I'm supposed to solve for x
#help-39
ripe jay
pearl pondBOT
I'm supposed to solve for x
midnight haven
there is a nice trig identity
uh
[
\m \cos a + \m \cos b = 2\m\cos{\f{a+b}2}\cdot \m \cos{\f{a-b}2}
]
jolly parrotBOT
midnight haven
you can use that and make Ur thing more tolerable
ripe jay
yeah I used that on the cos2x + cosx, but I didn't really find any way to continue after that
midnight haven
use it on cos3x and cosx
pearl pondBOT
toxic lichen
$\sum_{m=1}^{1999} \paren{1+\frac{1}{m^2}+\frac{1}{(m+1)^2}}^{1/2}$
jolly parrotBOT
AnnGhost
toxic lichen
this, right?
midnight haven
ya
toxic lichen
does combining the fractions into one do anything
gonna close and reopen the channel rq so the name updates, hold on
.close
pearl pondBOT
toxic lichen
.reopen
pearl pondBOT
✅
toxic lichen
there
anyway let's see
we get $\frac{\sqrt{m^2(m+1)^2 + m^2 + (m+1)^2}}{m(m+1)}$ in the sum don't we
jolly parrotBOT
AnnGhost
toxic lichen
,w simplify m^2 (m+1)^2 + m^2 + (m+1)^2
jolly parrotBOT
toxic lichen
i see something
it's (m^2 + m + 1)^2
so we have $\sum_{m=1}^{1999} \frac{m^2+m+1}{m(m+1)}$
jolly parrotBOT
AnnGhost
toxic lichen
which is more promising, i think.
chrome crescent
how do i do this again
pearl pondBOT
how do i do this again
@chrome crescent Has your question been resolved?
Available help channel!
fickle finch
pearl pondBOT
fickle finch
wouldnt v'(pi)= <0,0,0> they are all constants no?
makes no sense they are deriving pi^3
inland ivy
they are differentiating t^3 to get 3t^2 and then plugging in t=pi
fickle finch
oh, so would you not be able to plug it in then derive it? would that not be the same
inland ivy
then, as you said, you would just be differentiating a constant, which would result in 0
you have to differentiate the function and then plug the value in, not the other way around
cerulean granite
pearl pondBOT
grave thistle
!status
pearl pondBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
glass meadow
You just need to express it as F = something
How would you do it if let's say C = 4?
glass meadow
Yes but can you list the steps you took?
cerulean granite
-9 times 4 is -36
And then you isolate the 5F so you plus 36 to both sides
160 + 36 is 196
So I’m left with 5F = 196
glass meadow
Ok so when you manipulate equations, you can also add and multiply variables
So for example a + b = 4
Add -b on both sides: a + b - b = 4 - b
That gives you a = 4 - b
You just expressed a in terms of b
Here you need to express F in terms of C
cerulean granite
Umm okay I think I’m following
Do I need to isolate F on the left side of the equation?
glass meadow
Yes
cerulean granite
Sooo plus +9C to both sides
glass meadow
Yes
cerulean granite
Okay, what’s my next step?
glass meadow
What do you have now?
cerulean granite
5F = 160 + 9C
glass meadow
Ok so we only want F, not 5F
cerulean granite
Ah okay, so I’d just -5 from both sides?
glass meadow
No that would give you 5F - 5
glass meadow
And then divide by 5?
You had the answer earlier
glass meadow
Almost
If you divide both sides by something, you need to divide the whole sides
It's fine if it doesn't give you a value straight away
glass meadow
$5F = 160 + 9C \implies \frac{5F}{5} = \frac{160 + 9C}{5}$
jolly parrotBOT
Nel
glass meadow
Yes, the whole thing
glass meadow
Well you basically have the answer already
cerulean granite
How so?
glass meadow
$\frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}$
jolly parrotBOT
Nel
cerulean granite
So 160/5 + 9/5 ?
glass meadow
You forgot the C
cerulean granite
I thought 5 was the C
glass meadow
No no I mean the big C in 160 + 9C, not the small c above
cerulean granite
I’m sorry, I’m embarrassed. I feel like an idiot. I’m not understanding what your saying
glass meadow
$\frac{160+9C}{5} = \frac{160}{5} + \frac{9C}{5}$
jolly parrotBOT
Nel
glass meadow
Keep the "9C" as is, don't drop the C
inland ivy
$5F-9C=160 \implies 5F = 160+9C \implies F = \frac{160+9C}{5} = \frac{160}{5} + \frac{9C}{5} = 32 + \frac{9}{5}C$
jolly parrotBOT
kheerii
inland ivy
see if you can interpret this'
cerulean granite
I think I understand it and what they did to get to 32+9/5C
We were on the last step, right?
So 9C/5 is just 9/5C ?
inland ivy
yep
glass meadow
Yes
cerulean granite
So in totality, you isolate the 5F, and then got divide the WHOLE sides by 5.
And then those fractions equal into 32 + 9/5C , thus giving us our answer?
glass meadow
Yes
inland ivy
yes
glass meadow
And usually you put the term with the variable in front so F = 9/5C + 32, but it's the same thing
cerulean granite
Thank you guys!
glass meadow
I just realized this is the conversion between Fahrenheit and Celsius
cerulean granite
Oh
You know, as someone who’s major is meteorology, I should’ve realized that
ornate drum
pearl pondBOT
ornate drum
why is there an exclamation point
toxic lichen
these are called factorials
judging by your reaction, i suppose you've never encountered those before?
ornate drum
no I have not
toxic lichen
well, now you know the name of that thing.
pearl pondBOT
@ornate drum Has your question been resolved?
ornate drum
.close
thorny stream
if there is a curve that is defined by a 2d circle on the xy-plane, the torsion must be zero, right?
