linear-algebra

[vast torrent]

dim(M)

[restive shuttle]

it is the rows?

so 3 rows, 3 dimensions?

[vast torrent]

No

Do you know the dimension of a space?

[restive shuttle]

no i dont

we werent introduced that

[vast torrent]

It's the minimum number of vectors you need to span the space

What's the dimension of R^n?

[restive shuttle]

n?

[vast torrent]

Yes

The dimension of a matrix

Is the dimension of the span of its columns

This is a 3x3 matrix

If its dimension is 3

Then the vector colums span R3

Which is bad

Do you see why it's bad?

[restive shuttle]

no i dont

[vast torrent]

Because the problem wants vectors that don't span R3

[restive shuttle]

my prof said that if a 3x3 spans r3, it just has to be consistent

[vast torrent]

That's absolutely false

Ill give a counterexample

[restive shuttle]

consistent for any (a, b, c)

[vast torrent]

$\begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$

[stoic python]

gfauxpas:

[vast torrent]

is this a consistent system

[restive shuttle]

no

[vast torrent]

why not

[restive shuttle]

0x cannot equal some b

[vast torrent]

if you dont like that example

$\begin{bmatrix} 1 & 1 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} a \ b \ c \end{bmatrix}$

[half ice]

It is true that the matrix the vectors make should be non-singular

That is, will reduce to identity

[restive shuttle]

so this has 2 free variables, so it's not consistent?

[stoic python]

gfauxpas:

[vast torrent]

why isn't this consistent?

[restive shuttle]

because there are 2 free variables?

[vast torrent]

what's your definition of consistent

[half ice]

It is worth knowing exactly why you'd care about rref to identity

[restive shuttle]

oh uh, having a solution

[vast torrent]

sure, what do you have a solution for

[restive shuttle]

you have a solution for a, 0, 0?

wait

b and c can't be anything

other than 0

[vast torrent]

why not

[restive shuttle]

since there are 0s

there's no (x,y,z) that makes b something other than 0

[vast torrent]

I think we learned different definitions of consistent

[restive shuttle]

what's your defn of consistent

[half ice]

I'll use (2,-1,3), (1,2,2), (-4,1,1) as an example.

If these are linearly dependent, then there's a non-trivial way to choose a,b,c such that:
a(2,-1,3) + b(1,2,2) + c(-4,1,1) = (0,0,0)

But that can be rewritten with a matrix

[2 1 -4][a] [0]
[-1 2 1][b] = [0]
[3 2 1][c] [0]

[vast torrent]

um, you have the order switched

[half ice]

Oop I do

Thx

Basically, the way that reduces decides what a,b,c need to be

[vast torrent]

okay so I learned inconsistent was a statement about a particular equation, not a statement about all equations with a matrix

[half ice]

If it reduces to identity, then a = 0, b = 0, c = 0 and that is the definition of independence

[vast torrent]

that seems to be the source of confusion

anyway matrices aren't equations

and knowing that a set of vectors doesnt span R2 doesnt require an equation

you just need to find the dimension of the matrix of columns of your vectors and make sure the dimension is < 3

that's where the row reduction comes in

[half ice]

If it doesn't reduce to identity, then there's a way to make a,b,c something other than 0 but still solve the equation, which is the definition of lin dep

[vast torrent]

the question isnt asking to show the vectors aren't lin. ind. baytrex it sounds like they haven't learned that phrase yet

have you?

[restive shuttle]

ive learned linear independence

[vast torrent]

oh good, can you restate your problem in terms of linear dependence?

[restive shuttle]

so i want to understand some things conceptually first. Would you agree linear independence for a set of vectors {v1,...vm} is when a1v1+...+amvm = 0 vector, has no non-trivial solutions?

[vast torrent]

that's the definition, yes

[restive shuttle]

so span of vectors, is the set of all their linear combinations

[nimble egret]

Sure

[restive shuttle]

and if all the vectors in r^n, are in the span{v1...vm} we can that that set spans R^n

[vast torrent]

bzzzzt

oh

R^n you mean

[restive shuttle]

yes

[vast torrent]

you mean if

[gray dust]

m & n subscripts

[vast torrent]

span{v1,...,vn}=R^n

is that what you mean

[restive shuttle]

yeaaa i think

[vast torrent]

but surely span{v1,...,vn,0}=R^n too so keep that in mind

[restive shuttle]

right

[vast torrent]

anyway if you want to use the answer key's solution my way is the way to go

through row reduction

but there's many ways to do it, all equivalent

[restive shuttle]

my row reduction has the answer key solution in it, and i am not sure why. : (

[vast torrent]

because it's turning the solution into a problem of choosing z1 and z2 such that the dimension of the matrix is less than 3

[restive shuttle]

OH

wait

so if the bottom row is all 0s

does that reduce the dimension

[vast torrent]

You need a tHeOrEm

[restive shuttle]

ok lemme look at all my theorems ;-;

[rotund jetty]

@vast torrent if you have a sec, I have a follow up question about the problem I'm working on.

So, I'm asked to prove that every element is its own inverse. My approach rn is

x + x = [1]x + [1]x = ([1] + [1])x = [0]x

however, [0]x isn't necessarily 0 right? like an earlier problem in this assignment actually had a scalar multiplication where [0]x wasn't 0. So I'm kinda like... completely stuck

[vast torrent]

The dimension of a matrix is the number of pivot rows in row echelon form

[restive shuttle]

pivot rows

what's that mean

leading variables?

leading terms?

[half ice]

You can also think of linear independence as "you can't write one of the vectors as a combination of the other vectors"

[vast torrent]

0 scalar times vector is always 0 vector in a vector space

[rotund jetty]

wait

[restive shuttle]

is the pivot rows = to the dimension

[vast torrent]

Doesn't say these algebraic structures are vector spaces

In fact they aren't

[half ice]

A basis of a vector space is a linearly independent set that spans the space

[rotund jetty]

it's a vector space

[vast torrent]

Oh

[half ice]

Important theorem: Any basis of a vector space has the same number of vectors in it

[rotund jetty]

it has a 0 element (which is 3), 1x = x, and it satisfies all the axioms

[vast torrent]

Yes okay but "0" is then not the zero vector

Since its not even in V

[rotund jetty]

OH WAIT YES

SCALAR 0 TIMES X GIVES 3

WHICH IS 0

FFS

[half ice]

So this gets us interested in this "special number". We call it the dimension of the space. The dimension is the number of vectors in any basis of the space

[rotund jetty]

I'm SO sorry that was a major dumbass moment

I should sleep

thanks again for the help :)

[vast torrent]

Uh this isnt reduced but

Pretend entry 2,1 was 0

The pivots are the first non zero entries in each row

After the matrix is put into row echelon form

It still isnt reduced one sec

Better

The pivots are the circled numbers

Pivot rows are rows with pivots

The theorem is that the dimension of a matrix is the number of pivot rows

In row echelon form

That's what your answer key is doing

The dimension of the matrix is what dimension you get from the space spanned by the columns

And since you want the span to be <3, you want less than 3 pivot rows in row echelon forn

[restive shuttle]

oh so, if i make the last row 0 0 0, then ive reduced the pivots and also the dimensions

[vast torrent]

You got it

[restive shuttle]

oh ok. thanks

[vast torrent]

You also see from.that theorem that no choice will have the dimension 1

Because you have 2 pivot rows no matter what

Right?

