anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Is there even a solution? The third row doesn't make a lot of sense 0 + 0 + 0 + 0 + 1 = 0?
ganeshie8
  • ganeshie8
is that the augmented matrix in row echelon form ?
anonymous
  • anonymous
I'm not sure. The problem doesn't say.

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anonymous
  • anonymous
Wait! okay, it is the augmented matrices.
ganeshie8
  • ganeshie8
good, how many pivots are there ?
anonymous
  • anonymous
I don't know. How do I tell if they are pivots?
ganeshie8
  • ganeshie8
in echelon form, pivot is the first "non zero" number in a row
anonymous
  • anonymous
So there would be three pivots then.
ganeshie8
  • ganeshie8
Right, `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
anonymous
  • anonymous
Okay.
ganeshie8
  • ganeshie8
just 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
  • ganeshie8
x1 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
  • ganeshie8
notice that the corresponding system is ``` x5 = 0 x2 + 4x3 - x4 = 6 x1 - 5x3 - 8x5 = 3 ``` let x3 = x4 = 0 and solve remaining variables
anonymous
  • anonymous
Okay, so how do I solve for the pivot points?
ganeshie8
  • ganeshie8
plugin x3=x4=0 in above system
ganeshie8
  • ganeshie8
then you will have 3 equations and 3 unknowns in triangular form which is a piece of cake to solve by back substitution
anonymous
  • anonymous
so.. x2 = 5 x5 = 0 x1=0
ganeshie8
  • ganeshie8
try again
ganeshie8
  • ganeshie8
the 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 ?
anonymous
  • anonymous
Yes.
ganeshie8
  • ganeshie8
solve it
ganeshie8
  • ganeshie8
its trivial, but you have to do it
anonymous
  • anonymous
I'm trying to do it right now, give me a moment.
anonymous
  • anonymous
x5 = 0 x2 + 4(0) - 0 = 6 -> x2 = 6 x1 - 5(0) - 8(0) = 3 -> x1 = 3
ganeshie8
  • ganeshie8
so \(\begin{pmatrix} 3\\6\\0\\0\\0 \end{pmatrix}\) is a particular solution we still need to find the null solution, familiar with the process ?
anonymous
  • anonymous
Nope
ganeshie8
  • ganeshie8
null solution is the solution to following system : ``` x5 = 0 x2 + 4x3 - x4 = 0 x1 - 5x3 - 8x5 = 0 ```
ganeshie8
  • ganeshie8
Notice that I have put 0s on the right hand side now
ganeshie8
  • ganeshie8
again 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
  • anonymous
Do I use the values for the system that we came up with earlier?
ganeshie8
  • ganeshie8
let x3 = 0, x4 = 1 and find one null solution let x3 = 1, x4 = 0 and find the other null solution
ganeshie8
  • ganeshie8
null 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
anonymous
  • anonymous
x2 - 1 = 0 x2 = 1
ganeshie8
  • ganeshie8
yes, keep going find the remaining two variables
anonymous
  • anonymous
x1 = 0
anonymous
  • anonymous
Unless you wanted me to use "let x3 = 1, x4 = 0 and find the other null solution" to solve x1 - 5x3 - 8x5 = 0
anonymous
  • anonymous
Because then x1 would equal 5
ganeshie8
  • ganeshie8
one thing at a time
ganeshie8
  • ganeshie8
null 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 ?
anonymous
  • anonymous
Yeah.
ganeshie8
  • ganeshie8
save it, find the other null solution also similarly
ganeshie8
  • ganeshie8
null 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
  • anonymous
Okay, so then x1 = 1 and x2 = -1
ganeshie8
  • ganeshie8
try again
anonymous
  • anonymous
Thanks 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
  • ganeshie8
we're mostly done just wait one minute
ganeshie8
  • ganeshie8
null 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
  • ganeshie8
so 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
  • ganeshie8
then 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
  • anonymous
Closing now as it is no longer helpful.

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