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anonymous
 one year ago
Find the general solutions of
1 0 5 0 8 3
0 1 4 1 0 6
0 0 0 0 1 0
0 0 0 0 0 0
Last question I'm going to ask, I promise.
anonymous
 one year ago
Find the general solutions of 1 0 5 0 8 3 0 1 4 1 0 6 0 0 0 0 1 0 0 0 0 0 0 0 Last question I'm going to ask, I promise.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is there even a solution? The third row doesn't make a lot of sense 0 + 0 + 0 + 0 + 1 = 0?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7is that the augmented matrix in row echelon form ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure. The problem doesn't say.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait! okay, it is the augmented matrices.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7good, how many pivots are there ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't know. How do I tell if they are pivots?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7in echelon form, pivot is the first "non zero" number in a row

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So there would be three pivots then.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7Right, `1` 0 5 0 8 3 0 `1` 4 1 0 6 0 0 0 0 `1` 0 0 0 0 0 0 0 you have 3 pivot columns and 2 free columns lets find a particular solution by letting free variables = 0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7just for definiteness, call the variables : x1, x2, x3, x4, x5 so to find a particular solution you let the free variables x3 = x4 = 0 and solve for the pivot variables x1, x2, x5

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7x1 x2 x3 x4 x5 b `1` 0 5 0 8 3 0 `1` 4 1 0 6 0 0 0 0 `1` 0 0 0 0 0 0 0 ^ ^ free columns

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7notice that the corresponding system is ``` x5 = 0 x2 + 4x3  x4 = 6 x1  5x3  8x5 = 3 ``` let x3 = x4 = 0 and solve remaining variables

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so how do I solve for the pivot points?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7plugin x3=x4=0 in above system

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7then you will have 3 equations and 3 unknowns in triangular form which is a piece of cake to solve by back substitution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so.. x2 = 5 x5 = 0 x1=0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7the corresponding system is ``` x5 = 0 x2 + 4x3  x4 = 6 x1  5x3  8x5 = 3 ``` letting x3 = x4 = 0, the system becomes ``` x5 = 0 x2 = 6 x1  8x5 = 3 ``` yes ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7its trivial, but you have to do it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm trying to do it right now, give me a moment.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x5 = 0 x2 + 4(0)  0 = 6 > x2 = 6 x1  5(0)  8(0) = 3 > x1 = 3

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7so \(\begin{pmatrix} 3\\6\\0\\0\\0 \end{pmatrix}\) is a particular solution we still need to find the null solution, familiar with the process ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7null solution is the solution to following system : ``` x5 = 0 x2 + 4x3  x4 = 0 x1  5x3  8x5 = 0 ```

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7Notice that I have put 0s on the right hand side now

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7again we mess with the free variables to cookup the null solution : since we have two free variables, we expect to get two independent null solutions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do I use the values for the system that we came up with earlier?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7let x3 = 0, x4 = 1 and find one null solution let x3 = 1, x4 = 0 and find the other null solution

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7null solution is the solution to following system : ``` x5 = 0 x2 + 4x3  x4 = 0 x1  5x3  8x5 = 0 ``` plugin x3 = 0, x4 = 1 and solve the other variables

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7yes, keep going find the remaining two variables

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Unless you wanted me to use "let x3 = 1, x4 = 0 and find the other null solution" to solve x1  5x3  8x5 = 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Because then x1 would equal 5

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7null solution is the solution to following system : ``` x5 = 0 x2 + 4x3  x4 = 0 x1  5x3  8x5 = 0 ``` plugin x3 = 0, x4 = 1 and solve the other variables this gives a null solution \(\begin{pmatrix} 0\\1\\0\\1\\0\end{pmatrix}\), yes ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7save it, find the other null solution also similarly

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7null solution is the solution to following system : ``` x5 = 0 x2 + 4x3  x4 = 0 x1  5x3  8x5 = 0 ``` plugin x3 = 1, x4 = 0 and solve the other variables

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so then x1 = 1 and x2 = 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for your help ganeshie8, but I have to go to sleep, it's really late here and I have to wake up for class. Have a wonderful evening, and I'm sorry for not being about to finish the problem.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7we're mostly done just wait one minute

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7null solution is the solution to following system : ``` x5 = 0 x2 + 4x3  x4 = 0 x1  5x3  8x5 = 0 ``` plugin x3 = 1, x4 = 0 and solve the other variables that gives the null solution \(\begin{pmatrix} 5\\4\\1\\0\\0\end{pmatrix}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7so the nullspace is given by \[\large c_1\begin{pmatrix} 0\\1\\0\\1\\0\end{pmatrix} + c_2\begin{pmatrix} 5\\4\\1\\0\\0\end{pmatrix}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.7then the final general solutions is given by \(x_p + x_{null}\) : \[x_{\text{general }} = \begin{pmatrix} 3\\6\\0\\0\\0 \end{pmatrix} +\large c_1\begin{pmatrix} 0\\1\\0\\1\\0\end{pmatrix} + c_2\begin{pmatrix} 5\\4\\1\\0\\0\end{pmatrix} \]

anonymous
 11 months ago
Best ResponseYou've already chosen the best response.0Closing now as it is no longer helpful.
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