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Find all the solutions possible of the system ...using matrices

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system is hidden LOl
wait a second.. she must be writing it !!

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Other answers:

anxhela , you want it to be done by cramer's rule or gauss elimination
i did |dw:1334243367057:dw| i know wassiq...but u can't use cramer when determinative equals 0
ahhh yes, determinant is zero . we must go for gauss elimination
do you know gauss, i can do that
i did |dw:1334243615536:dw| krocker-capelli but i got
i know Gauss but we have a k in there
good :), i havnt done krocker one til yet, but yes k would have been a headache in gauss
k has to be 1
if you want a solution
how did u find it
Zarkon is the only one I won't chastise for just giving answer since they are so provocative ;)
-2 times row one = row 3
Zarkon if k=1 , then game is over :D
no...if \(k\ne 1\) then there is no soluition.
u see..i tried Gauss..and i got at the row3 0 0 0 2 where did i got wrong O.o
\[\left.\begin{matrix}x-y+z=k \\ -2x+2y-2z=-2\end{matrix}\right.\] multiply first by 2 \[\Rightarrow\] \[\left.\begin{matrix}2x-2y+2z=2k \\ -2x+2y-2z=-2\end{matrix}\right.\] add you get \(0=2k-2\Rightarrow k=1\)
NIce work zarkon
  • phi
If you put the augmented matrix in reduced row echelon form you get \[\left[\begin{matrix}1 &0 & 1&1 +k/2 \\ 0 & 1 &0 & 1-k/2 \\ 0 & 0 & 0&2k-2 \\\end{matrix}\right]\]
  • phi
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 non-zero null space
the system would have infinite solutions for r(A)=r(Ab) and no solutions for r(A)
  • phi
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

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