Geometrically, because torsion is the twisting force applied to an object. But you can't twist something that has no height, so it must be 0, right?
thorny stream
am i correct in how i reasoned this?
midnight haven
For me it’s got no tortion because it’s in xy so it can’t get out to “twist”
And for the <= part , whenever you have a tortion of 0 => planar
thorny stream
For me it’s got no tortion because it’s in xy so it can’t get out to “twist”
well wait a second. Let's say there's a curve that starts out like a spring, and then as you get closer to the bottom (or top) of the spring, the revolutions become more frequent and thus the spring eventually approaches a circle
the tortion at the bottom (or top) has to be 0 because it becomes a 2d shape, so there's nothing to twist since twisting requires depth
midnight haven
the tortion at the bottom (or top) has to be 0 because it becomes a 2d shape, so...
Right but your original shape was not a 2D
You’re considering the 2D object of a 3D object if I read right
thorny stream
yes. It starts out 3D and then becomes a 2D circle
midnight haven
yes. It starts out 3D and then becomes a 2D circle
Based on the perspective you chose yes
thorny stream
okay awesome. But how is the original shape not 2d
midnight haven
But tu be fair I am not really good at this so do check w someone else as well lol
I wouldnt want to mislead you
midnight haven
okay awesome. But how is the original shape not 2d
Because you chose a 3D object lmao it’s kinda like if you considered the « shadow » of an object
You get a 2D shape out of it
Fun fact this is a whole field of mathematics in which there’s been some movement recently (kinda like all fields but anyways lmao)
thorny stream
okay, thank you so much
pearl pondBOT
@midnight haven Has your question been resolved?
pearl pondBOT
pearl pondBOT
Available help channel!
broken burrow
Hello guys, in the diagram I'm confused about the fact that how are we getting the distances(represented with the two blue lines). Why are we using x-x1 and why x2-x? isnt x had already been used as the name of the x axis? What are we referring to when using x as a value?
broken burrow
ohh
I was confused that why is it called x(if the x had already been used as the axis), but I assume that would be a general form where the first value of the pair is a value from the x axis.
mental hinge
the x there is referring to a variable x. when we just say "x", we don't usually refer to the axis
pearl pondBOT
thanks man
If you are done with this channel, please mark your problem as solved by typing .close
@broken burrow Has your question been resolved?
pearl pondBOT
Available help channel!
vast holly
pearl pondBOT
vast holly
d) A matrix B is called symmetrical if B^T = B
- Let A be a n x m matrix. Show that the matrix A * A^T is symmetrical
- Assume that B is symmetrical and invertible. Is B^-1 also symmetrical? Explain your answer.
I don't know how to prove either of these
fluid axle
well what does it mean for A A^T to be symmetrical ? @vast holly
vast holly
well what does it mean for A A^T to be symmetrical ? <@608728365268205642>
Im not sure
fluid axle
B is symmetrical when B^T = B
vast holly
mhm
fluid axle
A^T A is symmetrical when (A^T A)^T = A^T A
that's what you want to prove
(A^T A)^T = A^T A, for any matrix A
vast holly
okay so
We know that a matrix A is symmetrical if A = A^T.
For the vector A * A^T to be symmetrical, it must hold true that A * A^T = (A * A^T)^T
To prove that A * A^T = (A*A^T)^T ...
cursive wraith
what is (AB)^T if A and B are matrices?
vast holly
hm?
vast holly
you gotta actually prove it at some point tho
how should I go about that?
fluid axle
start with (A^T A)^T, and try to simplify that with the hint rafilou gave
vast holly
ah
vast holly
start with (A^T A)^T, and try to simplify that with the hint rafilou gave
Wait wait
you mean (A*A^T)^T right?
vast holly
(A * A^T)^T = A^T * A^T^2
cursive wraith
unfortunately no
first of, T^2?
and second of all, T is not distributive like that with multiplication
(AB)^T = B^T A^T
transposing a product swaps the order
vast holly
the original
cursive wraith
yes
vast holly
okay, so!
A * A^T should be (A * A^T) ^T to be symmetrical.
(A * A^T)^T = (A^T)^T * A^T . We know that (A^T)^T = A, so: (A^T)^T * A^T = A*A^T
is that the end of the proof?
cursive wraith
yes
vast holly
Great!
cursive wraith
A * A^T should be (A * A^T) ^T to be symmetrical.
