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BenBlackburn
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how do you form a particular solution in linear algebra when you have more unknowns then equations?
 2 years ago
 2 years ago
BenBlackburn Group Title
how do you form a particular solution in linear algebra when you have more unknowns then equations?
 2 years ago
 2 years ago

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TuringTest Group TitleBest ResponseYou've already chosen the best response.2
you mean a single solution? not possible... can you give an example?
 2 years ago

BenBlackburn Group TitleBest 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
 2 years ago

TuringTest Group TitleBest 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
 2 years ago

BenBlackburn Group TitleBest ResponseYou've already chosen the best response.0
ok i did that and i got
 2 years ago

BenBlackburn Group TitleBest ResponseYou've already chosen the best response.0
w+2x+5y5z=1 1x+5y+2z=1 1y+14z=2
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
reduce it all the way
 2 years ago

BenBlackburn Group TitleBest ResponseYou've already chosen the best response.0
see that is where i get stuck
 2 years ago

TuringTest Group TitleBest 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....
 2 years ago

BenBlackburn Group TitleBest 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?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
and it's true, you can't type it unless you know latex
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
the equation editor is too limited
 2 years ago

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

TuringTest Group TitleBest 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
 2 years ago

BenBlackburn Group TitleBest ResponseYou've already chosen the best response.0
alright so how do you get from reduced row echelon to a general solution?
 2 years ago

TuringTest Group TitleBest 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 \(\mathbb R\)
 2 years ago

TuringTest Group TitleBest 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
 2 years ago

BenBlackburn Group TitleBest ResponseYou've already chosen the best response.0
ah ok i see Thank you
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
welcome :)
 2 years ago

datanewb Group TitleBest 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: \[\left[\begin{matrix}0\\2\\9\\9\end{matrix}\right] + c\left[\begin{matrix}79\\72\\14\\1\end{matrix}\right]\]
 2 years ago

datanewb Group TitleBest 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.
 2 years ago

TuringTest Group TitleBest 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 :)
 2 years ago

datanewb Group TitleBest ResponseYou've already chosen the best response.3
No problem. Thank you. Also, following your explanation helped round out my understanding.
 2 years ago

BenBlackburn Group TitleBest ResponseYou've already chosen the best response.0
Should the particular be [2,9,9,0]?
 2 years ago

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

datanewb Group TitleBest 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.
 2 years ago

BenBlackburn Group TitleBest ResponseYou've already chosen the best response.0
ah alright thank you
 2 years ago
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