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Nyoxis
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Hey all, I've got some homework that I'm having troubles with, so I thought I'd seek advice from the intellectuals @ Gameamp. =)
'Prove that the matrix:
(k+1) | (k)
(-k) | (k+1)
is non-singular for all real values of k.'
Answer: I found the determinant to be 2(k^2)+2k+1, but how do I actually use this as evidence? Also, is a singular matrix a coloumn/row matrix?
The next question:
'Prove that the equations, (x-y=ax) and (x+y=ay) have a unique solution for all possible values of the real constant 'a'.'
Answer, in matrix form, the determinant of the equations is (a^2)-2a+2 which cannot equal zero for any real values.
But the problem's with the next part, 'solve the equations for all real values of 'a'.'
Thanks guys! =)
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| 02/03/08 13:51 |
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Skittle Cody
GameAmp Staff
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O_o
4?
Buy me a giraffe <3
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| 02/03/08 14:07 |
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Nyoxis
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Flicks through answer book, searches down page...WTF! He's right! =O
...but not quite. =P
Oh btw, Cody, a friend keeps telling me to go see 'Megadeth' with him, and he says Job for a Cowboy's supporting.
I keep telling him I'm not going. *grin*
***THIS POST HAS BEEN EDITED***
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| 02/03/08 14:10 |
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Skittle Cody
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| QUOTE |
Flicks through answer book, searches down page...WTF! He's right! =O
...but not quite. =P
Oh btw, Cody, a friend keeps telling me to go see 'Megadeth' with him, and he says Job for a Cowboy's supporting.
I keep telling him I'm not going. *grin* |
YOU WILL GO!
At least go for me!
Take pictures.
Buy me a giraffe <3
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| 02/03/08 14:21 |
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itsthatemochild
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even if you dont like the music, you NEED to go for megadeth. those guys are *kitten* legends. and imho job for a cowboy arent half bad either
"The most merciful thing in the world... is the inability of the human mind to correlate all its contents." -H.P. Lovecraft
"He who is not radical at twenty has no heart,
while he who is still radical at forty has no brain."
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| 02/03/08 14:32 |
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Liquide
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That isn't very hard.
The equivalent of k=.5, so the equivalent of .5=k. Normally the - in front of "k" means it is negative, but in here it doesn't mean it's negative because k=.5. If "k" would be negative (for example if "k" would be -.5) the - in front of "k" WOULD be negative and the - in front of the .5 would be useless. Hopefully this helped you.
thaskippy: | QUOTE | | omg... you should be the new dalai lama 0.o |
Mai Englisch is'nt verry good, so plaese keep taht in you're minds.
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| 02/03/08 14:42 |
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Nyoxis
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| QUOTE | That isn't very hard.
The equivalent of k=.5, so the equivalent of .5=k. Normally the - in front of "k" means it is negative, but in here it doesn't mean it's negative because k=.5. If "k" would be negative (for example if "k" would be -.5) the - in front of "k" WOULD be negative and the - in front of the .5 would be useless. Hopefully this helped you. |
Sorry this doesn't help. =S
I've realised how to do part of it now though, thanks for the effort, +cred.
***THIS POST HAS BEEN EDITED***
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| 02/03/08 14:57 |
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jaffar_al_kahyet
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You wins tha internetz.
Age of Conan: Aeson ~ Battlescar [PvP-En] ~ Cimmerian Barbarian
Guildwars: Tydus of Ascalon ~ Proud Leader of The Preschool Teaparty Massacre [Opal]
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| 02/03/08 15:08 |
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Brynden
Posts: 45
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| QUOTE | Hey all, I've got some homework that I'm having troubles with, so I thought I'd seek advice from the intellectuals @ Gameamp. =)
'Prove that the matrix:
(k+1) | (k)
(-k) | (k+1)
is non-singular for all real values of k.'
Answer: I found the determinant to be 2(k^2)+2k+1, but how do I actually use this as evidence? Also, is a singular matrix a coloumn/row matrix?
The next question:
'Prove that the equations, (x-y=ax) and (x+y=ay) have a unique solution for all possible values of the real constant 'a'.'
Answer, in matrix form, the determinant of the equations is (a^2)-2a+2 which cannot equal zero for any real values.
But the problem's with the next part, 'solve the equations for all real values of 'a'.'
Thanks guys! =) |
A non-singular matrix is the same as saying it is invertible, and invertible matrices' determinants cannot equal zero. So for the first part you have to show that there are no values of k that make the determinant equal to zero, which is true, since solutions to that quadratic equation are -1/2 +/- i/2, which is an imaginary number.
For the second part, you just do the same thing, because systems of linear equations are solvable if their matrices are invertible. So just show that there is no value of a that makes the determinant equal zero. The thing about solving it for all real values of a just means that if you solve for a and get an imaginary number, then there are no real solutions to the equation. If you don't know what a real number is, it's simply a number that doesn't depend on the imaginary number i, which is the square root of negative 1. So if you end up taking square roots of negative numbers, your answer is imaginary and you have proven what you set out to prove.
