Question

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.

Answer 1

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

Answer 2

is that the augmented matrix in row echelon form ?

Answer 3

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

Answer 4

Wait! okay, it is the augmented matrices.

Answer 5

good, how many pivots are there ?

Answer 6

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

Answer 7

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

Answer 8

So there would be three pivots then.

Answer 9

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

Answer 10

Okay.

Answer 11

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

Answer 12

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

Answer 13

notice that the corresponding system is
```
x5 = 0
x2 + 4x3 - x4 = 6
x1 - 5x3 - 8x5 = 3
```

let x3 = x4 = 0
and solve remaining variables

Answer 14

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

Answer 15

plugin x3=x4=0 in above system

Answer 16

then you will have 3 equations and 3 unknowns in triangular form
which is a piece of cake to solve by back substitution

Answer 17

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

Answer 18

try again

Answer 19

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 ?

Answer 20

Yes.

Answer 21

solve it

Answer 22

its trivial, but you have to do it

Answer 23

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

Answer 24

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

Answer 25

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 ?

Answer 26

Nope

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

x2 - 1 = 0
x2 = 1

Answer 34

yes, keep going find the remaining two variables

Answer 35

x1 = 0

Answer 36

Unless you wanted me to use "let x3 = 1, x4 = 0 and find the other null solution" to solve x1 - 5x3 - 8x5 = 0

Answer 37

Because then x1 would equal 5

Answer 38

one thing at a time

Answer 39

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 ?

Answer 40

Yeah.

Answer 41

save it, find the other null solution also similarly

Answer 42

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

Answer 43

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

Answer 44

try again

Answer 45

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.

Answer 46

we're mostly done
just wait one minute

Answer 47

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 \(\begin{pmatrix} 5\\-4\\1\\0\\0\end{pmatrix}\)

Answer 48

so the nullspace is given by
\[\large c_1\begin{pmatrix} 0\\1\\0\\1\\0\end{pmatrix} + c_2\begin{pmatrix} 5\\-4\\1\\0\\0\end{pmatrix}\]

Answer 49

then the final general solutions is given by \(x_p + x_{null}\) :

\[x_{\text{general }} = \begin{pmatrix}
3\\6\\0\\0\\0
\end{pmatrix} +\large c_1\begin{pmatrix} 0\\1\\0\\1\\0\end{pmatrix} + c_2\begin{pmatrix} 5\\-4\\1\\0\\0\end{pmatrix} \]

Answer 50

Closing now as it is no longer helpful.