Samenvatting
Cheatsheet and summary maths 2 conversion class
- Vak
- Maths2
- Instelling
- Vrije Universiteit Amsterdam (VU)
Cheatsheet for exam and summary maths 2 conversion class.
[Meer zien]Voorbeeld 1 van de 2 pagina's
Voorbeeld 1 van de 2 pagina's
In winkelwagengesponsord bericht van onze partner
Voorbeeld van de inhoud
Intro §1Gaussian elimination = row echelon form Exercise 95: Exercise 99: Solve for x & y, and
Example 46 Gauss-Jordan elimination = reduced row-
write column format & solve
echelon form
graphically.
Remark The solution to Take-home message: a “solution” may be
the linear equation ax = b an empty set, a single number, multiple
is
(countably many) numbers, or a set of Solution:
infinitely many numbers. In other words, we
generalize the concept of a solution to a set
a)echelon form of this sys eq is:
of objects that is not necessarily countable.
Exercise 100:
Solution:
Example 45 The solution set of this system of b) sol u,v,w are: u =3, v = -2 & w = 1
§§
equations is the intersection of The solution set of these two equations
c) suppose u = lnx , v =lny, w = lnz,
the two individual solution sets: then x = eu, y = ev, z = ew.
seen in figure above.
b) single solution exist = NO Matrix notation §3:
The intersection of the two solution sets
Calculate intersection by substituting the c) RHS of 3 eq = 0? = YES
1
equation is the tuple (x,y) = (2, /2) d) sys eq in lineair combination of 2 column
x = 3 in x = 5−2y. Obtaining y = 1. vectors:
à read in graph.
à So, whereas y was
. à
Then we obtain:
a free variable (i.e. y ∈ R) in the individual solution sets, 1 1 1 Column vectors
there is only a single valid value for y in the 2×2−3× 2 = 4−1 2 = 2 /2
intersection of the two sets. The solution set is a set 1 e) Non-zero choice of RHS that allows the three
with a single tuple: {(3, 1)}. Geometrically, this 2+4× /2= 2+2 = 4
lines to intersect at same point:
corresponds to one point in the plane. The à so, this solution is correct
intersection of solution sets is represented
geometrically by the intersection of the lines that Obtaining solution with algebraic X=2 & y=3, gives col vector
mapped the solution sets of the individual equations, manipulations: format:
as shown in fig. 30. So, we see a consistent 2nd eq à x = 4 − 4y
Substitution this for x : Example:
correspondence between the algebraic, set-theoretic, . .
Examples matrix multiplication
and geometric objects. 𝑥 + 𝑦 + 3𝑧 = 12. 𝑥 + 𝑦 + 3𝑧 = 12 1) exercise 97
2𝑥 + 2𝑦 + 𝑧 = 9 𝑅2
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&⃗
= 𝑅2 − 2𝑅1 −5z = −15 /////////////////////⃗
𝑅3 = 𝑅3 − 𝑅1
Graphically solution
𝑥− 𝑦+ 𝑧=2 x−y+ z= 2
.
𝑥 + 𝑦 + 3𝑧 = 12 1𝑥 + 𝑦 + 3𝑧 = 12
Substitution this for y in 2nd eq: −5z = −15 /////////////⃑
𝑅2 ⟺ 𝑅3 − 2𝑦 − 2𝑧 = −10
x = 4−2 = 2. −2𝑦 − 2𝑧 = −10 − 5z = −15 Solution:
Exercise 88: Solve w/ Gaussian el, Solution set:
determine unique solution or infinite eq3: z = 3
Exercise 83: Solve graphically and algebraically solution set. fill in eq3 in eq2: –2y – (2*3) = –10, met y = 2
fill in eq3 & eq 2 in eq 1: 1x + 2 + (3*3)= 12, met x = 3 .
So, solution is: (x,y,z) = (1,2,3). 2) exercise 96
.
Same but with matrix:
.
Solution:
Solution 1:
à then write equations and get the solution as seen above.
Remark A system of linear equations with n
variables and m equations:
• has at least n −m free variables.
Exercise 90 (same as 88): • after Gaussian elimination has at most m pivot
elements. Solution 2:
Algebraic solution: from the first equation we Nullspace of a full column rank
• has n−#pivots free variables in its solution set.
derive y = x. Substitution in the second equation
Exercise 93: matrix:
yields 3x = 6, or, x = 2. Since y = x, also y = 2
1 When it is a square matrix:
equal # of equations and
Exercise 87: Solve graphically Solution: variables. Nullspace is equal
and give solution set. to the null vector 0.
Solution set: 114 c) Find the conservation relations 2 When it is a non square
Solution:
by performing this gaussian elimination matrix: # of rows must be
1st eq = 2nd eq (if both sides larger then # of columns,
* by -2) Last eq gives z = 7,5. Subst this in 2nd otherwise matrix could not be
full column rank. So, no free
So: 2nd eq does not add eq gives:
“extra info”, so solution set variables & null space will be
Subst this in 1rst eq gives:
can by specified using eq1 the null vector, also in this
only. à gives z=4, w =3, v = 2 & u =1. Last row LHS =0, same as vector(ȧ , ḃ, ċ ) case.
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