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Kreshnik
 2 years ago
Best ResponseYou've already chosen the best response.0wait a second.. she must be writing it !!

angela210793
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1334243246880:dw

wasiqss
 2 years ago
Best ResponseYou've already chosen the best response.0anxhela , you want it to be done by cramer's rule or gauss elimination

angela210793
 2 years ago
Best ResponseYou've already chosen the best response.0i did dw:1334243367057:dw i know wassiq...but u can't use cramer when determinative equals 0

angela210793
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1334243466671:dw

wasiqss
 2 years ago
Best ResponseYou've already chosen the best response.0ahhh yes, determinant is zero . we must go for gauss elimination

wasiqss
 2 years ago
Best ResponseYou've already chosen the best response.0do you know gauss, i can do that

angela210793
 2 years ago
Best ResponseYou've already chosen the best response.0i did dw:1334243615536:dw krockercapelli but i got

angela210793
 2 years ago
Best ResponseYou've already chosen the best response.0i know Gauss but we have a k in there

wasiqss
 2 years ago
Best ResponseYou've already chosen the best response.0good :), i havnt done krocker one til yet, but yes k would have been a headache in gauss

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0Zarkon is the only one I won't chastise for just giving answer since they are so provocative ;)

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.52 times row one = row 3

wasiqss
 2 years ago
Best ResponseYou've already chosen the best response.0Zarkon if k=1 , then game is over :D

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.5no...if \(k\ne 1\) then there is no soluition.

angela210793
 2 years ago
Best ResponseYou've already chosen the best response.0u see..i tried Gauss..and i got at the row3 0 0 0 2 where did i got wrong O.o

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.5\[\left.\begin{matrix}xy+z=k \\ 2x+2y2z=2\end{matrix}\right.\] multiply first by 2 \[\Rightarrow\] \[\left.\begin{matrix}2x2y+2z=2k \\ 2x+2y2z=2\end{matrix}\right.\] add you get \(0=2k2\Rightarrow k=1\)

phi
 2 years ago
Best ResponseYou've already chosen the best response.0If you put the augmented matrix in reduced row echelon form you get \[\left[\begin{matrix}1 &0 & 1&1 +k/2 \\ 0 & 1 &0 & 1k/2 \\ 0 & 0 & 0&2k2 \\\end{matrix}\right]\]

phi
 2 years ago
Best ResponseYou've already chosen the best response.0as zarkon noted, k must be 1 for there to be a solution. However, note that there are an infinite # of sols if k=1, because you have a nonzero null space

angela210793
 2 years ago
Best ResponseYou've already chosen the best response.0the system would have infinite solutions for r(A)=r(Ab) and no solutions for r(A)<r(Ab) how do u find the range of Ab when u have k in there..r(A)=r(Ab) if k wasn't 1....sorry mozilla froze _

phi
 2 years ago
Best ResponseYou've already chosen the best response.0The complete solution is \[\left(\begin{matrix}3/2 \\ 1/2 \\ 0\end{matrix}\right)+c\left(\begin{matrix}1 \\ 0 \\ 1\end{matrix}\right)\] where c is arbitrary
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