Signature by Mimori
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| 02/03/08 15:42 |
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Jamnog
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| QUOTE |
'Prove that the equations, (x-y=ax) and (x+y=ay) have a unique solution for all possible values of the real constant 'a'.'
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umm first insides outsides last :/ no thats not it
you make them = to 0 err... both sides = 0.... umm graph for values of x and y... oh wait I think its substituting a value for a on a scale say a=10 and a=20
so a=10 then x-y=10x so take the x acros to the other side by adding it to both sides I think... no wait divide both sides by x to get rid of the x leaving 10 ok so...
10=(x-y)/x where a = 10 now what number could x be that comes out as 10 hmm not forgetting that 10 also in the other equasion.
becomes
10=(x+y)/y so the two have to be true at the same time only one number can fit x and only one number can fit y.
you could say that (x+y)/y=(x-y)/x
perhaps we can get all those x and y on the same side :/
however that is a proof of sorts that the same thing is true no matter the constant value of a... something doesnt seem right here
ok lets say x=4 why not, who cares its a guildwars forum and I want to know if this works. Its also been a long time since I tried to figure out these weird puzzles.
(4+y)/y=(4-y)/4
both sides x 4
4(4+y)/y = 4-y ok then lets add y to both sides
[4(4+y)/y]+y = 4 hmm this isn't really going anywhere.
oh wait didn't I say that x=4 oh that means x=[x(x+y)/y]+y
not much use if you want x on its own :/ umm or x-x(x+y)/y-y=0
x-y=ax -> x-y-ax=0
x+y=ay -> x+y-ay=0
next I think I remember making a graph but its all hazy sry
oh right if you substitute in values of x and y and plot each one on a graph I think the point at which it crosses 0 is important for some reason.
edit : I cant believe your going to miss out on megadeath :/
edit 2 : hey bry seems to know what he is talkign about read his post mine will just fry peoples brains. i failed on so many levals in that post.
***THIS POST HAS BEEN EDITED***
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| 02/03/08 16:14 |
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Nyoxis
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Joined: 12/31/1969
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| QUOTE | A non-singular matrix is the same as saying it is invertible, and invertible matrices' determinants cannot equal zero. So for the first part you have to show that there are no values of k that make the determinant equal to zero, which is true, since solutions to that quadratic equation are -1/2 +/- i/2, which is an imaginary number.
For the second part, you just do the same thing, because systems of linear equations are solvable if their matrices are invertible. So just show that there is no value of a that makes the determinant equal zero. The thing about solving it for all real values of a just means that if you solve for a and get an imaginary number, then there are no real solutions to the equation. If you don't know what a real number is, it's simply a number that doesn't depend on the imaginary number i, which is the square root of negative 1. So if you end up taking square roots of negative numbers, your answer is imaginary and you have proven what you set out to prove. |
Oh that's brilliant, explained it perfectly, I've used imaginary numbers before, so I'm cool with recognising when they come into play, but didn't think of using them as good evidence here.
And thanks for the effort Jamnog ;) I may, possibly consider Megadeth, maybe. =)
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| 02/03/08 16:32 |
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Liquide
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| QUOTE | | QUOTE | That isn't very hard.
The equivalent of k=.5, so the equivalent of .5=k. Normally the - in front of "k" means it is negative, but in here it doesn't mean it's negative because k=.5. If "k" would be negative (for example if "k" would be -.5) the - in front of "k" WOULD be negative and the - in front of the .5 would be useless. Hopefully this helped you. |
Sorry this doesn't help. =S
I've realised how to do part of it now though, thanks for the effort, +cred. |
Well, this comment wasn't really serious... I didn't even know what you meant with the | between the k's... I'm on high school and we haven't had those |'s yet, we are at the "difficult" 1\k=k^2 or something...
well thanks for the cred anyway :D
thaskippy: | QUOTE | | omg... you should be the new dalai lama 0.o |
Mai Englisch is'nt verry good, so plaese keep taht in you're minds.
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| 02/03/08 16:58 |
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Nyoxis
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| QUOTE | Well, this comment wasn't really serious... I didn't even know what you meant with the | between the k's... I'm on high school and we haven't had those |'s yet, we are at the "difficult" 1k=k^2 or something...
well thanks for the cred anyway :D |
Those |'s were used incorrectly, as my way of breaking up the expressions, don't know if there's a way of doing matrices in html...?
Though had I put a | either side of an expression it would mean 'modulus of'. e.g. |x| means modulus of x, or how far away it is from 0, so |6|=6, |-6|=6, |(3,4)|=5. =)
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| 02/03/08 17:32 |
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