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

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- anonymous

Is there even a solution? The third row doesn't make a lot of sense
0 + 0 + 0 + 0 + 1 = 0?

- ganeshie8

is that the augmented matrix in row echelon form ?

- anonymous

I'm not sure. The problem doesn't say.

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## More answers

- anonymous

Wait! okay, it is the augmented matrices.

- ganeshie8

good, how many pivots are there ?

- anonymous

I don't know. How do I tell if they are pivots?

- ganeshie8

in echelon form, pivot is the first "non zero" number in a row

- anonymous

So there would be three pivots then.

- 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

Okay.

- 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

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

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

Okay, so how do I solve for the pivot points?

- ganeshie8

plugin x3=x4=0 in above system

- 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

so..
x2 = 5
x5 = 0
x1=0

- ganeshie8

try again

- 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

Yes.

- ganeshie8

solve it

- ganeshie8

its trivial, but you have to do it

- anonymous

I'm trying to do it right now, give me a moment.

- anonymous

x5 = 0
x2 + 4(0) - 0 = 6 -> x2 = 6
x1 - 5(0) - 8(0) = 3 -> x1 = 3

- 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

Nope

- ganeshie8

null solution is the solution to following system :
```
x5 = 0
x2 + 4x3 - x4 = 0
x1 - 5x3 - 8x5 = 0
```

- ganeshie8

Notice that I have put 0s on the right hand side now

- 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

Do I use the values for the system that we came up with earlier?

- ganeshie8

let x3 = 0, x4 = 1 and find one null solution
let x3 = 1, x4 = 0 and find the other null solution

- 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

x2 - 1 = 0
x2 = 1

- ganeshie8

yes, keep going find the remaining two variables

- anonymous

x1 = 0

- 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

Because then x1 would equal 5

- ganeshie8

one thing at a time

- 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

Yeah.

- ganeshie8

save it, find the other null solution also similarly

- 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

Okay, so then x1 = 1 and x2 = -1

- ganeshie8

try again

- 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

we're mostly done
just wait one minute

- 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

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

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

Closing now as it is no longer helpful.

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