angela210793
  • angela210793
Find all the soulutions possible ...
Mathematics
schrodinger
  • schrodinger
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angela210793
  • angela210793
using matrices |dw:1333988132875:dw|
angela210793
  • angela210793
since |A"=0 can qe use KRONECKER-CAPELLI THEOREM i did smth like |dw:1333988210313:dw|
angela210793
  • angela210793
|dw:1333988330791:dw| but latter using gauss i got |dw:1333988378631:dw|

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angela210793
  • angela210793
did i do smth wrong?? |dw:1333988426922:dw|
experimentX
  • experimentX
seems like the first and the last plane are parallel.
angela210793
  • angela210793
which???? show me please
experimentX
  • experimentX
when ever two planes are parallel, they do not have solution.
experimentX
  • experimentX
first and the last.
experimentX
  • experimentX
and two planes will give you infinite solution.
experimentX
  • experimentX
a line to be exact.
angela210793
  • angela210793
wht will b the soluyions??? can u find them Experiment?
angela210793
  • angela210793
solutions****
anonymous
  • anonymous
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.
experimentX
  • experimentX
you will have infinite solutions, substitute z from second and third.
experimentX
  • experimentX
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.
angela210793
  • angela210793
noo..k is a costant
experimentX
  • experimentX
because you are free to choose any value of k, ... and for particular value of k, you will have particular solution.
experimentX
  • experimentX
your solution WOULD have been in terms of k (that what i meant by infinite).
angela210793
  • angela210793
lemme think a lil bit please
angela210793
  • angela210793
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
experimentX
  • experimentX
says that A must have same rank as [A|b]
experimentX
  • experimentX
use google translate http://lavica.fesb.hr/mat1/predavanja/node38.html
angela210793
  • angela210793
yes...since r(A)=2 r(Ab) must be 2 too right
experimentX
  • experimentX
A has rank 2 i guess
angela210793
  • angela210793
what language is that Experiment??? O.o
experimentX
  • experimentX
it has rank two ... checked from soft. Croatian ... i don't understand too. i used google to translate.
experimentX
  • experimentX
can you tell me what is [A|b] ??
angela210793
  • angela210793
r(Ab)=4-4k at least tht's wht i found
angela210793
  • angela210793
|dw:1333989679050:dw|
angela210793
  • angela210793
r u getting anything?
experimentX
  • experimentX
Nop .. I am not really quite into mathematics ... I am just experimental
angela210793
  • angela210793
hmmm.....ok...:) Thanks anyway :D I appreciate it :))
experimentX
  • experimentX
i couldn't understand what does [A|b] represent.
angela210793
  • angela210793
Ab is the matrice formed by the constants before variables and the solutions |dw:1333990018496:dw|
angela210793
  • angela210793
not soultions sorry..they r not called solutions..I'm just not finding the right word -_-
experimentX
  • experimentX
oh .. yeah i remember doingnthat (thanks to you). let's see what can i do for any arbitrary value .
angela210793
  • angela210793
@hoblos any idea???
experimentX
  • experimentX
for k=5, we get rank 3
experimentX
  • experimentX
okay i will check this value from -1000 to 1000
angela210793
  • angela210793
no actually u get rang 3 for every k different from 1
experimentX
  • experimentX
it seems that we get rank 2 for k = -1
angela210793
  • angela210793
oh wait...did i go wrong by equaling 4-4k=2?????????? O.o...i think i shouldn't have done that
experimentX
  • experimentX
I think i understand what's going on.
experimentX
  • experimentX
LOL ... this is quite funny.
anonymous
  • anonymous
I have never heard of KRONECKER-CAPELLI THEOREM.
experimentX
  • experimentX
it's amazing to know that @ffm does not know something. http://lavica.fesb.hr/mat1/predavanja/node38.html
angela210793
  • angela210793
my google doesn't translate tht -_-
anonymous
  • anonymous
\[ x = \frac 3 2 -z \quad y = \frac 1 2 \] z can be any number
experimentX
  • experimentX
use translate feature of chrome.
angela210793
  • angela210793
Can u show me how u got that please @eliassaab ???
anonymous
  • anonymous
write x + y = 2 - z -2 x + 2 y = 2 z - 2 and solve the two equations with repect to x and y
hoblos
  • hoblos
|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
anonymous
  • anonymous
x + y = 2 - z -x + y = z - 1 add the two equation to get y=1/2 and then find x
hoblos
  • hoblos
sorry... instead of (-2-2k) it is (-2+2k)
angela210793
  • angela210793
But Hoblos don't u get |dw:1333991218301:dw|
angela210793
  • angela210793
i got -2-2k O.o cause i tried this too
hoblos
  • hoblos
mm i cant see the whole drawing!! always different from what?
angela210793
  • angela210793
zero
angela210793
  • angela210793
u know what....take the system and try solving it urself and ignore everything i've said
hoblos
  • hoblos
it must be equals to zero in order to have solutions!
experimentX
  • experimentX
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)
angela210793
  • angela210793
i think the solutions r depended on k for k=1 are the soluions tht u find wht abt for kdifferent from 1?/
anonymous
  • anonymous
Can't we use the rank method?
hoblos
  • hoblos
for k different than 1 we dont have solutions..
experimentX
  • experimentX
@hoblos it's -1 i think
angela210793
  • angela210793
tht's wht we're doing Fool...Kronecker-capelli :P u don't have solutions??? how??????
anonymous
  • anonymous
If \[ k\ne 1 \], there is no solutions. If \[ k= 1 \], there are infinite number of solutions that I listed before,
hoblos
  • hoblos
why would it be -1 @experimentX ??!!
anonymous
  • anonymous
Oh god! I only didn't knew the name
angela210793
  • angela210793
@FoolForMath tht makes Anxhela happy :) now start solving it....I'm going crazyyyy!!!!!!!!! :P
experimentX
  • experimentX
because when k=1, we will have two lines ... for system of three planes.
anonymous
  • anonymous
But then I am not a Russian :P So I guess I can't blame my teacher
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Rouch%C3%A9%E2%80%93Capelli_theorem
angela210793
  • angela210793
lol :D
hoblos
  • hoblos
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
anonymous
  • anonymous
You really do not need heavy Theorems to do a rather simple problem.
experimentX
  • experimentX
i guess .. so i might have made computation mistake.
angela210793
  • angela210793
@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 :)
experimentX
  • experimentX
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.
hoblos
  • hoblos
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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