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# Matrices (Generalised Inverse 2) watch

1. Consider the solution of the matrix system M x = r, where the matrix and right-hand side are given by

;

The previous parts to this question was proving what the eigenvectors were (simple dot product) and writing down associated eigenvalues.

The question I'm stuck on:

We can write the vector x as a linear combination of the eigenvectors.

Writing ; , derive a formula for in terms of and the eigenvalues

Edit: Is something up with latex? Wow...
2. (Original post by Jagwar Ma)
Consider the solution of the matrix system M x = r, where the matrix and right-hand side are given by

;

The previous parts to this question was proving what the eigenvectors were (simple dot product) and writing down associated eigenvalues.

The question I'm stuck on:

We can write the vector x as a linear combination of the eigenvectors.

Writing ; , derive a formula for in terms of and the eigenvalues

Edit: Is something up with latex? Wow...
Yes, something appears to be up in latex land...anyway, press on...

How far have you got? Have you derived the relation ?

If is the eigenvector corresponding to the zero eigenvalue, what does that imply about ?
3. (Original post by Gregorius)
Yes, something appears to be up in latex land...anyway, press on...

How far have you got? Have you derived the relation ?

Now you should notice that the are not linearly independent (as the matrix is singular). If is the eigenvector corresponding to the zero eigenvalue, what does that imply about ?
v3 is not the eigenvector that corresponds to the zero eigenvalue though?

I did manage to solve it, but now I'm trying to use the result to determine the solution vector x
4. (Original post by Jagwar Ma)
v3 is not the eigenvector that corresponds to the zero eigenvalue though?
I gave them an arbitrary ordering for convenience: v3 = (2,-2,1)
5. (Original post by Gregorius)
I gave them an arbitrary ordering for convenience: v3 = (2,-2,1)
Ah, yeah that's my V1

I'm trying to solve to find the solution vector x but don't really understand my lecturers working (to a similar problem).
mx = r
6. (Original post by Jagwar Ma)
Ah, yeah that's my V1

I'm trying to solve to find the solution vector x but don't really understand my lecturers working (to a similar problem).
mx = r
OK, so I take v1 = (4,5,2); v2 = (-1,0,2) and v3 = (2,-2,1). Then by inspection, r = v1 + 2 v2. So you now have to solve

7. (Original post by Gregorius)
OK, so I take v1 = (4,5,2); v2 = (-1,0,2) and v3 = (2,-2,1). Then by inspection, r = v1 + 2 v2. So you now have to solve

r = (2, 5, 6) I've already completed that stage.

Trying to find the x in Mx = r

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