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Samenvatting Lineaire Algebra

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Dit document bevat een samenvatting met voorbeelden van het vak Lineaire Algebra, behorende tot jaar 1 van de studie Farmaceutische Wetenschappen.

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  • October 26, 2023
  • 6
  • 2020/2021
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Samenvatting Lineaire Algebra
Lengte van een vector: |(𝑎𝑏)| → √𝑎2 + 𝑏²

Scalar is een getal wat je met de vector vermenigvuldigd: 𝑐 (𝑎𝑏) = (𝑐𝑎
𝑐𝑏
)

Van a naar b = b – a
Van b naar a = a – b

Punt (a, b, c) → ai + bj + ck



Matrices vermenigvuldigen
𝑔 𝑗
(𝑑𝑎 𝑏 𝑐
𝑒 𝑓) ∙ (ℎ 𝑘 ) = (𝑚
𝑜
𝑛
𝑝)
𝑖 𝑙
2x3 3x2

m = ag + bh + ci
n = aj + bk + cl
o = dg + eh + fi
p = dj + ek + fl



Inverse
(10 0
1
= 𝐼2
)
1 0 0
(0 1 0) = 𝐼3
0 0 1
A ∙ I = 𝐴−1 → 𝐴−1 ∙ A = I

Ax = b → x = 𝐴−1 ∙ b



Getransponeerde matrix
𝐴𝑇 is de getransponeerde matrix. Dat wil zeggen 𝐴𝑗𝑖 → 𝐴𝑖𝑗 . Oftewel spiegelen in de hoofddiagonaal
1 4
A = (14 2 3
5 6
) 𝐴𝑇 = (2 5)
3 6
Een matrix heet symmetrisch als A = 𝐴𝑇

, Hoeken in een driehoek
𝑎2 = 𝑏 2 + 𝑐 2 - 2𝑏𝑐 cos 𝛼
𝑏 2 = 𝑎2 + 𝑐 2 - 2𝑎𝑐 cos 𝛽
𝑐 2 = 𝑏 2 + 𝑎2 - 2𝑏𝑎 cos 𝛾



In product
a∙b (𝑎𝑎1 ) ∙ (𝑏𝑏1 )
2 2
inproduct: 𝑎1 𝑏1 + 𝑎2 𝑏2
𝑎∙𝑏
Hoek tussen twee vectoren met behulp van in product: cos 𝛼 = |𝑎| |𝑏|
→ Er is een hoek van 90° als het inproduct 0 is.



Determinant
𝑎1 𝑏1
Determinant is → |𝑎 | = 𝑎1 𝑏2 − 𝑏1 𝑎2
2 𝑏2
𝑎 𝑑 𝑔 + − +
𝑒 ℎ 𝑑 𝑔 𝑑 𝑔
|𝑏 𝑒 ℎ| = a | |− + −|
𝑓 𝑖| – b | 𝑓 𝑖 | + c | 𝑓 𝑖 |
𝑐 𝑓 𝑖 + − +

Een matrix heeft een inverse als de determinant niet gelijk is aan 0
𝑎1 𝑏1 𝑥
(𝑎 ) ( ) = (𝑐𝑐1 ) → oplossen met de regel van Cramer
2 𝑏2 𝑦 2
𝑐1 𝑏1
|𝑐 |
2 𝑏2
→x= 𝑎1 𝑏1
|𝑎 |
2 𝑏2
𝑎1 𝑐1
|𝑎 𝑐2 |
→y= 𝑎1
2
𝑏1
|𝑎 |
2 𝑏2




Uitproduct
𝑎1 𝑏1 𝑐1
𝑎 𝑐
( 2 ) x (𝑏2 ) = ( 2 )
𝑎3 𝑏3 𝑐3

𝑖 𝑗 𝑘
𝑎
Oplossen van het uitproduct: | 1 𝑎2 𝑎3 |
𝑏1 𝑏2 𝑏3
𝑎2 𝑎3
→i = |𝑏 𝑏 |
2 3
𝑎1 𝑎3
→ j = |𝑏 𝑏 |
1 3
𝑎1 𝑎2
→ k = |𝑏 𝑏 |
1 2

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