(A * A^T)^T = (A^T)^T * A^T ...
be careful you wrote "(A * A^T)^T = (A^T)^T * A" missing a T at the end
vast holly
ah, yes
Rafilou can you check something for me btw? Me and a friend are disagreeing on whether 4 vectors are linearly independent
"Decide whether the 4 vectors in R^3 are linearly independent"
fluid axle
"4 vectors in R^3" you don't need 2 pages of work to know whether they are lin indep or not
vast holly
Here's what I did:
- Set it up as a matrix
- Get rref
- Count pivot elements
- If the matrix m * n has the rank n then they are linearly independent
vast holly
"4 vectors in R^3" you don't need 2 pages of work to know whether they are lin i...
please do not tell me they are linearly dependent
cuz then my method is all out of the window
cursive wraith
Here's what I did:
1. Set it up as a matrix
2. Get rref
3. Count pivot elements...
- should have given you that the resulting 3*4 matrix has a rank of 3
vast holly
4. should have given you that the resulting 3*4 matrix has a rank of 3
well this matrix is 4x3
4 pillar + 3 row
vast holly
oh
cursive wraith
to remember this fact, a vector of R^n is a n*1 matrix
vast holly
Right, okay
Thanks
lets do the next questiont hen x)
- Assume that B is symmetrical and invertible. Is B^-1 also symmetrical? Explain your answer.
lemme repost this image so i remember
cursive wraith
what is (AB)^T if A and B are matrices?
use this result with B and another useful matrix
vast holly
use this result with B and another useful matrix
so basically
if B is symmetrical then B = B^T. For B^-1 to be symmetrical then (B^-1)^T has to be symmetrical
cursive wraith
so basically
if B is symmetrical then B = B^T. For B^-1 to be symmetrical then ...
"For B^-1 to be symmetrical then (B^-1)^T has to be symmetrical" that's not the real point, the point is that (B^(-1))^T = B^(-1) if B^(-1) is symmetrical
vast holly
oh wait
if B is symmetrical then B = B^T. For B^-1 to be symmetrical then B^-1 = (B^-1)^T
jolly parrotBOT
rafilou2003
vast holly
thats the tranpose of the identity matrix
vast holly
the transpose of the identity matrix is...
shit i dont remember
haha
i just looked through my book
but transpose of id matrix should be equal to id matrix
i think
cursive wraith
yes it is
id matrix is symmetric
diagonal matrices are symmetric
since when you swap rows and columns, the terms on the diagonal stay in the exact same spot
so $(B\cdot B^{-1})^\top = I$
jolly parrotBOT
rafilou2003
cursive wraith
what is (AB)^T if A and B are matrices?
but using this...
vast holly
**rafilou2003**
where did you get (B * B^1)^T from?
cursive wraith
where did you get (B * B^1)^T from?
you'll understand when you do
cursive wraith
but using this...
this
vast holly
I dont fully get it
So far I wrote:
For B^-1 to be symmetric, it should be such that B^-1 = (B^-1)^T.
vast holly
mhm
cursive wraith
This result was further up if you want to check
vast holly
(MN)^T = N^T * M^T
pearl pondBOT
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
vast holly
This result was further up if you want to check
was I right?
vast holly
(MN)^T = N^T * M^T
.
cursive wraith
Yes it's correct
cursive wraith
**rafilou2003**
So I'm asking the same question again, but I expect a different answer
vast holly
but with the information that B^-1 = (B^-1)^T, where do you get (B * B^-1)^T
(B * B^-1)^T = (B^-1)^T * B^T
cursive wraith
(B * B^-1)^T = (B^-1)^T * B^T
Yes correct
So joining both results together
$(B^{-1})^\top \cdot B^\top = I$
jolly parrotBOT
rafilou2003
vast holly
but with the information that B^-1 = (B^-1)^T, where do you get (B * B^-1)^T
why do you just times with B inside the ()
thats veyr random
cursive wraith
thats veyr random
It's not, the only link between B and B^-1 is that their product is I
vast holly
maybe you can write the next line for what I wrote so I can get it:
"For B^-1 to be symmetric, it should be such that B^-1 = (B^-1)^T."
cursive wraith
Since we know something about B and we want to know something about B^-1, using B • B^-1 = I is very logical
And since we know what the transpose of product of matrices is...
vast holly
hmm could you maybe continue with what i started writing and then until end? that way maybe I can get fully what you mean
because I am still unsure how its logical to times B^-1 and B if we want to prove that B^-1 also is symmetrical
cursive wraith
We want to prove B^-1 = (B^-1)^T
So we want to prove I = B^-1 • B = (B^-1)^T • B
Clearer ?
vast holly
I dont see the connection between those 2 unfortunately.
cursive wraith
I dont see the connection between those 2 unfortunately.
We literally just multiplied by B
vast holly
We want to prove B^-1 = (B^-1)^T so we multiply by B ?
cursive wraith
Yes
vast holly
whats the idea behind multiplying by B. what basis do we have for that
cursive wraith
(MN)^T = N^T * M^T
As was said, we have a result on the transpose of a product of matrices
So it would be nice for a product to appear
cursive wraith
Either that or directly use the result that $(A^\top)^{-1} = (A^{-1})^\top$
jolly parrotBOT
rafilou2003
cursive wraith
**rafilou2003**
If you dont have that result then you have to do what I did
vast holly
Could you try to write out your whole thought process
cursive wraith
I know this but its super lengthy to explain
If you know it then use it
Literally what I'm doing is the exact same as explaining why this is true
It's almost re-proving it
cursive wraith
**rafilou2003**
So if you knew this already then use it instead of wasting your time reproving it like we are almost doing
vast holly
just checked my notes
i dont remember how i proved this lmao
I wrote:
(A^-1)^T * A^T = (A^-1*A)^T = I^T
I^T / A^T = (A^T)^-1
cursive wraith
Yes, that's it
Althoug wacky
Corrected version :
(A^-1)^T * A^T = (A*A^-1)^T = I^T = I (remember the inversion)
So A^T is invertible and its inverse is (A^-1)^T
So (A^T)^-1 = (A^-1)^T
vast holly
Right..
