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Find all the solutions possible of the system ...using matrices
 2 years ago
 2 years ago
Find all the solutions possible of the system ...using matrices
 2 years ago
 2 years ago

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

angela210793Best ResponseYou've already chosen the best response.0
dw:1334243246880:dw
 2 years ago

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

angela210793Best ResponseYou've already chosen the best response.0
i did dw:1334243367057:dw i know wassiq...but u can't use cramer when determinative equals 0
 2 years ago

angela210793Best ResponseYou've already chosen the best response.0
dw:1334243466671:dw
 2 years ago

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

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

angela210793Best ResponseYou've already chosen the best response.0
i did dw:1334243615536:dw krockercapelli but i got
 2 years ago

angela210793Best ResponseYou've already chosen the best response.0
i know Gauss but we have a k in there
 2 years ago

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

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

ZarkonBest ResponseYou've already chosen the best response.5
2 times row one = row 3
 2 years ago

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

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

angela210793Best ResponseYou've already chosen the best response.0
u see..i tried Gauss..and i got at the row3 0 0 0 2 where did i got wrong O.o
 2 years ago

ZarkonBest 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 \(0=2k2\Rightarrow k=1\)
 2 years ago

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

phiBest ResponseYou've already chosen the best response.0
as 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
 2 years ago

angela210793Best ResponseYou've already chosen the best response.0
the 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 _
 2 years ago

phiBest ResponseYou've already chosen the best response.0
The 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
 2 years ago
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