@restive shuttle i just remembered that the dimension of the matrix's space is called rank

I forgot the term, sorry

[pulsar turret]

I am working on a problem and have to show that a polynomial f could be rewritten as a linear combination of p1, p2 and p3. So what you do is (ap1 + bp2 + cp3 = f), I found one solution (a,b,c) but how do you know there are not more solutions?

is it because it is "Linear" that it only has 1 solution if so, my question still stands, how are you sure there is only 1 solution. And is there a way to visualize this? I am more of a visual learner so I'm having difficultys with an abstract subject as this itself

[dusky epoch]

is {p1, p2, p3} known to be linearly independent?

[pulsar turret]

it doesnt say

[dusky epoch]

then i'm gonna need to see what p1, p2 and p3 are

[dusky epoch]

,rccw

[dusky epoch]

ok at a glance this looks linearly independent to me

[pulsar turret]

linearly independent meaning the only way to get to the solution f is by changing (a,b,c) in my example?

[dusky epoch]

no

linearly independent means linearly independent

it means that $c_1 p_1 + c_2 p_2 + c_3 p_3 = 0$ only if $c_k = 0$ (where $k = 1, 2, 3$)

[stoic python]

Ann:

[dusky epoch]

i mean ok suppose you have your expression $f = ap_1 + bp_2 + cp_3$

[stoic python]

Ann:

[dusky epoch]

and suppose there's a different expression of $f$ as a linear combination of $p_1, p_2$ and $p_3$: $$f = a'p_1 + b'p_2 + c'p_3$$

[stoic python]

Ann:

[dusky epoch]

subtract these two, and you'll get $(a - a')p_1 + (b - b')p_2 + (c - c')p_3 = 0$

[stoic python]

Ann:

[dusky epoch]

and by linear independence, you'll get that $a-a'=0$, $b-b'=0$ and $c-c'=0$ so the two expressions were actually the same all along

[stoic python]

Ann:

[pulsar turret]

idk why I'm having difficultys with this

so the difference of f and f' is the coeficient?

[dusky epoch]

there is no f'

you found one solution, (a,b,c)

i said

what if there was another

and called it (a',b',c')

but then it turned out mine and yours were actually one and the same

[pulsar turret]

how do you know?

[dusky epoch]

reread what i wrote lmao

i don't want to go full broken-record here

[pulsar turret]

ah ok

so you only have 1 solution if your p1, p2 and p3 is linearly independent?

[dusky epoch]

yes that's the whole point

[pulsar turret]

def of linear independence is when you sum of all vectors equals zero vector when only your coeff equal zero. Is it still linear independent its not the zero vector?

for example like (a) this is a polynomial function of degree 2

[dusky epoch]

you're missing the point

[pulsar turret]

it is right? because zero vector doesn't neccerally mean 0

[dusky epoch]

ugh

[pulsar turret]

def of linear independence is when you sum of all vectors equals zero vector when only your coeff equal zero. Is it still linear independent its not the zero vector?
@pulsar turret well I ment if it's not zero instead of zero vector at the end.

[warm wedge]

Any good videos that can teach me how a projection matrix, view matrix and a model matrix work?

I'm having a hard time understanding them

I just started learning linear algebra

I've recently finished 3blue1brown's playlist

[dusky epoch]

maybe it's not worth your time to jump headfirst into that stuff? you gotta spend some time with more basic linear algebra concepts to build up a better intuition

[warm wedge]

What are the things that I'll need to know to be fluent with 3d graphics

Can you tell me what should i do?

Where should i start?

I understand some of it

From 3blue1brown's playlist

[dusky epoch]

pick up a linear algebra textbook, read it and do the exercises. 3b1b's playlist is good but it's not meant as a standalone course.

are you self-studying linalg?

[warm wedge]

Yep

I wanna make video games

And i wanna be as perfect as possible

[quartz compass]

making videogames is an entirely different task

[warm wedge]

Not entirely

[quartz compass]

entirely

[warm wedge]

3d games

They make use of linear algebra

Depends on what video game you're making honestly

[dusky epoch]

what merosity is trying to say is that there's a lot more to video game development than linear algebra

[warm wedge]

2d games need very little linear algebra

Oh

So, what do i start with, any suggestions?

[quartz compass]

start learning unity or blender

[warm wedge]

I wanna be really thorough with this subject

Blender?

[quartz compass]

it's a 3d modelling software

[warm wedge]

I know

But why blender

[quartz compass]

cause you seem clueless and it's free

[warm wedge]

Okay

[quartz compass]

houdini is cool though but is hardcore math

[warm wedge]

And what about the mathematical part? Any good textbooks you recommend?

[quartz compass]

I already gave you my recommendation

[warm wedge]

Okay

[quartz compass]

drop the math if you want to make a videogame

you can learn math but it's just procrastination at this point

[warm wedge]

Okay but I'm working with opengl

It's a low level api

Meaning a lot of math

[quartz compass]

don't

no reason to use opengl to make your first game

[warm wedge]

Okay

I needed to learn blender anyway

[quartz compass]

finishing a game is a huge task

[warm wedge]

And i guess i can pick opengl back up when I'm a pro

I know

[quartz compass]

I would almost say don't really even learn blender

and just use unity or something to that effect

[warm wedge]

I kinda wanted to make my own engine and start working from that

[quartz compass]

no no no

this is exactly what I was talking about

making a game engine and making a game are entirely different things

[warm wedge]

I see

[quartz compass]

if you want to make a game engine, better off trying to help contribute to Godot

[warm wedge]

You actually made me realise a big part of my course

[quartz compass]

rather than reinventing the wheel

[warm wedge]

