angela210793
Find all the solutions possible of the system ...using matrices
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wasiqss
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system is hidden LOl
Kreshnik
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wait a second.. she must be writing it !!
angela210793
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|dw:1334243246880:dw|
wasiqss
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anxhela , you want it to be done by cramer's rule or gauss elimination
angela210793
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i did |dw:1334243367057:dw| i know wassiq...but u can't use cramer when determinative equals 0
angela210793
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|dw:1334243466671:dw|
wasiqss
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ahhh yes, determinant is zero . we must go for gauss elimination
wasiqss
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do you know gauss, i can do that
angela210793
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i did |dw:1334243615536:dw| krocker-capelli but i got
angela210793
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i know Gauss but we have a k in there
wasiqss
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good :), i havnt done krocker one til yet, but yes k would have been a headache in gauss
Zarkon
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k has to be 1
Zarkon
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if you want a solution
angela210793
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how did u find it
TuringTest
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Zarkon is the only one I won't chastise for just giving answer since they are so provocative ;)
Zarkon
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-2 times row one = row 3
wasiqss
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Zarkon if k=1 , then game is over :D
Zarkon
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no...if \(k\ne 1\) then there is no soluition.
Zarkon
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solution
angela210793
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u see..i tried Gauss..and i got at the row3 0 0 0 2 where did i got wrong O.o
Zarkon
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\[\left.\begin{matrix}x-y+z=k \\ -2x+2y-2z=-2\end{matrix}\right.\]
multiply first by 2
\[\Rightarrow\]
\[\left.\begin{matrix}2x-2y+2z=2k \\ -2x+2y-2z=-2\end{matrix}\right.\]
add
you get \(0=2k-2\Rightarrow k=1\)
wasiqss
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NIce work zarkon
phi
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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
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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
angela210793
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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 -_-
angela210793
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ok...thanks
phi
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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