Rafilou can you write all of it out like you would to answer the question? Its still hard for me to piece it together. Im just stuck at that part where I said that it should be true that B^-1 = (B^-1)^T
@cursive wraith
like what exact sentence would you write from there
fluid axle
the theorem you just reproved w/ rafilou makes the proof super easy
vast holly
Maybe. But I just don't know what to write in the next sentence
vast holly
I wanna show B^-1 = (B^-1)^T
fluid axle
try rewriting (B^-1)^T using that theorem
vast holly
Rafilou says (B^1)^T * B^T
but I just dont understand why he got B^T
so thats why I keep asking to get all of it written out
with the thoughts behidn it
fluid axle
I'm not going that way at all
or at least it's very carefully hidden for you behind that theorem
fluid axle
try rewriting (B^-1)^T using that theorem
^
fluid axle
So (A^T)^-1 = (A^-1)^T
this one
vast holly
okay..
vast holly
"We know that (B^-1)^T = (B^T)^-1. Therefore B is symmetric, which means that B^T = B"
now how does that prove that B^-1 = (B^-1)^T
vast holly
Wait lemme try somethign
For B^-1 to be symmetrical, it should hold that B^-1 = (B^-1)^T. We know that B is symmetrical and invertible, which means B^T = B. We can now times with B on each side:
B^-1 * B = (B^-1)^T * B
We can now for simplicity replace B on the right side with B^T since they are the same:
B^-1 * B = (B^-1)^T * B^T
B^-1 * B = I
(B^-1)^T * B^T = (B^-1 * B)^T = I^T. And since we know the identity matrix I is symmetric, it holds that I^T = I.
Therefore we get I = I on each side, which is true.
@fluid axle @cursive wraith
cursive wraith
yes sure, however you don't use "therefore" or "this implies that" in proofs like this. You say "this is equivalent to"
vast holly
in which parts?
cursive wraith
because "what we want to prove => true" does not mean that what we want to prove is true
However, "what we want to prove <=> true" is the correct way to do things
vast holly
wait wait wait
cursive wraith
For B^-1 to be symmetrical, it should hold that B^-1 = (B^-1)^T. We know that B ...
In order:
"which means that" replaced by "which is equivalent to".
"By multiplying by B on each side, this is equivalent to:"
vast holly
this proof doesnt make sense
or
normally A * B != B * A
But I guess this does not matter here since its symmetric?
cursive wraith
normally A * B != B * A
what part in the proof exactly bothers you with this?
cursive wraith
For B^-1 to be symmetrical, it should hold that B^-1 = (B^-1)^T. We know that B ...
"(B^-1)^T * B^T = (B^-1 * B)^T" remember to SWITCH the order of multiplication
"(B^-1)^T * B^T = (B B^-1)^T"
This is correct
vast holly
I wrote that we times with B on each side and get:
B^-1 * B = (B^-1)^T * B, then B^-1 * B = (B^-1)^T * B^T
vast holly
"(B^-1)^T * B^T = (B B^-1)^T"
yea
cursive wraith
the rest is fine
vast holly
lemme rewrite
vast holly
what part in the proof exactly bothers you with this?
how do I know that B^-1 * B = I
I know that its the case for B * B^-1
but idk when you times from the right
?
<@&286206848099549185>
lilac lake
Commutative property; order doesn't matter
vast holly
Commutative property; order doesn't matter
well this is matrix multiplication so order certainly matters
lilac lake
Okay, didn't realize it was matrix
cursive wraith
well this is matrix multiplication so order certainly matters
A matrix is left invertible <=> it is right invertible <=> it is invertible
the inverse of A verifies $A\cdot A^{-1} = A^{-1}\cdot A = I$
jolly parrotBOT
rafilou2003
vast holly
After multiplying with B and doing calculations we get the identity matrix. So the conclusion should be that because the identity matrix is symmetrical, it means that B^-1 is also symmetrical? @cursive wraith
cursive wraith
the conclusion "I=I, which is true" only works if you worked with equivalences up until this point
as in B^-1 = (B^-1)^T <=> B^-1 * B = (B^-1)^T * B
<=> B^-1 * B = (B^-1)^T * B^T
<=> ...
vast holly
rigjht
cursive wraith
since your original statement is equivalent to (and not just implies) a true statement, the original statement is true
vast holly
B^-1 = (B^-1)^T <=> B^-1 * B = (B^-1)^T * B
<=> B^-1 * B = (B^-1)^T * B^T
<=> B^-1 * B = (B^T * B^-1)^T
<=> B^-1 * B = I
<=> I = I
cursive wraith
exactly
vast holly
but how would you word the conclusion? original question is to prove that if B is inversible and symmetric then B^-1 also is
cursive wraith
"suppose B is invertible and symmetric.