Well

I guess that's fine

[quartz compass]

Don't get me wrong, understanding all this stuff and wanting to do everything is nice but

you'll never actually finish anything if you go about it this way

at least if you can finish a project, you can refine it and deepen it each time you go through the process from start to finish

you're just not going to be able to understand all the little details of every single little thing

[warm wedge]

Well i guess

[quartz compass]

you'll be like 90 years old by the time you finish your first game

[warm wedge]

That sounds like quite a big hinge breaker

[quartz compass]

a big part of making a game is marketing it, promoting it, actually playtesting it to make sure people enjoy it and it's fun and designed well, etc

[warm wedge]

Yeah

[quartz compass]

you could easily spend years just learning about design

[warm wedge]

Hmm

I guess I'll start working from unity then

[quartz compass]

good luck and have fun

[warm wedge]

Thanks

But I'm still a bit curious

About one thing

Might be kinda pretentious

But one thing that keeps coming to my mind is minecraft

Notch made everything in a week after understanding it all

Yeah he may have taken some years to understand it

All the things that went into jt

It

But it is definitely possible

[dusky epoch]

don't compare yourself to others

[warm wedge]

Yeah

You're right

I'll start with unity

[wintry steppe]

Hi, I have a question. I'm a first year taking linear algebra for the first time. So how do I know if a two vectors with any linear combination can represent the entire R^2?

[indigo cloak]

@wintry steppe is that peepeepoopooman in your pic lmao

[feral grove]

this depends on what you've done, the answer ends up being that they need to be linearly independent, but for a class you'd probably need to prove that

[obsidian jackal]

I have a short question about this equality

Shouldnt be AA^-1 = I?

And with the I the same size with A

How is Ix=x shown in the next image?

Shouldnr the product of Ix the same size with A?

[wintry steppe]

@indigo cloak yea it is LOL

[half ice]

@obsidian jackal
Let's say you have the matrix equation
Ax = b

You can multiply both sides on the left by A':
x = A'b
Giving you the solution

[dusky epoch]

Ix = x is kind of the point

that's the fundamental property of the identity matrix

it's the matrix equivalent of the number 1

[obsidian jackal]

So the size of I doesn't matter?

[dusky epoch]

wdym

clearly I here is I_2 bc no other identity matrix would make sense size-wise

[obsidian jackal]

1_2 bc?

[dusky epoch]

what

no i said I_2 not 1_2

[obsidian jackal]

What does that mean?

[dusky epoch]

identity matrix of size 2????

[obsidian jackal]

Oh

[dusky epoch]

why would i even say $1_2$. $1_2$ doesn't make any sense.

[stoic python]

Ann:

[obsidian jackal]

I am not familiar of typing with matrices

Sorry

What I mean was with A with the size of m_n and AA'=I_n

Shouldn't the I_n transform the column vector x to a matric of m_n?

[dusky epoch]

m_n

no

m by n

or $m \times n$

[stoic python]

Ann:

[dusky epoch]

not "m_n"

[obsidian jackal]

Oh okay

[dusky epoch]

anyway, $I_n$ is $n \times n$ and $x$ is $n \times 1$

[stoic python]

Ann:

[dusky epoch]

$(n \times n)(n \times 1) = (n \times 1)$

[stoic python]

Ann:

[obsidian jackal]

Oh yeah when I multiply it from the left

Okay i was just confused

Thank you

[heady eagle]

Is the ∈ symbol the same as < or related?

[sonic osprey]

no

[dusty vale]

How to find area of a triangle with vertices A(-1,2,3) B(3,0,2) C(1,4,5) ?

[south wadi]

Try the #preschool channel

[dusty vale]

@south wadi where is this channel?

[south wadi]

i guess they dont have one

[gray dust]

#geometry-and-trigonometry fits @dusty vale

[pale shell]

What is the signifigance of the dot prpduct

[kindred oxide]

can someone help me out with this problem?

[vast torrent]

@kindred oxide what have you tried, where are you stuck, etc

[kindred oxide]

ive tried simplifying the 3 vectors (ie dividing the left vector by 4, the middle by 3 and the right by 2) and ive tried to get h to equal the other x values. it didnt work. i feel like im missing something. I tried entering "none" but that was also wrong

[vast torrent]

when you say you were trying to get h to equal the other x values

how would that work, if there are 2 other x values?

You're not completely off, but your approach is unfocused

what are you trying to do

[kindred oxide]

if you are trying to find the value when h doesnt span R3, dont you want to get 2 of the lines to equal each other?

idk im kind of lost

[vast torrent]

these are vectors, not lines. a span of one vector is a line

it is true that

[kindred oxide]

my bad i meant vectors

[vast torrent]

if a1, a2, a3 don't span R3

then one of the vectors is a linear combination of the others

but here it's not easy to do that

if you happen to find an h that works, then great

but I recommend another approach if theres no obvious h. And to me, there isn't an obvious h

let's define a matrix

[kindred oxide]

whats your approach?

[vast torrent]

$\mathbf A : = \begin{bmatrix} & & \ \mathbf{a}_1 & \mathbf{a}_2 & \mathbf{a}_3\ & & \end{bmatrix}$

the columns of A are the 3 vectors

what's the rank of A?

[stoic python]

gfauxpas:

[vast torrent]

whats A's rank?

[kindred oxide]

i dont understand what you mean

[vast torrent]

have you ever encountered the term "rank" of a matrix

[kindred oxide]

no

[vast torrent]

okay

how about "column space"

[kindred oxide]

is it the n x m thing>

?

[vast torrent]

the column space of a matrix is the span of its columns

so the column space of A is the span(a1,a2,a3)

[kindred oxide]

that makes sense

[vast torrent]

the rank of A is the dimension of the column space

so, I ask you

what is rank(A)?

[kindred oxide]

R3?

[vast torrent]

it's a number

the dimension of a vector space is a number

[kindred oxide]

3?

[vast torrent]

well it's either 3 or 2

[kindred oxide]

because you have 3 columns

[vast torrent]

but if the vectors a1,a2, and a3 span a 2 dimension space

then rank(A) = 2

[kindred oxide]

since we want an h where its not in R3, would you use the rank 2?

[vast torrent]

the question is asking for which h is the dimension of span(a1,a2,a3) equal to 2?

agree or disagree

[kindred oxide]

i agree

[vast torrent]

okay

know what row reduction is?

[kindred oxide]

yea

[vast torrent]

Theorem

the rank of a matrix is the same as the rank of the row reduced form of the matrix

ALSO

it's the same as the number of pivot rows when row reduced

[kindred oxide]

so would you set the vector with h as the augmented row?

and then reduce the matrix?

