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how do you form a particular solution in linear algebra when you have more unknowns then equations?
 one year ago
 one year ago
how do you form a particular solution in linear algebra when you have more unknowns then equations?
 one year ago
 one year ago

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TuringTestBest ResponseYou've already chosen the best response.2
you mean a single solution? not possible... can you give an example?
 one year ago

BenBlackburnBest ResponseYou've already chosen the best response.0
yep so i have three equation w+2x+5y5z =1 w+3x+10y3z =2 x+4y+16z=1 and im told to find the general solution. I know that this is the particular and the homogenious solution but i am unsure how to get them in this case
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
oh I think you are mixing up terminology go gaussjordan elimination first and see what you can reduce this too
 one year ago

BenBlackburnBest ResponseYou've already chosen the best response.0
ok i did that and i got
 one year ago

BenBlackburnBest ResponseYou've already chosen the best response.0
w+2x+5y5z=1 1x+5y+2z=1 1y+14z=2
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
reduce it all the way
 one year ago

BenBlackburnBest ResponseYou've already chosen the best response.0
see that is where i get stuck
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
you have to keep going do you know how to do this from matrix form? that is the most efficient....
 one year ago

BenBlackburnBest ResponseYou've already chosen the best response.0
ya i do i just dident know the best way for me to type it. do you mean reduced row echelon form?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
and it's true, you can't type it unless you know latex
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
the equation editor is too limited
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
\[\begin{array}9w+2x+5y5z =1\\w+3x+10y3z =2\implies\\~~~~~~~~~x+4y+16z=1\end{array}\left[\begin{matrix}1&2&5&5&1\\1&3&10&3&2\\0&1&4&16&1\end{matrix}\right]\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
\[\left[\begin{matrix}1&2&5&5&1\\1&3&10&3&2\\0&1&4&16&1\end{matrix}\right]\implies\left[\begin{matrix}1&2&5&5&1\\0&1&5&2&1\\0&1&4&16&1\end{matrix}\right]\]\[\left[\begin{matrix}1&0&5&11&1\\0&1&5&2&1\\0&0&1&14&2\end{matrix}\right]\implies\left[\begin{matrix}1&0&0&81&9\\0&1&0&72&9\\0&0&1&14&2\end{matrix}\right]\]I hope I did that right
 one year ago

BenBlackburnBest ResponseYou've already chosen the best response.0
alright so how do you get from reduced row echelon to a general solution?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
so we got all the way to\[\left[\begin{matrix}1&2&5&5&1\\1&3&10&3&2\\0&1&4&16&1\end{matrix}\right]\implies\left[\begin{matrix}1&2&5&5&1\\0&1&5&2&1\\0&1&4&16&1\end{matrix}\right]\]\[\left[\begin{matrix}1&0&5&11&1\\0&1&5&2&1\\0&0&1&14&2\end{matrix}\right]\implies\left[\begin{matrix}1&0&0&81&9\\0&1&0&72&9\\0&0&1&14&2\end{matrix}\right]\](I fixed the minus sign so now this is complete) translate this back into letters and we have\[\begin{array}0w81z=9\\x+72z=9\\y14z=2\end{array}\]so everything is dependent on \(z\) (note we could have done this so that it was a different variable, but we will get the same solution space) so what we do is we say that this represents a solution space depending on the value of the parameter \(z\) so we rename \(z\) as \(t\) and say that \[\begin{array}0w=81t+9\\x=72t9\\y=14t+2\end{array}\text{ where }t\in\mathbb R\]and that is our solution space in \(\mathbb R\)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
I seem to have made a mistake somewhere in the matrix, but the concept is right here is the solution as wolfram has it http://www.wolframalpha.com/input/?i=solve+w%2B2x%2B5y5z++%3D1+%2Cw%2B3x%2B10y3z+%3D2%2Cx%2B4y%2B16z%3D1+for+x%2Cy%2Cw though they do not rename z that is common practice to stress the dependence of each variable on a single real parameter not also that we could have solved for the other 3 variables as well instead
 one year ago

BenBlackburnBest ResponseYou've already chosen the best response.0
ah ok i see Thank you
 one year ago

datanewbBest ResponseYou've already chosen the best response.3
Ben, I think you are following the lectures from Strang's OCW. I wanted to point out that to get a particular solution, one easy way is to set the free variables to zero. This could be done as soon as the matrix A has been reduced to U... w+2x+ 5y 5z= 1 1x+ 5y+ 2z= 1 1y+14z=2 or in terms of matrices:\[\left[\begin{matrix}1&2&5&5\\ 0 &1&5&2\\0&0&1&14\end{matrix}\right]\left[\begin{matrix}w\\x\\y\\z\end{matrix}\right]=\left[\begin{matrix}1\\1\\2\end{matrix}\right]\] Setting the free variable (in this case z), you can use back substitution to calculate z = 0, y = 2, x = 9, w = 9... this is one particular solution. Strang calls it xparticular. Then adding all possible solutions to Ax = 0, N(A) gives all of the solutions to Ax = b. If you continue to reduce U to R, then the augmented matrix Ab looks like: \[\left[\begin{matrix}1&0&0&79&9\\0&1&0&72&\neg9\\0&0&1&\neg14&2\end{matrix}\right]\] And, you can simply read the solutions to xparticular off of b. Finally, setting the free variable, z, to 1, the N(A) is also easily discerned from R. z = 1, y = 14, x=72, w=79. So the full solution would look something like this in vector form: \[\left[\begin{matrix}0\\2\\9\\9\end{matrix}\right] + c\left[\begin{matrix}79\\72\\14\\1\end{matrix}\right]\]
 one year ago

datanewbBest ResponseYou've already chosen the best response.3
This answer is equivalent to the one provided by TuringTest, but presented differently: the way Strang presented it.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
sorry, I was wondering what they meant by "particular solution" in this case it clearly is asking for what @datanewb stated. thanks for finishing this problem correctly :)
 one year ago

datanewbBest ResponseYou've already chosen the best response.3
No problem. Thank you. Also, following your explanation helped round out my understanding.
 one year ago

BenBlackburnBest ResponseYou've already chosen the best response.0
Should the particular be [2,9,9,0]?
 one year ago

colorfulBest ResponseYou've already chosen the best response.0
just take the particular case of \(c=0\) and you get \[\left[\begin{array}00\\2\\2\\9\end{array}\right]\]
 one year ago

datanewbBest ResponseYou've already chosen the best response.3
colorful is right. \[\left[\begin{matrix}2\\9\\9\\0\end{matrix}\right]\] is just one of infinitely many particular solutions. It is not more or less correct than if we had chosen \[\left[\begin{matrix}79\\70\\5\\10\end{matrix}\right]\] as the particular solution. Choosing the particular solution in which all of the free variables are equal to zero, is the most efficient way to find a particular solution.
 one year ago

BenBlackburnBest ResponseYou've already chosen the best response.0
ah alright thank you
 one year ago
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