = ⎣ 1 2 x3 x2 x3 ⎤ ⎡ ⎦ = x2 ⎣ 0 1 0 ⎤ ⎤ ⎦ 1 0 −1 2 0 0 0 0 0 0 ⎡ ⎦ + x3 ⎣ 1 2 0 1 ⎤ ⎦
IX. Opgaver – 5. Januar 2005 597 Det ses, at egenrummet er ⎡ E2 = span( ⎣ 0 ⎤ ⎡ 1 ⎦ , ⎣ 0 Opgave 5. Betragt differentialligningen (for x > 0) dy 3y = + x. dx x Angiv den løsning y(x) (for x > 0), der opfylder y(1) = 2. 1 2 0 1 ⎤ ⎦) . Løsning. En analytisk løsning kan med notation fra [LA] 14 beskrives ved og stamfunktionerne a(x) = 3 , b(x) = x x 3 A(x) = a(x)dx = dx = 3ln x , x B(x) = e −A(x) b(x)dx = = x −3 x dx = x −2 dx = −x −1 . e −3 ln x x dx
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CALCULUS "SLIDES" TIL CALCULUS 1 +
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3. Generelle områder 201 4. Koordi
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Forord "Slides" til forelæsningern
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I. Differentiation - 1. Kontinuitet
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I. Differentiation - 2. Partielle a
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I. Differentiation - 3. Tangentplan
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I. Differentiation - 6. Maksimum/mi
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I. Differentiation - 7. Lagrangemet
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II. Integration - 1. Dobbelt integr
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III. Potensrækker - 4. Taylorpolyn
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1. Grafiske/numeriske metoder IV Di
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IV. Differentialligninger - 1. Graf
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1. Vektorer og matricer V Matricer
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V. Matricer - 1. Vektorer og matric
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V. Matricer - 4. Determinanter = (
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1. Egenvektorer VI Egenvektorer og
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VI. Egenvektorer og diagonalisering
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