[vast torrent]

no augmenting

you want to row reduce A in such a way that it only has 2 pivot rows

know what a pivot row is?

[kindred oxide]

yea

[vast torrent]

that's how I would do it

there are other ways but I think this way is easiest

[kindred oxide]

let me try

[vast torrent]

another way would be to take the determinant, it will be a function of h, and find the h where the determinant is 0

$\begin{vmatrix} 8 & h & 2 \ -20 & -12 & -18 \ 4 & 3 & 4 \end{vmatrix} = 0$

[stoic python]

gfauxpas:

[vast torrent]

that's another approach

that uses a theorem that if a square matrix doesn't have full rank, its determinant is zero. if and only if actually

there's an advantage to the determinant way

in that row reductions and column reductions are allowed, not just row

[formal pond]

I put... (a) REF, (b) RREF, (c) RREF, (d) RREF... correct?

[dusky epoch]

yes

[formal pond]

okay also are these RREF too?

[dusky epoch]

are they?

you seemed able to check the previous four matrices against the definitions of REF and RREF

what's giving you trouble with these two?

[formal pond]

i have something else written for these answers in my notes so was feeling conflicted if I wrote them down incorrectly in lecture. I wrote neither for the first one and forgot to write an answer for the second one.

[dusky epoch]

wrong bc those are both in rref

[formal pond]

the neither one is throwing me off. i may have just miswrote it earlier.

okay for sure

thanks

[formal pond]

So I used elementary row operations to change the original matrix into REF but I need to change it to RREF and am having some trouble because there is a 2 in the column in place of a 0 in the final matrix. Any tips? I also tried subtract R1 by 1.

[gray dust]

replace R1 w/ R1-2R2

[formal pond]

Okay thanks 👍

[gray dust]

[inland warren]

This is technically a question on my quantum homework but I figured it's more of a linear algebra question overall

Can this be done the same way a diagonal matrix is exponentiated by just applying the exponential across the diagonal?

It doesn't seem like it'd be that easy and I'm unsure if I can perform row operations on a matrix that is an exponent

[dusky epoch]

no it can't

[quartz compass]

that matrix satisfies the equation A^2=-I

[dusky epoch]

but you can make use of that ^

[quartz compass]

you can simplify it via power series fairly easily

[inland warren]

A being the matrix?

[dusky epoch]

yes

[inland warren]

When I square the matrix, I get a diagonal matrix with both entries being -1

How do I get it to satisfy the equation?

[vast torrent]

what's A^3 and A^4?

A^3 = A^2A

[inland warren]

A^3 is {0,-1}, {1, 0} and A^4 is {1,0}, {0,1}

[vast torrent]

so A^2 = -I, A^3 = -A, and A^4 = I

and then it cycles

[inland warren]

This is giving me complex number vibes lol

[vast torrent]

well sure, your matrix A is a square root of -I

[quartz compass]

more than a vibe I'd say

[vast torrent]

not a coincidence that it's giving you vibes

[inland warren]

Hehe

Very cool

[half ice]

The answer also will be very complex number-y

[vast torrent]

so because the exponential function is absolutely convergent\

you can split the series into parts

though I wonder if that's the easiest way

[quartz compass]

well if you know the power series for e^x, sinx, and cosx should be pretty straight forward

[inland warren]

Should I go up to 4th order to encapsulate the whole cycle?

Also for future reference, is this a somewhat well-known matrix that I should memorize? It's super neat that it has the same properties as a complex number

But I'm not sure if I would have been able to recognize the pattern without everyone's help here lol

[vast torrent]

Knowing the existence of matrices such that M²=-I is more important than knowing what they exactly look like

And you could brute force this with diagonalization if you didn't think of this

[inland warren]

Alrighty, I'll try to look into it

Would it end up like this?

[feral grove]

uh are those 1's I's?

[inland warren]

Unless it's got a dot over it, it's a 1 for this particular problem lol

[dusky epoch]

the first term should be I not 1

identity matrix

[inland warren]

ooooh

Is that specifically for expanding out series that involve matrices?

I guess because it would be A^0 and that ends up being I

[feral grove]

precisely

[inland warren]

Gotchaaa

It's still somewhat lost on me why I can say that a matrix is equal to a number or an imaginary number. That seems kind of illegal to me but I'm not the most knowledgeable on this type of thing

The matrices I get in the power series expansion do very closely resemble the Pauli matrices in QM but I think that the problem is now starting to delve more into the physics side of things

[vast torrent]

It's not that you're treating the matrix as a number

It's that since the entries of the matrix cycle you can examine each entry of the matrix as a power series of a number

The numbers being 0, 1, or -1

[dusky epoch]

It's still somewhat lost on me why I can say that a matrix is equal to a number or an imaginary number. That seems kind of illegal to me but I'm not the most knowledgeable on this type of thing
it's not quite formally correct

[chrome current]

I've tried for 2h to find the matrix X, please help

[vast torrent]

Well the matrix to the right of X is invertible

So it will probably be easier to right multiply both sides by [[-7,-3],[2,1]]-¹ so you're dealing with 3 matrices instead of 4

[chrome current]

I've done that but I get stuck after

[vast torrent]

So what's the new matrix equation after doing that

I used a computer, we get

[vast torrent]

good

[chrome current]

Is that right?

[vast torrent]

I like that column of zeros, I have a feeling it will make life easier

yes. I didn't feel like doing it by hand, the computer gave that answer

let's write X = [[a,b],[c,d]]

$\begin{bmatrix} 6 & -2 \ -3 & 1 \end{bmatrix} \begin{bmatrix} a & b \ c & d \end{bmatrix} = \begin{bmatrix} -4 & 0 \ 2 & 0 \end{bmatrix}$

[stoic python]

gfauxpas:

[vast torrent]

we should be able to get 4 equations in 4 unknowns for this

and the zero column should make 2 of those equations simple

what are the 4 equations?

[vast torrent]

okay that looks good

let's do the 0 entries first

$\begin{cases} 6b - 2d = 0 \ -3b + d = 0 \end{cases}$

[stoic python]

gfauxpas:

[vast torrent]

the other equations are

$\begin{cases} 6a - 2c = -4 \ -3a + c = 2 \end{cases}$

[stoic python]

gfauxpas:

[chrome current]

I know it sounds dumb, but I stuck there, I get answers like 6b-6b=0

[vast torrent]

it doesnt sound dumb, you're thrown off because the answer isn't unique

are you asked to find an X, or all possible X?

