100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
logo-home
Summary Linear Algebra and Its Applications, Global Edition, ISBN: 9781292092232 Linear Algebra $7.50   Add to cart

Summary

Summary Linear Algebra and Its Applications, Global Edition, ISBN: 9781292092232 Linear Algebra

 47 views  2 purchases
  • Course
  • Institution
  • Book

Summary chapter 1 to 4 and 6

Preview 4 out of 33  pages

  • No
  • Hoofdstuk 1 tm 4 en 6
  • November 13, 2021
  • 33
  • 2020/2021
  • Summary
avatar-seller
Lineaire algebra – blok 1
Week 1, les 1
1.2: systems of linear equations
X1 + 2x2 = 3 standard form
 Linear system:
R1 : 1 equation = line
2 equations = intersection point // parallel  n.a.
R2: x=0 R2
x=1 line
x=2 point
x=3 n.a.
R3: x=0 R3
x=1 plane
x=2 line
x=3 point
x= 4 n.a.
many solutions
consistent
1 solution
inconsistent
no solutions
- Example
X1 – 2x2 = -1
-x1 + 3x2 = 3
van elkaar afhalen
x2 = 2  invullen in eerste formule voor x1
Een andere manier van noteren is in een augmented matrix (variables):
[ 1 -2 -1
-1 3 3 ]
Coefficient matrix = hetzelfde als een augmented matrix, alleen dan zonder de dingen achter het
‘’=’’-teken  naam van de matrix moet/staat er altijd bij.
[ 1 -2 -1 [1 -2 -1 +2row2 [1 0 3 x1 = 3
-1 3 3 ] +row1 012 ] 012] x2 = 2
Op het moment dat er een 0 bovenaan de matrix staat, wissel deze dan om zodat er (het liefst een
1) een ander getal links bovenin staat. DUS:
1. Eliminate 1e variabel in 2 van de 3 rijen – alleen 1 e rij overhouden
2. Eliminate 2e variabel in 2 van de 3 rijen – alleen 2 e rij overhouden
3. Eliminate 3e variabel in 2 van de 3 rijen – alleen 3 e rij overhouden

,1.2: row reduction and echelon forms
Two linear equations are called equivalent if they have the same solution set.
 Elementary row operations
1. Replacement: replace one row by the sum of itself and a multiple of another row
2. Interchange: interchange 2 rows
3. Scaling: multiply all entries in a row by a non-zero constant
- Row equivalent matrixes = als je de ene kan verbouwen/omschrijven tot de andere  row
operations are reversible  if the augmented matrixes of two linear systems are row
equivalent, then the two systems have the same solution set. SCHEMATISCH:
X X X X
O X X X
O O X X
O O O X
O O O O

Pivots ^^^^^^^^^^^^ also non-zero number
1 0 0 0 0
O 1 0 0 0
O O 1 0 0
O O O 1 0
O O O O 1


- A matrix is an echelon form if:
1. Non-zero rows above rows of all zeros
2. Each leading entry of a row is in a column tot the right of the leading entry of the row above it.
3. All entries in the column below a leading entry are zeros.

- A matrix is a reduced echelon form if:
1. Non-zero rows above rows of all zeros
2. Each leading entry of a row is in a column to the right of the leading entry of the row above it.
3. All entries in the column below a leading entry are zeros.
4. All leading entries are 1
5. Each leading entry is the only non-zero entry in its column.
Linear system  augmented matrix  (by row reduction) matrix in echelon form  (by row
reduction)
0 solutions: pivot in last column
1 solution: no pivot in last column
: pivots in all other columns
∞ solutions: no pivot in last column
: at least 1 other column without a pivot
Matrix in reduced echelon form (to determine solution set precisely)
- Theorem: every matrix is row equivalent to one and only one reduced echelon form. PROOF:
A matrix is row equivalent to many matrixes in echelon form but they have the same pivot position.

different pivot positions so it can’t be row equivalent.

,Week 1, les 2
1.3: vector equations
R2 has a set of points : (2,1)
R2 has a set of vectors : [21]  eind van de vector = coordinaat
- Vector space: (R2, +, x) kan je doen met vector, niet met coordinaat
Definition:
U = [ab] & v = [cd]
u + v = [a+c b+d]
(when R2 = vector space ^^^)
u = v if a=c en b=d
o = [00] in R2  R3 [000] etc. zero vector
e1 = [10] e2 = [01] in R2  unit vector
1 – [11]
- Algebraic properties of (Rn, +, x)
1. U+v=v+u
2. (u + v) + w = u +(v + w)
3. U+0=u
4. U + (-u) = 0
5. C (u + v) = cu + cv
6. (c+d)u = cu + du
7. C(du) = (cd)u
8. 1u = u
Given vectors v1 and v2, Vp e Rn we can make other vectors using addition and scalar
multiplication. We get vectors y of the form:
Y = c1v1 + … + cpvp for certain C’s
Y is called a linear combination of v1 … vp with weights c1…cp
- Definition
 Given vectors v1…vp
 Span {v1, v2, v3} = {c1v1 + cpvp}
All linear combinations
 Span {[2 1 ]} – all vectors on the line through 0 and [2 1 ]
 Span {[ 2 1 ] [ 1 2 ]} = R2
 R3 = span {e1, e2, e3}
 Span {o} = {o}
 Span o doorgehaald = {o}

, 1.4: the matrix equation
- Definition
Let A ben a mxn matrix (m = boven naar beneden). N columns from Rm
Let x =[x1 … xn]
Ax = [a1, a2, an] [x1, x2, xn]  same size!
Ax = b has a solution x b e span {a1, a2, an}  system with an augmented matrix [a1, a2, anb]
has a solution.
- Algebraic properties
1. A(u + v) = Au + Av
2. A(cu) = c (Au)
3. Ao = o
4. Ae1 = A [1 0] : a1
5. Ae2 = a2 etc
6. A1 = a1 + an = [row sum of row 1 of A , row sum of row n of A]
Under which condition does Ax = b have a solution for every b?  A has a pivot in every row 
no pivot in last column = solution
Ax = b has a solution for every b  b e span {a1, an}  span {a1, an} = Rm  [a1, a2, anb] 
echolon form
- Equivalent are (theorem)
1. Ax = b has a solution for every b
2. Span {a1, …, an} = Rm
3. A is row equivalent with an echelon form with a pivot in every row.

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

Guaranteed quality through customer reviews

Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.

Quick and easy check-out

Quick and easy check-out

You can quickly pay through EFT, credit card or Stuvia-credit for the summaries. There is no membership needed.

Focus on what matters

Focus on what matters

Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!

Frequently asked questions

What do I get when I buy this document?

You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.

Satisfaction guarantee: how does it work?

Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.

Who am I buying this summary from?

Stuvia is a marketplace, so you are not buying this document from us, but from seller daniquevangeest. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy this summary for $7.50. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

67474 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy summaries for 14 years now

Start selling

Recently viewed by you


$7.50  2x  sold
  • (0)
  Buy now