anonymous
  • anonymous
how do you form a particular solution in linear algebra when you have more unknowns then equations?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
TuringTest
  • TuringTest
you mean a single solution? not possible... can you give an example?
anonymous
  • anonymous
yep so i have three equation w+2x+5y-5z =1 w+3x+10y-3z =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
TuringTest
  • TuringTest
oh I think you are mixing up terminology go gauss-jordan elimination first and see what you can reduce this too

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anonymous
  • anonymous
ok i did that and i got
anonymous
  • anonymous
w+2x+5y-5z=1 1x+5y+2z=1 -1y+14z=-2
TuringTest
  • TuringTest
reduce it all the way
anonymous
  • anonymous
see that is where i get stuck
TuringTest
  • TuringTest
you have to keep going do you know how to do this from matrix form? that is the most efficient....
anonymous
  • anonymous
ya i do i just dident know the best way for me to type it. do you mean reduced row echelon form?
TuringTest
  • TuringTest
yes
TuringTest
  • TuringTest
and it's true, you can't type it unless you know latex
TuringTest
  • TuringTest
the equation editor is too limited
TuringTest
  • TuringTest
\[\begin{array}9w+2x+5y-5z =1\\w+3x+10y-3z =2\implies\\~~~~~~~~~x+4y+16z=-1\end{array}\left[\begin{matrix}1&2&5&-5&|1\\1&3&10&-3&|2\\0&1&4&16&|-1\end{matrix}\right]\]
TuringTest
  • TuringTest
\[\left[\begin{matrix}1&2&5&-5&|1\\1&3&10&-3&|2\\0&1&4&16&|-1\end{matrix}\right]\implies\left[\begin{matrix}1&2&5&-5&|1\\0&1&5&2&|1\\0&1&4&16&|-1\end{matrix}\right]\]\[\left[\begin{matrix}1&0&-5&-11&|-1\\0&1&5&2&|1\\0&0&-1&14&|-2\end{matrix}\right]\implies\left[\begin{matrix}1&0&0&-81&|9\\0&1&0&72&|-9\\0&0&-1&14&|-2\end{matrix}\right]\]I hope I did that right
anonymous
  • anonymous
alright so how do you get from reduced row echelon to a general solution?
TuringTest
  • TuringTest
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\[\begin{array}0w-81z=9\\x+72z=-9\\y-14z=2\end{array}\]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 \[\begin{array}0w=81t+9\\x=-72t-9\\y=14t+2\end{array}\text{ where }t\in\mathbb R\]and that is our solution space in \(\mathbb R\)
TuringTest
  • TuringTest
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%2B5y-5z++%3D1+%2Cw%2B3x%2B10y-3z+%3D2%2Cx%2B4y%2B16z%3D-1+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
anonymous
  • anonymous
ah ok i see Thank you
TuringTest
  • TuringTest
welcome :)
datanewb
  • datanewb
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 x-particular. 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 A|b 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 x-particular 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: \[\left[\begin{matrix}0\\2\\-9\\9\end{matrix}\right] + c\left[\begin{matrix}79\\-72\\14\\1\end{matrix}\right]\]
datanewb
  • datanewb
This answer is equivalent to the one provided by TuringTest, but presented differently: the way Strang presented it.
TuringTest
  • TuringTest
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 :)
datanewb
  • datanewb
No problem. Thank you. Also, following your explanation helped round out my understanding.
anonymous
  • anonymous
Should the particular be [2,-9,9,0]?
anonymous
  • anonymous
just take the particular case of \(c=0\) and you get \[\left[\begin{array}00\\2\\-2\\9\end{array}\right]\]
datanewb
  • datanewb
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.
anonymous
  • anonymous
ah alright thank you

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