[rose bay]

the det from the matrix - can anyone tell me how this happend?

[chrome current]

an X

[vast torrent]

mag, the chat is in use

you're confused I think because you're looking to find a unique answer and the answer isn't unique

$\begin{cases} 6b - 2d = 0 \ -3b + d = 0 \end{cases}$

[stoic python]

gfauxpas:

[vast torrent]

you just need to find any solution to this

and note they're effectively the same equation, the second one times -2 is the first equation

so all you have are d = 3b

[chrome current]

my book says to find the X, but that might be wrong

[vast torrent]

and since b and d aren't in the other 2 equations, you can choose them arbitrarily

so let's introduce a parameter then

[rose bay]

@vast torrent so? this is after all a public chat why would I not be allowed to ask my question as well? if you wanna have a 1o1 conversation you should go into a private chat why so toxic?

[vast torrent]

if b = t, then d = 3t, where t is any real number

$X = \begin{bmatrix} ? & t \ ? & 3t \end{bmatrix}$

[stoic python]

gfauxpas:

[vast torrent]

what about the first column then

[chrome current]

really don't know

[vast torrent]

do the same thing

$\begin{cases} 6b - 2d = 0 \ -3b + d = 0 \end{cases}$

[stoic python]

gfauxpas:

[vast torrent]

relate b and d

err thats what we just did

$\begin{cases} 6a - 2c = -4 \ -3a + c = 2 \end{cases}$

[stoic python]

gfauxpas:

[vast torrent]

relate a and c

[rose bay]

pls tell me when you are done so that I know when I am allowed to post my question again

[vast torrent]

just solve for a in terms of c or solve for c in terms of a

if I told you x + y = 3

and asked you to solve for y

you would tell me y = 3 - x

does that make sense?

[vast torrent]

exactly

a and c are related by c = 2 + 3a

and a doesnt have a constraint

so if a is any real number s

c = 2 + 3s

$X = \begin{bmatrix} s & t \2+3s & 3t \end{bmatrix}$

[stoic python]

gfauxpas:

[vast torrent]

any real numbers s and t will make X a solution to your matrix equation

[chrome current]

ohhh snap, I think I understand now

[vast torrent]

you were tripped up, I'm pretty sure, because the answer isn't unique

you were trying to solve for a b c and d when in fact you can only reduce it to two variables, you cant solve for 4 constants

[chrome current]

yee I havn't seen that before

[vast torrent]

so I think that's why you were confused

[chrome current]

yeh

[dusky epoch]

$X = \begin{bmatrix} 0&0\2&0 \end{bmatrix} + \begin{bmatrix} 1\3 \end{bmatrix} \begin{bmatrix} s&t \end{bmatrix}$

[chrome current]

Big thanks man!!

[stoic python]

Ann:

[dusky epoch]

uwu

[vast torrent]

oh that's a nice way to say X, do you see why that's the same Den

[chrome current]

noo

[vast torrent]

@rose bay you're up at bat

[rose bay]

ty

[vast torrent]

try multiplying it out using matrix multiplication and you'll see

it's a 2x1 matrix times a 1x2 matrix so you're adding two 2x2 matrices together

[chrome current]

oh ye

[rose bay]

I get log(6+root(x-1)-(x-1)) and not what my prof got in his solution can anyone tell me how to do this correctly? :)

[dusky epoch]

you got what

[chrome current]

I'll spend some time getting good at it now, big thanks again!

[vast torrent]

[rose bay]

btw I got it just in case someone also needs this at some point

the task was to find the x when the result is 2

[limpid osprey]

Is this for linear equations as well

[wintry steppe]

yes

solving for A

[limpid osprey]

Okay so

I was solving 2 linear equations using the substitution method.

But got different values when I change the values I try to find first.

It should be
A=60
B=40

Intended answer

But when I try to find B first, I get B=60

My teacher tried it too.

But thats also what she got.

Problem =
You need 5% solution in 100ML
You only have 2% and 7% solutions, how much ML of 2% and 7% do you need to get 5% solution in 100ML.

[junior nacelle]

40 ml of 2%, 60 mls of 7%

@limpid osprey

[limpid osprey]

It should be

60ml is 2%

[junior nacelle]

@limpid osprey that is 4% final solution, not 5% right?

[limpid osprey]

Yeah

I just checked as well

[dry spear]

https://cdn.discordapp.com/attachments/227519788946161664/670355934899601421/unknown.png https://cdn.discordapp.com/attachments/227519788946161664/670356078713765929/image0.jpg

hey I've been trying to solve this for 2 days now I don't know if I'm even doing it right

tag me if you can help

[quartz compass]

not sure how you mean "within the matrix"

but yeah, a multiple of the first column will be added to a multiple of the second column to get your resulting vector

linearity is really just the property of saying if you know where your basis vectors transform then you know where all vectors transform

which is pretty handy

[wintry steppe]

anyone knows this? http://immersivemath.com/ila/index.html

[gilded bobcat]

@dry spear do you still need help?

[dry spear]

@gilded bobcat yes

also how do I solve it with so many variables

[gilded bobcat]

You don't have to,

just solve it for a and b

and your expression for V in the 2nd image is not correct

[dusky epoch]

can i see the original problem

[gilded bobcat]

https://cdn.discordapp.com/attachments/227519788946161664/670355934899601421/unknown.png its this one
@dry spear

[dusky epoch]

2a or 2b?

[gilded bobcat]

both I guess

[dry spear]

a and b

[dusky epoch]

ok let's start with a

what's giving you trouble in a

[dry spear]

there's so many variables in the equations I got

[dusky epoch]

how

can you show your equations

[dusky epoch]

...

👻

[thorn prairie]

How i do the multiplication?

its 2 different ones

[pliant harbor]

The coefficients of the vectors become column coefficients in a sum.

So you have 5(5, -2)-(1,-7)+3...

[dusky epoch]

@thorn prairie do you know how matrix multiplication works?

[thorn prairie]

@dusky epoch i multiply each line with the column?

[dusky epoch]

...that's one way of putting it, i guess

however, horizontals in matrices are usually called rows rather than lines in english

[thorn prairie]

ok nice

@dusky epoch i do linear algebra in greek so I dont know the terms

so il have a 2x1 matrix?

with the sum of all on the first row

and sum of all on 2nd row?

[dusky epoch]

the first one? it's a 2x1 matrix yes

not sure if "sum of all" is an appropriate description of the content

[thorn prairie]

it will be 5 ( 5 -2 ) + (-1)( 1 -7 ) + ....