B^-1 = (B^-1)^T <=> B^-1 * B = (B^-1)^T * B
<=> B^-1 * B = (B^-1)^T * B^T
<=> B^-1 * B = (B^T * B^-1)^T
<=> B^-1 * B = I
<=> I = I
Since "I=I" is true, B^-1 = (B^-1)^T "
we've already done the hard work
vast holly
nice, okay
thanks
next question is about a matrix
Let n be a natural number and A and B two n x n matrices. Assume that both A and B er invertible matrices. Show that (A * B)^-1 = B^-1 * A^-1
cursive wraith
so what do you need to show for B^-1 * A^-1 to be the inverse of AB?
cursive wraith
what does the inverse M^-1 of any matrix M verify?
vast holly
that M * M^-1 = I
vast holly
if M * N = I
vast holly
hmm im not 100% sure
cursive wraith
so N is the inverse of M if... ?
so B^-1 A^-1 is the inverse of AB if... ?
vast holly
if B * B^-1 * A * A^-1 = I * I
cursive wraith
no
M N = (AB) (B^-1 A^-1)
N is the inverse of M if MN = I
so B^-1 A^-1 is the inverse of AB if (AB) (B^-1 A^-1) = I
vast holly
right, okay
cursive wraith
and (AB) (B^-1 A^-1) = ?
vast holly
I * I ?
vast holly
I_A * I_B
vast holly
AB * B^-1 * A ^-1
vast holly
AB * (B * A) ^-1
cursive wraith
no, as we saw that's not how inverses work
cursive wraith
you're even asked to show that the inverse of a product SWAPS the product
AB * B^-1 * A ^-1, where do we put the parentheses
vast holly
no idea then
vast holly
i dont know where to put in the middle
vast holly
AB * (B^-1) * A^-1
cursive wraith
what worth is there putting a SINGLE matrix in parentheses?
vast holly
well its the middle x)
cursive wraith
bruv
vast holly
i dont know what else
cursive wraith
the middle is the way
and we don't want a single matrix in the parentheses
come on
vast holly
A * (B * B^-1) * A^-1 ?
cursive wraith
keep going
cursive wraith
a matrix times identity, you don't know???
vast holly
no, i dont remember
cursive wraith
what is 195637295*1 ?
vast holly
okay
so the result is I
for that whole equation
A * I * A^-1
I said for the whole equation
A * I = A
A * A^-1 = I
cursive wraith
ok sorry
cursive wraith
so B^-1 A^-1 is the inverse of AB if... ?
so?
vast holly
Right, we want to show that (A * B) ^ - 1 = B^-1 * A ^-1
B^-1 * A ^-1 is the inverse of AB if (AB) (B^-1 A^-1) = I
vast holly
yes
so we say:
We wish to show that (A * B) ^-1 = B^-1 * A ^ -1.
For B^-1 * A^-1 to be the inverse of AB, it should hold that (AB)(B^-1 * A^-1) = I, just like how a matrix A with an inverse holds that A * A^-1 = I.
So:
(AB)(B^-1 * A^-1) = AB * B^-1 * A^-1 = A * (B * B^-1) * A^-1 = A * I * A^-1 = A * A^-1 = I
As we can see, (AB)(B^-1 * A^-1) = I, which means that indeed (A * B) ^-1 = B^-1 * A ^ -1.
@cursive wraith is this correct?
cursive wraith
yes
vast holly
and we can only put a parenthesis around B * B^-1 because they are right next to each other, right? @cursive wraith
cursive wraith
yes
vast holly
Thanks!
I have one last
well actually two
"Given is a certain inhomogenous linear system of equations over R with four equations and three unknowns. It is shown that the vector ... is a particular solution to the linear system of equations. Is the vector 2 * V_p a solution to the sytem?"
My answer to this one is no, because its the same as saying:
x1+x2+x3 = k
2x1+2x2+2x3 = k
which is wrong
Then theres this one, where I have to find the determinant. My idea here is to laplace transform where I transform after column 4 and then get 4 different 3x3 matrices where I laplace again ?
or is there a simpler way?
sorry I meant laplace expansion
@cursive wraith
cursive wraith
or is there a simpler way?
look at the first 2 columns
@vast holly did you get the hint?
pearl pondBOT
@vast holly Has your question been resolved?
vast holly
<@608728365268205642> did you get the hint?
Hmm no
Well it’s all 1s
Hahah I don’t know more than that
But am I right about solving the question before that also?
tired delta
pearl pondBOT
tired delta
Need help
blissful cloak
,rotate
jolly parrotBOT
blissful cloak
can you explain some notation here?
tired delta
Description says [a]- a’s whole part {a}= a’s fraction part
Idk how to do log in this [],{}
blissful cloak
log_2(3) is irrational, what does it mean fraction part?
tired delta
Its just 5.8
blissful cloak
yeah ok, that makes sense
blissful cloak
well log_2(3)=1.158..., right?
blissful cloak
,calc log_2(3)
jolly parrotBOT
The following error occured while calculating:
Error: Undefined function log_2
blissful cloak
,w log_2(3)
jolly parrotBOT
blissful cloak
yep log_2(3)=1.158 
tired delta
It says 1.584?
tired delta
Idk man
blissful cloak
1.58, right?
tired delta
Yes
blissful cloak
so then the decimal part of that is just 1-log_2(3), right?
tired delta
U mean {log2(3)}=1-log2(3)?
blissful cloak
tired delta
so then the decimal part of that is just 1-log_2(3), right?
