The Original Gay Porn Community - Free Gay Movies and Photos, Gay Porn Site Reviews and Adult Gay Forums

  • Welcome To Just Us Boys - The World's Largest Gay Message Board Community

    In order to comply with recent US Supreme Court rulings regarding adult content, we will be making changes in the future to require that you log into your account to view adult content on the site.
    If you do not have an account, please register.
    REGISTER HERE - 100% FREE / We Will Never Sell Your Info

    PLEASE READ: To register, turn off your VPN (iPhone users- disable iCloud); you can re-enable the VPN after registration. You must maintain an active email address on your account: disposable email addresses cannot be used to register.

  • Hi Guest - Did you know?
    Hot Topics is a Safe for Work (SFW) forum.

Anyone good with Linear Algebra?

EtherealMystic

On the Prowl
Joined
Nov 23, 2010
Posts
52
Reaction score
0
Points
0
Location
Richmond
I'm doing my homework and have these 3 problems left, but I have no idea how to approach them...

Any help would be appreciated!
 

Attachments

  • #1.jpg
    #1.jpg
    7.4 KB · Views: 38
  • #2.jpg
    #2.jpg
    17.3 KB · Views: 58
  • #3.jpg
    #3.jpg
    21.1 KB · Views: 38
I got my degree in linear algebra/differential equations.
In 1992.
And haven't looked at it since.
Sorry.

Lex
 
you should have done similar problems before. at least for 2 and 3. 2 can also be solved with any mathematica like program.
 
I probably have but the different ways it is asked just throws me off.

This class I have to use MapleSoft and as I've never used it before I wouldn't really know where to begin. This summer class is quiet fast paced so he doesn't really go over every type of questions he could ask...

Hopefully I will figure it out sooner or later.
 
Be patient. There are a few math wizard here. They'll show up and help you, I'm sure.
 
2 is really easy, it's just a system of 3 unknown in three equations :

you need to search alpha (a), beta (b) and gamma (c) so that :

a+b+c = 0 (1)
2a +4c = 0 (2)
3a + 2b +4c = 0 (3)

(2) gives you a = -2c
you replace a in (1) and (3) by -2c you got
b - c = 0 (1')
2b - 2c = 0 (3')

so, any real numbers can be b and c and a is -2c
 
for #1 and #3 you need to search for definitions. You apply calculus rules to the matrix and you will find the solutions. It's really only definition applications here.

for #3, a) and b) are definition application and calculus. Once you have established a) and b) you get c) trivially
as c) = a) + b) :
1/2(A + At) + 1/2(A - At) = A (duh !)
d) is just an example of what you just have done, all you need to do is calculate A (duh ! it's A !) and At (so the transform form of A). Again just application of definition.

good luck
 
Thanks oakpope. I got the same answers as you for #2 and #3. I'm just having a hard time explaining #1, I think I'm making it harder than it should be.
 
for #1 and #3 you need to search for definitions. You apply calculus rules to the matrix and you will find the solutions. It's really only definition applications here.

for #3, a) and b) are definition application and calculus. Once you have established a) and b) you get c) trivially
as c) = a) + b) :
1/2(A + At) + 1/2(A - At) = A (duh !)
d) is just an example of what you just have done, all you need to do is calculate A (duh ! it's A !) and At (so the transform form of A). Again just application of definition.

good luck

That's some sexy Algebra-ing!
 
Thanks oakpope. I got the same answers as you for #2 and #3. I'm just having a hard time explaining #1, I think I'm making it harder than it should be.

you're welcome ! :) for #1 I am not familiar with the English expression, so it's more difficult for me to help you (I'm French, sorry :) )
 
for #1 this code (from wiki) will give you the result, but you should explain yourself the calculus, so understand the program and search in your books the validation for it :

function ToRowEchelonForm(Matrix M) is
nr := number of rows in M
nc := number of columns in M

for 0 ≤ r < nr do
allZeros := true
for 0 ≤ c < nc do
if M[r, c] != 0 then
allZeros := false
exit for
end if
end for
if allZeros = true then
In M, swap row r with row nr - 1
nr := nr - 1
end if
end for

p := 0
while p < nr and p < nc do
label nextPivot:
r := 1
while M[p, p] = 0 do
if (p + r) <= nr then
p := p + 1
goto nextPivot
end if
In M, swap row p with row (p + r) <-- bug. nr < p+r at this point
r := r + 1
end while
for 1 ≤ r < (nr - p) do
if M[p + r, p] != 0 then
x := -M[p + r, p] / M[p, p]
for p ≤ c < nc do
M[p + r, c] := M[p, c] * x + M[p + r, c]
end for
end if
end for
p := p + 1
end while
end function
 
Back
Top