[dusky epoch]

the first entry will be 5*5 + 1*(-1) + (-8)*3 + 4*(-2)

[thorn prairie]

ok nice i got it

[dusky epoch]

the second entry will be (-2)*5 + (-7)*(-1) + 3*3 + (-5)*(-2)

[thorn prairie]

so the final answer will be (8 16)

[dusky epoch]

assuming you didn't fuck up your arithmetic, yes

[thorn prairie]

@dusky epoch tbh looking at my theory they didnt explain it to mee

im trying to guess kind of based on other stuff we wrote in class

and the 2nd will be something like ( 1 -9 12 -4 )

[gilded bobcat]

shouldn't it be transpose of (8 16){assuming it is correct}?

[thorn prairie]

8 16 as a column

8

16

[gilded bobcat]

ok then

[dusky epoch]

here

[thorn prairie]

and 1
-9
12
-4

i see

i dont think we will do with the 2nd matrix being more than a column

Thanks tho

saved the picture haha

[gilded bobcat]

can someone pls explain why will we be able to remove one of w's here?

[thorn prairie]

honestly the theory is crap

[dusky epoch]

can you show the statement of the linear dependence lemma @gilded bobcat

[thorn prairie]

Find the linear system corresponding to the equation Ax = b and find the solution x as a vector.

its basically doing the multiplication again?

[dusky epoch]

yes

[thorn prairie]

the answer of the multiplication is the x vector?

[dusky epoch]

Ax = b, where x = [x1; x2; x3]

you'll have three equations

[gilded bobcat]

@dusky epoch here

[dusky epoch]

well

[thorn prairie]

so finding the linear system is just writing those numbers again just putting x1 , x2 ,x3 up front?

[dusky epoch]

this lemma tells you that, since {w_1, ..., w_n} is LD, you can find a vector in there that is in the span of what came before it and removing which from the list won't diminish the span @gilded bobcat

@thorn prairie ...yes?

bad wording but yes

[thorn prairie]

hm ok i see

i see that the exercises are easy but they are saying same things differently and i dont really understand the difference

[gilded bobcat]

yes, but let us assume that w's form a basis, then w's are LI too right? and since u belongs to V then we can write u as a linear combination of w's,

[dusky epoch]

no??

wait

ok wait

misread, sorry

yeah so the list with the u1 is LD

[thorn prairie]

does u derive from the columns of A?

[dusky epoch]

so it's from THAT list that we can throw out a vector

[gilded bobcat]

rather in a linear sum of u and w, u will be the one to have a non 0 coefficient

[dusky epoch]

but it can't be u1

well

okay hang on

ugh

kinda hard talking to two people at once

[thorn prairie]

haha sorry

[gilded bobcat]

i can wait tho

[thorn prairie]

i can go do calculus haha

[dusky epoch]

the vector that the LD lemma guarantees you can throw out

can't be the first in the list

unless the first vector in the list is 0

[gilded bobcat]

can't be the first in the list
@dusky epoch why is that?

[dusky epoch]

bc the span of what comes before the first element in your list is {0}

the span of an empty list is the zero subspace

[gilded bobcat]

dose the order really matter?

[dusky epoch]

as per the statement of the lemma

it does

with the way it is stated

order does matter

[gilded bobcat]

but the lemma says that j = 0(1)m

[dusky epoch]

no it doesn't

ok look

without symbols

the lemma says the following

IF you have a LD list of vectors
THEN there is a vector in that list satisfying the following:

• it is in the span of what comes BEFORE it in the list
• throwing it out of the list will not affect the list's span

do you understand this or not

[gilded bobcat]

yes, now I understand, I think, I've misinterpreted the 1st statement of the lemma

[dusky epoch]

so what i was gonna continue with

...unless the first vector in the list is 0, which u1 is not, by virtue of being part of the LI list of u's.

[gilded bobcat]

yes now I get it

thank you

makes perfect sense

[thorn prairie]

does u derive from the columns of A?

it asks if i can get the left column on any of the right ones?

for example doing 2r1 + r2 >> r2 , r1 + r3 >>>r3

and then -3/4r3 + r1 >>> r1

the left column of A gives 0 4 4 if im not mistaken

this is the solution?

By doing Ax = b and finding the solution (c) what does c represent?

[thorn prairie]

what does it mean when it asks me to find the span of a matrix?

[dry spear]

https://cdn.discordapp.com/attachments/227519788946161664/670355934899601421/unknown.png https://cdn.discordapp.com/attachments/227519788946161664/670356078713765929/image0.jpg

@dusky epoch sorry I went to bed

I really need help with this though, it's due tn

my equations are there too

[sinful quiver]

If you keep moving your basis vectors closer to each other until the orientation shifts and you keep moving it in that direction will the determinant ever become positive again? Will it only be positive up until the initial orientation is disturbed and never again or when it returns to the initial orientation it becomes positive again?

Like if you were to graph determinants based on that sort of movement of your basis vectors is it a wave or a curve that never changes direction

[pale shell]

Notation to show two equations are the same

???????

[slow scroll]

= ?????

[south wadi]

if there is infinite solutions than b is a linear combination of a1,a2,a3 right?

[nimble egret]

what do you mean?

[south wadi]

i solved the matrixc

[nimble egret]

ok

[south wadi]

and got 0 0 0 0

i would send u a pic but the wifi is total garbadge at this university and wont connect to my phone

@nimble egret

it laoded in

see i got all zeros on the bottom

[nimble egret]

which is a true statement

so yes

[south wadi]

ok so yea

[sharp merlin]

Does anyone know how to do this

[nimble egret]

what do you think the answer should be?

[sharp merlin]

nvm got it

[novel sapphire]

How is linalg compared to multi?

[half ice]

Lin alg is the easier class imo

[novel sapphire]

That’s good

[half ice]

Also the more useful class

[novel sapphire]

Is it more theory or applications?

[half ice]

Take Lin alg if you haven't yet, which is advice for everybody

[novel sapphire]

Ok thanks

[slow scroll]

some LA courses focus more on theory, some focus more on applications

[half ice]

I'm guessing you're looking at applied classes? You'll see some theory, but mostly applications

[dusky epoch]

@dry spear welp, now i was asleep when you posted that

[gray dust]

dw spectex's hw has been dealt with

[dry spear]

^

I have another q though

an integral

I'll post in one of the question channels

[wintry steppe]

why is this problem not in standard form?

is it because it says minimize not maximize?