I think its log2(3)-1
blissful cloak
oh yeah sorry its midnight over here and I'm real tired 😅
{log_2(3)}=log_2(3)-1 yes :)
that is to say {c}=c-[c]
tired delta
It depends on c’s decimal is >5 or not idk
If we take 1.58 as a example
{1.58} becomes 0.58
glass meadow
Why though
"x's whole part" doesn't mean "x rounded"
The whole point of "whole part" and "fractional part" is that you can split any number into those and add them back up to form the number back
tired delta
Why though
Idk it just says that in example
glass meadow
Then show the example
tired delta
glass meadow
,rccw
jolly parrotBOT
tired delta
If we takes example on 1.42 [sqrt27]+5*{3.4}+[-3.9]-18*{-4.5}= 5+2-4-9=-6
Idk if im wrong
glass meadow
Only way I see this working is if [-a.b] = -a-1 if b is not 0 which is weird to me
Are the circled answers checked or are they just your guesses?
tired delta
Checked answers are correct answers (i saw from back of the book)
Book might be wrong. Can be -5 with ur assumption
glass meadow
Book might be wrong. Can be -5 with ur assumption
It can't because of the possible answers to 1.41
tired delta
It can't because of the possible answers to 1.41
Yes
glass meadow
I mean it's fine to define it that way, but to me that's not what "whole part" means
tired delta
Its foreign book. Maybe i translated wrong and i dont understand {log2(3)}
In 1.43 D is the correct answer
glass meadow
Ok so {log2(3)} = log2(3) - 1, [log5(47)] = 32, [log4(9)] = 5
So log2(3) - 1 +32 - 5.8 - 5 + 2*0.4 = log2(3) + 21
Hm
tired delta
Are u sure [log5(47) =32
glass meadow
,w log5(47)
jolly parrotBOT
glass meadow
My bad
So log2(3) - 1 + 2 - 5.8 - 5 + 2*0.4 = log2(3) - 9
Still weird
,w log4(9)
jolly parrotBOT
tired delta
-1+2-5-2=-6
glass meadow
Yeah google does not like logarithms
So log2(3) - 1 + 2 - 5.8 - 1 + 2*0.4 = log2(3) - 5
So yes, D
glass meadow
[log4(9)] = 1, and you subtract that in the expression
glass meadow
No?
We established that [x] is x rounded toward -inf
In other words it's the nearest integer less than x
tired delta
Then i should become 2 since its 1.58?
glass meadow
Is 2 the nearest integer less than 1.58??
tired delta
Yes? I mean 1.58 is closer to 2 than 1
glass meadow
You're reading half of the sentence...
tired delta
Lemme read it again
tired delta
Is 2 the nearest integer less than 1.58??
No?
glass meadow
Why? And then what is it if it's not 2?
tired delta
I think somehow got wrong in {log2)3)}
glass meadow
... I'm still not sure you understand what "the nearest integer less than x" means
glass meadow
Ok, so then do you agree we have log2(3) - 1 + 2 - 5.8 - 1 + 2*0.4 = log2(3) - 5?
tired delta
Ok, so then do you agree we have log2(3) - 1 + 2 - 5.8 - 1 + 2*0.4 = log2(3) - 5...
Yes
glass meadow
Great, then you should be able to do the rest
Just remember what [x] is and that [x] + {x} = x
pearl pondBOT
pearl pondBOT
Available help channel!
vast holly
pearl pondBOT
midnight haven
row reduce and spam Laplace expansion
vast holly
row reduce and spam Laplace expansion
RREF?
real scarab
i would try to reduce the first two columns as they are the same
then develop
and then use the 3*3 determinant
vast holly
reduce the first two columns? wdym?
vast holly
We haven't been taught that you can just subtract one column from another
real scarab
can you do it with rows?
vast holly
yes
real scarab
ok
so use the fact that a matrix and its transpose have the same determinant
and done
vast holly
aha, smart
real scarab
now you know everything you can do with rows you can do with columns if its to calculate a determinant
vast holly
im tryna remember
certain row operations change determinant
others dont
R1 + 1 * R2 doesnt change determinant right?
fluid axle
indeed
fluid axle
and you get a row of 0s yes
fluid axle
you can do laplace expansion right now
vast holly
oh okay.
fluid axle
you have a row of 0s after all
vast holly
Should I transpose?
vast holly
so what should the determinant be then ?
0
fluid axle
yeah
the det of the transposed matrix is 0
so the det of the original matrix is also 0
vast holly
so the det of the original matrix is also 0
where in this defintion can i show that a zero row would equal to a determinant of 0 if expanded based on that
i guess a_i1
fluid axle
all the a_i1 are 0 essentially in this case yes
you can do laplace expansion on any column or any row also, it's not limited to the 1st column
vast holly
actually i forgot what the 1 represents here
fluid axle
1 means the 1st col here
vast holly
oh its column
lmao
so I guess I should word it differently
but wtf is i then
hmm
fluid axle
you can do laplace expansion on any column or any row also, it's not limited to ...
although some additional negative signs will pop up if you don't do 1st row or 1st col
fluid axle
but wtf is i then
i goes through all the column
vast holly
aha
fluid axle
a11, a21, ..., an1
fluid axle
well we are not doing first row, we are doing second row
then you also negate the whole result
it comes from the fact that if you swap two rows/cols, the determinant gets negated
fluid axle
instead of doing 1st row, you're doing second row
if we want to get in line with your def
vast holly
yes
vast holly
my definition is for first column x)
fluid axle
so you've reduced computing the expansion on second row, to computing the expansion on the first row
and since if you transpose your matrix you get the same determinant
expanding on the first row or the first column gives the same result
so if you know how to compute the expansion on 1st row, you know how to compute it for any row/any col
fluid axle
so if you know how to compute the expansion on 1st row, you know how to compute ...
