Summary
Cheatsheet and summary maths 2 conversion class
- Course
- Institution
Cheatsheet for exam and summary maths 2 conversion class.
[Show more]
Seller
Followarunthaskanagasabai
Reviews received
Content preview
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.
The benefits of buying summaries with Stuvia:
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
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
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 these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller arunthaskanagasabai. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $17.14. You're not tied to anything after your purchase.
Can Stuvia be trusted?
4.6 stars on Google & Trustpilot (+1000 reviews)
67866 documents were sold in the last 30 days
Founded in 2010, the go-to place to buy study notes for 14 years now