[dusky epoch]

what's your definition of standard form

and what's your definition of canonical form

[wintry steppe]

is it because the last inequality is >= not <=?

[sand zenith]

Hello i have a very quick Question: Are the eigenvalues of a diagonal-matrix the values on the diagonal?

(i think yes)

[jagged pendant]

yes, the eigenvectors are standard basis vectors (0,...,0,1,0,...,0)

[sand zenith]

Thanks

i thought so ^^

[dusky epoch]

is it because it says minimize not maximize?
yes

is it because the last inequality is >= not <=?
no, the last inequality is a sign constraint

@wintry steppe

[vast torrent]

@spice storm

Are you familiar with the vector space axioms

[spice storm]

No

[vast torrent]

what class is this for then

calc 3?

[spice storm]

Vector analysis

[vast torrent]

do you know what a "vector" is

I ask becausew

they're usually defined as "objects satisfying the vector space axioms"

[spice storm]

Well we haven’t covered anything that is related to axioms. When I took Intro to real Analysis my professor skipped that section

[vast torrent]

okay,. so what's a "vector"?

[spice storm]

Is an object that has both magnitude and direction.

[vast torrent]

okay, we'll go with that for now

that's part of the definiton but it's not enough

you also need, for example

the existence of a 0 vector with no magnitude such that for any vector v, v+0 = 0 + v = v

and you also need the existence of a negative vector -v for every v such that v + (-v) = (-v) + v = 0

follow?

without these two properties/axioms, the objects aren't vectors

[spice storm]

I mean I do but I would rather have someone tell me in person what it really means so I can ask questions or see it in person. Visual person

[vast torrent]

those are called tutors and they generally charge per hour

so you have these points in space

(x,y,z)

such as (-1, 1/2, 3)

and I'm considering them all as a group, called a vector space

I'm telling you that in addition to magnitude or direction you also need

a zero vector , with the above property

and a negative vector for each vector, such as the above property

your assignment is asking

for any arbitrary vector a = (a1,a2,a3)

the negative vector -a such that a + (-a) = 0

is the vector (-a1,-a2,-a3)

in other words

(a1,a2,a3) + (-a1,-a2,-a3) = 0

make sense or no?

so you just have to prove those vector add to 0

equivalently, you can prove that

-(a1,a2,a3) = -1(a1,a2,a3)

because if you prove that you can distribute the -1 inside the vector

you may find that easier or harder to prove than the first way

[wintry steppe]

I'm a bit confused by how many solutions this would be

[dusky epoch]

what do you think

[wintry steppe]

unique soluion

I tried to write down everything, but my answer for this question is wrong. And i'm not sure where i'm going wrong

[dusky epoch]

if you claim there is a unique solution here

then what would x_1 be

[paper sparrow]

good evening ,guys

i need help with some questions and i will be really grateful for anyone who will help me

[sand zenith]

For matrices is $Q := (Q_1 \cdot .... \cdot Q_n) * b = (Q_n \cdot (... \cdot (Q_1 \cdot b))$ or the other way round?, e.g. $Q_1 \cdot ...$

[stoic python]

oswald:

[sand zenith]

its really simple question but me really dumb

[elder robin]

I struggled with this one for a bit, but I think I got it now. I have to find the equation of the plane between two parallel lines. Can anyone check my work to see if it is right?

I think it make sense, but i'm not sure if my ways of checking are actually verifying anything

[sharp merlin]

Anyone know this

[feral mountain]

t

should be $\theta?$

[stoic python]

EpicGuy4227:

[boreal crescent]

could someone explain what the notation on the left implies

i get that overall it means that F is a transformation which is being applied to a function g and is evaluated at a value x

and that C(inf) implies its differentiable infinitely many times

but what does [0,1] --> [0,1] imply

only for functions defined on the [0,1] domain or what

[gray dust]

infinitely d/dx-able on the interval [0,1]

[boreal crescent]

like interval in the sense

a restriction on the domain?

so the function and its image function will only be infinitely many times differentiable for x between [0,1]?

[gray dust]

all the elements in F's domain are functions that are for sure infinitely d/dx-able on [0,1]. outside that interval, no one can say what happens there

[boreal crescent]

ohh gotcha

so if that statement is false and i have to provide a counter example

i must choose a value for x from [0,1] to prove it to be incorrect

[gray dust]

what statement

[boreal crescent]

the whole question is i need to prove if F is a linear transformation, or provide a counter example if its not

i know its a LT, but just assuming it wasnt, i would need to take a value for x and y between 0 and 1 and show that values differ

[gray dust]

if you did that, you'd actually show that F somehow fails to work under its own definition, which i don't think will happen

i'm quite sure the hw wants you to focus on showing the linear part of the linear transform

oh if you're talking about showing F isn't linear by plugging specific x values into F(a+b) & F(a)+F(b) for some elements a,b in F's domain, then that'd work

[boreal crescent]

yea

thanks

[gray dust]

yw

[ionic dust]

Is there some form of failing the horizontal line test for planes?

[dusky epoch]

wdym

[ionic dust]

a one to one function

[dusky epoch]

can you be more precise about what exactly you're looking for

[ionic dust]

so

when we have more than one input map to the same output

in R2 you can see it on a graph failing the horizontal line test

so in R3 what would that look like geometrically?

[dusky epoch]

by "in R^2", do you mean "considering the graph of a function R -> R"?

if so, then what functions are you considering whose graphs lie in R^3?

R^2 -> R or R -> R^2?

[ionic dust]

I think I have to draw it out really

like f(x^2)

it fails the horizontal line test

[dusky epoch]

i don't think "f(x^2)" means what you think it means.

[formal pond]

any idea how i can start this off to convert to rref? doing elementary row operations.

I was thinking to add row 1 and 3 and put into row 3

to get rid of the -1

[dusky epoch]

that sounds like a good idea

[formal pond]

cool deal

[dusky epoch]

btw rref conversion steps aren't unique

like. you could convert the same matrix into rref in many different ways

[formal pond]

okay thanks for the heads up

[wintry steppe]

Can someone please help me with this?

[gray dust]

Post in #prealg-and-algebra

[pliant harbor]

@ionic dust You would see z=f(x,y) fail the horizontal plane test for some plane z=c.

What the hell is the "trains" method?

[pliant harbor]

$w_n = 4w_{n-2}+2w_{n-1}$ with $w_0=1.$

[stoic python]

aadfg:

[vast torrent]

Googling trains method doesnt give anything

[wintry steppe]

Got it thanks!