did your prof not talk about that ?
fluid axle
And you can tell me if I understood correclty
sure np
vast holly
To get the determinant of the following matrix we first use the knowledge that the transposed of a matrix has the same determinant as the matrix. The tranposed would be:
1111
1111
1246
1357
We can now eliminate the first row, as we know it is one of the row operations that does not change the determinant and get:
0000
1111
1246
1357
We can now get the transposed again as it does not change the determinant and get:
0111
0123
0145
0167
From here, we can look at definition 8.2, which defines the determinant of a nxn matrix. From this definition we can see, that we need to multiply with the a_i1 and the other elements in the first column, which in this case would be 0. Therefore the determinant of this matrix is 0.
Please be super critical of wording and notation so im sure to fully get it
fluid axle
I'd describe a bit more clearly what row operation you do on the second step
otherwise it's very good
vast holly
I'd describe a bit more clearly what row operation you do on the second step
hmm you mean from:
1111
1111
1246
1357
to :
0000
1111
1246
1357
vast holly
Alright
Get the rank of the following matrix
@fluid axle I did gaussian elemination and got:
1 - 1/2 0 0
0 1 1 1/2
0 0 0 0
which should be rank 2
Can you verify ?
my actions were:
R_2 <- R2 - R1
R3 <- R3 - R2
R1 <- R1 * 1/2
R2 <- R2 * 1/2
north trench
<@177516015033384960> I did gaussian elemination and got:
1 - 1/2 0 0
0 1 1 1/2...
I get the same rank, seems good
edgy stone
that's correct
vast holly
Thanks! Did you also look at actions ?
north trench
yeah those actions look right
vast holly
great
"Given is a certain inhomogenous linear system of equations over R with four equations and three unknowns. It is revealed that the vector v_p ... is a particular solution to the system. Is the vector 2 * v_p also a solution to the system"
My answer:
For this to be true, it would have to be the case that:
c1 * x1 + c2 * x2 + c3 * x3 = k
while also:
2 * (c1 * x1 + c2 * x2 + c3 * x3) = k
This cannot be true, as it would have to be 2 * k.
Therefore 2 * v_p is not a solution to the linear system of equations
Not sure if this is a good answer for this kind of thing ?
or even if its correct
<@&286206848099549185>
pearl pondBOT
@vast holly Has your question been resolved?
north trench
My answer:
For this to be true, it would have to be the case that:
c1 * x1 + c...
well, that tells you you need k = 2*k. when does that happen?
vast holly
well, that tells you you need k = 2*k. when does that happen?
That happens when you multiply the entire row by 2
pearl pondBOT
That happens when you multiply the entire row by 2
vast holly
But since we only multiply one side of the equation it doesn’t add up
vast holly
for which vectors k do you have k = 2k
For 2 * v_p
north trench
let's ignore the rest of the problem. you claimed "this cannot be true, as it would have to be 2*k". what stops k from being equal to 2*k?
vast holly
let's ignore the rest of the problem. you claimed "this cannot be true, as it w...
The fact that we have multiplied the vector by 2
north trench
there's a vector k where k = 2k is perfectly fine
k = the zero vector
k =/= 2k for any other vector. here, the problem is stated to be "inhomogeneous", which means we're assuming k is nonzero, so we can't have k = 2k.
vast holly
k =/= 2k for any other vector. here, the problem is stated to be "inhomogeneous...
right?
north trench
Yeah
so in total, i would write the answer like
"For this to be true, it would have to be the case that:
c1 * x1 + c2 * x2 + c3 * x3 = k
while also:
2 * (c1 * x1 + c2 * x2 + c3 * x3) = k
If this is true, then 2k = k. Since the system is inhomogeneous, k =/= 0, and we can't have 2k = k. So this can't happen, and 2*v_k is not a solution to the linear system of equations."
(just adding a tiny bit to yours)
vast holly
so in total, i would write the answer like
"For this to be true, it would have ...
Okay, thanks a lot!
pearl pondBOT
Available help channel!
torn willow
can anyone help me on how to solve this quesion pls?
pearl pondBOT
can anyone help me on how to solve this quesion pls?
foggy carbon
can anyone help me on how to solve this quesion pls?
what have you tried?
foggy carbon
first, you know that there are r rows, and c columns
using that info, what is the total number of chairs in terms of r and c?
foggy carbon
yes
now, can you come up with an expressions for the total number of boys in terms of r and c?
torn willow
total no. of boys = 12r
torn willow
total no. of girls = 10c
foggy carbon
yes
now you know that there is one empty chair
can you create an equation from that?
foggy carbon
yes
torn willow
what do i do next?
foggy carbon
lets assume that the total no. of chairs are 'x'
12r*10c = x-1
this should be 12r + 10c = x - 1
torn willow
12r+10c = rc-1
foggy carbon
because you are adding number of boys + number of girls
torn willow
k
torn willow
whats the correct sollution?
