Write the complete solution as the particular solution plus any multip

[s3a]

Homework Statement

Could someone please help me do this problem?:

“Write the complete solution as x_p plus any multiple of s in the nullspace:
x + 3y + 3z = 1
2x + 6y + 9z = 5
–x – 3y + 3z = 5”

The answer is x = x_p + x_n = {{-2},{0},{1}} + x_2 {{-3, 1, 0}}.

Homework Equations

Ax = b

The Attempt at a Solution

I get the {{-2},{0},{1}} and {{-3},{1},{0}} but, I don't get how the x_2 was obtained there.

Any input would be really appreciated!

 

[LCKurtz]

Homework Statement

Could someone please help me do this problem?:

“Write the complete solution as x_p plus any multiple of s in the nullspace:
x + 3y + 3z = 1
2x + 6y + 9z = 5
–x – 3y + 3z = 5”

The answer is x = x_p + x_n = {{-2},{0},{1}} + x_2 {{-3, 1, 0}}.

Homework Equations

Ax = b

The Attempt at a Solution

I get the {{-2},{0},{1}} and {{-3},{1},{0}} but, I don't get how the x_2 was obtained there.

Any input would be really appreciated!

Did you try solving the system by row reduction?

 

[tiny-tim]

hi s3a!

(try using the X₂ button just above the Reply box :

I get the {{-2},{0},{1}} and {{-3},{1},{0}} but, I don't get how the x_2 was obtained there.

i've no idea why they've written x₂

we'd normally use C (for "constant"! )

(btw, why are you using so many curly brackets, instead of fewer, ordinary brackets?)

 

[LCKurtz]

Also, are you sure the problem is copied correctly? I don't think your solution works.

[Edit] Yes it does work, arithmetic mistake here.

 

[s3a]

Sorry. I double-posted. This website should have some sort of mechanism where you can't post twice until 10 seconds elapse or something like that.

 

[s3a]

LCKurtz, I did use row reduction. Also, thanks for telling me about the existence of the x_2 button and, yes, the solution was copied down correctly. Letting c = 1, for example, yields a solution to the system of equations so, it works. :)

tiny-tim, the abundance of curly brackets is Wolfram Alpha's notation for emphasizing that the vectors are vertical.

Okay, so basically x_2 = c is a constant. That pretty much helped me see through my confusion, lol!

Thanks!

 

[LCKurtz]

LCKurtz, I did use row reduction. Also, thanks for telling me about the existence of the x_2 button and, yes, the solution was copied down correctly. Letting c = 1, for example, yields a solution to the system of equations so, it works. :)

tiny-tim, the abundance of curly brackets is Wolfram Alpha's notation for emphasizing that the vectors are vertical.

Okay, so basically x_2 = c is a constant. That pretty much helped me see through my confusion, lol!

Thanks!

So you solved it OK with row reduction and all is well, right?

 

[s3a]

Yes.

To elaborate, I solved it augmented with b and that allowed me to get a particular solution and then took that fully-row-reduced matrix and took away the last column (=the augmented column) and replaced it with zeroes and that gave me the other part which, when all multiples of it are summed with the particular solution, all solutions are obtained.

(Sorry for the sloppy explanation, I am sleepy as I write this.)

(If anyone knows what that other part is called, I'd be interested in knowing.)

 

[Dick]

Yes.

To elaborate, I solved it augmented with b and that allowed me to get a particular solution and then took that fully-row-reduced matrix and took away the last column (=the augmented column) and replaced it with zeroes and that gave me the other part which, when all multiples of it are summed with the particular solution, all solutions are obtained.

(Sorry for the sloppy explanation, I am sleepy as I write this.)

(If anyone knows what that other part is called, I'd be interested in knowing.)

The part you got with the specific vector is called a 'particular solution'. The part with the parameter you got from Ax=0 is the 'homogeneous solution'.

 

[s3a]

Oh, you ... I remember that term! Thanks!

 

[LCKurtz]

Yes.

To elaborate, I solved it augmented with b and that allowed me to get a particular solution and then took that fully-row-reduced matrix and took away the last column (=the augmented column) and replaced it with zeroes and that gave me the other part which, when all multiples of it are summed with the particular solution, all solutions are obtained.

(Sorry for the sloppy explanation, I am sleepy as I write this.)

(If anyone knows what that other part is called, I'd be interested in knowing.)

Of course, you don't have to solve it twice like that. For example if your reduced augmented matrix came out$$
\begin{bmatrix}
1&3&0&-2\\
0&0&1&1\\
0&0&0&0
\end{bmatrix}$$that corresponds to the two equations
x + 3y = -2
z = 1
So you can let y be anything, say y = c and write the solution like this$$
\begin{bmatrix}
x\\
y\\
z\end{bmatrix}=
\begin{bmatrix}
-2-3c\\
c\\
1
\end{bmatrix}=\begin{bmatrix}
-2\\
0\\
1
\end{bmatrix}+c\begin{bmatrix}
-3\\
1\\
0
\end{bmatrix}
$$and you have it.