I screwed up because I thought w0 = 0

[stray minnow]

is what im doing in III even correct?

[wintry steppe]

@pliant harbor , if you check it for w8, the equation doesn't work...

It works

[stray minnow]

ok so i think its absolutely wrong

[stray minnow]

anyone know how?

😦

[pliant harbor]

@stray minnow You're claiming that the region that your constraints form is a rectangular prism. Check if this is really the case.

[stray minnow]

i did something different

but im going crazy after 4 hours trying to find stuff in my notes

[pliant harbor]

Maybe, but that's what your bounds for the integral imply.

[stray minnow]

that

idk what to do about z

[pliant harbor]

I see. So you're transforming by u, v, w = ...

[stray minnow]

but i have the jacobian thingamajig

u=2y-z
v=-x+z
w=3x-y-z

[pliant harbor]

The answer should be a multiple of your current answer, but that's not saying much.

[stray minnow]

so that way the limits would stay the same

but what do i do about the z?

[pliant harbor]

Either you incorrectly computed the determinant, or you incorrectly integrated.

[stray minnow]

is this new thing i've done even correct

[pliant harbor]

Right. And you have to solve for z, which you didn't do.

You integrated with the original dx Dy dz but the new bounds,

[stray minnow]

yeah i realised that wasnt smart

so i kept the boundaries and changed each inequality to just u,v and w

[pliant harbor]

Which doesn't work. You need to integrate with respect to du dv dw and find z in terms of u, v, w.

[stray minnow]

z=2y-u
z=x+v
z=3x-y-w

but how do i get rid of the other crap

[pliant harbor]

2y-z=u, v=-x+z, w=3x-y-z reduces to 1 equation for z because from A[x, y, z]=[u, v, w], you can invert the matrix A.

[stray minnow]

I feel like I know what to do but my mind is blocking it today 😢

[slow oar]

does anyone know how to simplify 6sqrt44/44

[pliant harbor]

You need to invert the matrix in your previous picture.

[stray minnow]

why

[pliant harbor]

@slow oar Wrong chat. Move to #prealg-and-algebra

[stray minnow]

ignore my previous picture

just look at the newer one

the previous one is a mess

[pliant harbor]

The picture posted at 3:54 contains exactly a matrix A such that A[x, y, z]=[u, v, w]. Now how do you solve for [x, y, z]?

[stray minnow]

what matrix

i have the determinant of the jacobian?

OHHHHHH

OMGGGG

is it

erm

x=Px'

I use that

to erm

get the new value of z

whew

1 sec

i think

@pliant harbor

this?

i got 25 lol

[pliant harbor]

Yes. You should be correct unless you made an algebra mistake.

[stray minnow]

omg

you are a lifesaver 😄

it had taken me way too long to do this lol

[pliant harbor]

The problem is that you just tried to solve for z. You can't solve for z without getting x and y as well, and that requires a matrix approach.

[stray minnow]

i would like to give a special shoutout to my university for not mentioning how to even do these questions 😄

[pliant harbor]

I suppose once the professor reviews the problem next class (was this homework), you could ask him if the method was ever covered.

[stray minnow]

this is for the exam 😉

next week

🙃

we havent had this lecturer since mid november

😄

[pliant harbor]

Who's been lecturing since then? When do your semesters start and end?

[stray minnow]

start end of sept, end week before christmas, thought technically these 3 weeks im currently in the middle of count as it, but are just exams

our module is split into 2 6 week blocks

2 different lectures

the first block was awful

@pliant harbor

[pliant harbor]

You made a mistake inverting the matrix. I just checked, and u+v+2w≠3z.

[stray minnow]

...

i cant have

that is the matrix they give

arggggh

maybe i was meant to triple it

[pliant harbor]

You wrote -3 0 3 for row 2 in the picture. Isn't it -1 0 1?

[stray minnow]

no

the inverse was already given in part I

i kept it tripled

OH WTF

ARGGGGHG

heads will roll for this

mine

[pliant harbor]

Alright. I see that all the problem parts are connected now. You should've realized that part I, ii is supposed to help you for iii.

[stray minnow]

yeah

i noticed that

but

-3 0 3 became -1 0 1?

[pliant harbor]

I was just looking at your scratchwork. The inverse they gave works. But now you also have to divide by the volume of the original region to find the "average value."

[stray minnow]

i was told that i didnt need to?

because i changed the coordinates

🙃 i dont know how to find the vol of the original region either

[pliant harbor]

Let $V, V'$ be the volume of the old, new region respectively. You want to find $\frac{1}{V} \int_V z.$ But change of coordinates just let you calculate $\int_V z.$ Luckily, the Jacobian is $V/V'$ and $V'$ is easy to find.

[stoic python]

aadfg:

[stray minnow]

is the integral i calculated correct?

i just need to divide it by the old volume though

[pliant harbor]

Wait. This is evil. You really can't use part 1 unless the matrix there has a typo. We almost have 3P^-1[x,y,z]=[u,v,w], but again the 2nd row is -3 0 3 instead of -1 0 1.

[stray minnow]

...

[pliant harbor]

So either it doesn't have typo and you have to compute an inverse that was not given, or it has a typo and you're supposed to know that.

[stray minnow]

ur joking

there's a typo?

I have spent 4 hours reading up trying to get this and my clapped ass university has typos in this

[pliant harbor]

Do you see what 3P^-1 is in the picture of the problem and how it's almost the matrix that transforms [x,y,z] to [u,v,w]?

You're lucky that the typo is not too far off. Change the 2nd constraint to 0≤-3x+3z≤6 and integrate with bounds 0 to 6 for v.

[stray minnow]

are you sure?

is there actually a typo

[pliant harbor]

Well none of us have actually typed the matrices into a calculator online to see what the result should be. I could try that.

[stray minnow]

what sort of calculator

the inverse given is correct

[pliant harbor]

Alright. The inverse of P is written correctly. I wouldn't call it a typo, but it sure seems like you have to jump through a lot of hoops to borrow the calculation of P^-1.

[stray minnow]

yeah

[pliant harbor]

Because you're not using P^-1 to solve for z, you're using P. But at the same time, you have to notice that using P wouldn't help unless you change the 2nd inequality to 0≤-3x+3z≤6 so that 3P^-1[x,y,z]=[u,v,w] is correct.

[stray minnow]

so...what I did...was right?

im getting so confused

[pliant harbor]

Yes, but you should have v=-3x+3z and then integrate from 0 to 6.

You can check that your current equation z=(u+2v+2w)/3 is false unless you change v like this.