foggy carbon
you know that 12r < 1000 and 10c < 1000, so the only valid solution is c = 23, r = 21
sonic rivet
seems tricky, but if you consider the prime factors of r and c you can eliminate a few possibilities
for example, neither r or c can be a multiple of 2 (why?)
torn willow
do u mean factorisation
torn willow
i dont get it
sonic rivet
starting from the equation rc = 12r + 10c + 1
torn willow
k
sonic rivet
can you tell whether 12r + 10c + 1 is even or odd?
sonic rivet
can 12r + 10c be odd ?
sonic rivet
ok but just so we agree, 12r + 10c + 1 is always odd, yes?
sonic rivet
2*(6r + 5c) is even, almost by definition
an even number is a number that can be written as 2 times k for some other number k
torn willow
anything multiplied by 2 is even
sonic rivet
yes
sonic rivet
yep
torn willow
whats next?
sonic rivet
now what happens if r were to be even?
torn willow
rc = an odd no.
sonic rivet
why?
sonic rivet
we established that 12r + 10c + 1 is odd, and now we focus just on the left hand size of the equality
so just looking at rc
torn willow
k
sonic rivet
is rc even or odd when r is even?
torn willow
even?
sonic rivet
why?
torn willow
if r=2(even) and c=3(odd)
then, 2*3=6(even)
sonic rivet
ok, its good to check on small examples like that to understand whats going on
torn willow
k
sonic rivet
but that's not really a "proof" or anything
sonic rivet
maybe for r = some much bigger even number and c = some much bigger odd number, we might get r*c odd
yeah ok
thats true
you just showed it for the particular case r = 2 and c = 3, so I wasnt sure if you were aware of the more general rule
torn willow
r = 1235648(even) c=6958733(odd)
rc= 12356486958733
=8.59854454110^12(even)
sonic rivet
haha
torn willow
you just showed it for the particular case r = 2 and c = 3, so I wasnt sure if y...
what rule?
sonic rivet
the important thing to take away is that an even number times anything will always be even
sonic rivet
so if r is even, rc is even
but we established earlier that the right hand side, 12r + 10c + 1 is always odd
torn willow
or if the lhs is even the rhs should be even
sonic rivet
yep
sonic rivet
IF r is even
big if
we essentially found a contradiction
r cannot be even, otherwise we would have even = odd
does that make sense?
sonic rivet
r cannot be even
if you come to me with a specific value of r and c claiming that they solve the problem
torn willow
even times even is even
sonic rivet
so whatever your specific r and c are that you claim solve the problem, I know that they cannot be correct because we wouldnt have rc = 12r + 10c + 1, since one side is even and the other odd
ok I think I wasnt clear enough
What we are doing here is a proof by contradiction
sonic rivet
If you want to prove that something is true, then you can argue by contradiction: you say "well what if hypothetically what I'm trying to prove is actually false?" (in our case we want to show that r is odd, so we say "well what if r were even?") And from that, you show that somewhere along the way you have contradicting statements (in our case, we found that even = odd) and so logically, the thing we're trying to prove HAS to be true
so r has to be odd, if it were even then we would get that an even number is equal to an odd one, which is impossible
torn willow
what does this have to do with the question?
torn willow
so r has to be odd, if it were even then we would get that an even number is equ...
yes
sonic rivet
well we found that r has to be odd
torn willow
so?
sonic rivet
we can do the same for c, and show that c must also be even
we're narrowing down possible values for r and c
sonic rivet
we can also show in sort of a simillar way that c cannot be a multiple of 3 and r cannot be a multiple of 5
torn willow
how do we calculate the possibilities tho?
sonic rivet
we're not going to get an exact answer out, but we're eliminating a whole bunch of numbers
sonic rivet
+1
sonic rivet
there's some info we haven't exploited yet
if there are 12 boys per row, then there must be at least 12 columns
similarly, at least 10 rows
torn willow
k
foggy carbon
fewer than 1000 boys and girls in the school
does that mean boys + girls < 1000 or are they separately under 1000?
sonic rivet
I think < 1000 kids in total
sonic rivet
why the -1?
midnight haven
1 sit is empty
midnight haven
there’s 500 boys and girls
midnight haven
there’s 500 boys and girls
minimum
midnight haven
2 is the factor here
torn willow
common factor?
sonic rivet
what do your b and g refer to?
if they refer to the number of boys and girls respectively you should reread the question
torn willow
question:
An auditorium has a rectangular array of chairs. There are exactly 14 boys seated in each row and exactly 10 girls seated in each column. If exactly 3 chairs are empty, prove that there are at least 567 chairs are in the auditorium.
answer:
sonic rivet
oh of course
torn willow
did u get it
sonic rivet
wow
that makes a lot of sense, I thought about factoring but I thought it wouldnt work
ok so from rc = 12r + 10c + 1
we have a sum here, which is a bit annoying for divisibility
torn willow
k
sonic rivet
so we try to factorize stuff, in particular the terms with r and c
rc - 12r - 10c = 1
torn willow
how?
sonic rivet
just subtracting 12r and 10c from both side of the equation
oh you mean how do we factor?
sonic rivet
well notice that rc - 12r - 10c kind of looks like what you would get if you foiled out (r-a)(c-b)
do you see it?
torn willow
whats the common factor tho?
sonic rivet
(r-a)(c-b) = rc - rb - ac -(-ab)
sonic rivet
whats the common factor tho?
well there isnt one
torn willow
how do u do without common factor?