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using matrices |dw:1333988132875:dw|
since |A"=0 can qe use KRONECKER-CAPELLI THEOREM i did smth like |dw:1333988210313:dw|
|dw:1333988330791:dw| but latter using gauss i got |dw:1333988378631:dw|

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did i do smth wrong?? |dw:1333988426922:dw|
seems like the first and the last plane are parallel.
which???? show me please
when ever two planes are parallel, they do not have solution.
first and the last.
and two planes will give you infinite solution.
a line to be exact.
wht will b the soluyions??? can u find them Experiment?
k=1, the last equation will be -2 times the first one The solutions lie on the line of intersections of the last two plane.
you will have infinite solutions, substitute z from second and third.
that will be a line, and line has infinite points. even if the first and the last line were not parallel, you were going to have infinite solution, because 4 variables and 3 equations---> that usually ends up that way.
noo..k is a costant
because you are free to choose any value of k, ... and for particular value of k, you will have particular solution.
your solution WOULD have been in terms of k (that what i meant by infinite).
lemme think a lil bit please
tht's wht i did by using Kroneker-capelli....but the prob is that i get 0+0+0=2 at the last equation.... maybe i am being too dumb but i am really not getting it
says that A must have same rank as [A|b]
use google translate
yes...since r(A)=2 r(Ab) must be 2 too right
A has rank 2 i guess
what language is that Experiment??? O.o
it has rank two ... checked from soft. Croatian ... i don't understand too. i used google to translate.
can you tell me what is [A|b] ??
r(Ab)=4-4k at least tht's wht i found
r u getting anything?
Nop .. I am not really quite into mathematics ... I am just experimental
hmmm.....ok...:) Thanks anyway :D I appreciate it :))
i couldn't understand what does [A|b] represent.
Ab is the matrice formed by the constants before variables and the solutions |dw:1333990018496:dw|
not soultions sorry..they r not called solutions..I'm just not finding the right word -_-
oh .. yeah i remember doingnthat (thanks to you). let's see what can i do for any arbitrary value .
@hoblos any idea???
for k=5, we get rank 3
okay i will check this value from -1000 to 1000
no actually u get rang 3 for every k different from 1
it seems that we get rank 2 for k = -1
oh wait...did i go wrong by equaling 4-4k=2?????????? O.o...i think i shouldn't have done that
I think i understand what's going on.
LOL ... this is quite funny.
I have never heard of KRONECKER-CAPELLI THEOREM.
it's amazing to know that @ffm does not know something.
my google doesn't translate tht -_-
\[ x = \frac 3 2 -z \quad y = \frac 1 2 \] z can be any number
use translate feature of chrome.
Can u show me how u got that please @eliassaab ???
write x + y = 2 - z -2 x + 2 y = 2 z - 2 and solve the two equations with repect to x and y
|dw:1333990759995:dw| this can be reduced to |dw:1333990839175:dw| to find solutions.. r(A) must be same as r(A|B) so -2+2k=0 ----> k=1 thus 2y=1 ---> y=1/2 x-1/2 + z = 1 x+z = 3/2
x + y = 2 - z -x + y = z - 1 add the two equation to get y=1/2 and then find x
sorry... instead of (-2-2k) it is (-2+2k)
But Hoblos don't u get |dw:1333991218301:dw|
i got -2-2k O.o cause i tried this too
mm i cant see the whole drawing!! always different from what?
u know what....take the system and try solving it urself and ignore everything i've said
it must be equals to zero in order to have solutions!
for other values of k other than -1, we will have two parallel planes and one non parallel. means two lines. for k=-1, we have first plane = last pane, so we will have one line (I think this is what the theorem is trying to tell .... I did not understand anything)
i think the solutions r depended on k for k=1 are the soluions tht u find wht abt for kdifferent from 1?/
Can't we use the rank method?
for k different than 1 we dont have solutions..
@hoblos it's -1 i think
tht's wht we're doing Fool...Kronecker-capelli :P u don't have solutions??? how??????
If \[ k\ne 1 \], there is no solutions. If \[ k= 1 \], there are infinite number of solutions that I listed before,
why would it be -1 @experimentX ??!!
Oh god! I only didn't knew the name
@FoolForMath tht makes Anxhela happy :) now start solving it....I'm going crazyyyy!!!!!!!!! :P
because when k=1, we will have two lines ... for system of three planes.
But then I am not a Russian :P So I guess I can't blame my teacher
lol :D
well.. considering the first equation and the last one.. x-y+z =k *(-2) -2x+2y-2z=-2 -2x+2y-2z=-2k -2x+2y-2z=-2 so k must be 1
You really do not need heavy Theorems to do a rather simple problem.
i guess .. so i might have made computation mistake.
@eliassaab this is how we solve it in my univ..... @experimentX @hoblos @FoolForMath @eliassaab thanks a lot for trying and willing to help me :) forget this question :) I'll think again and repost it :))) thanks :)
damn;; my mistake was computational ... LOL A = [1, -1, 1, i; 1, 1, 1, 2; -2, 2, -2, 2]; should have been -2 at the end.
i remembred something we used to do... when r(A|B) < number of variables... we have no exact solutions.. instead we use a parameter so we take z=m this gives x= 3/2 -m so the solution is (3/2 -m, 1/2 , m) for any m belong to R

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