# Matrices and determinant

1. Mar 25, 2012

### rohan03

please look at the attachement and my attempt at the solution - hope you can help.

Thanks

Last edited: Mar 25, 2012
2. Mar 25, 2012

### alfredska

I see no attachment.

3. Mar 26, 2012

### rohan03

sorry I amended the post and forgot to reattached my document - here it is please look at it.

Last edited: Mar 26, 2012
4. Mar 26, 2012

### alfredska

After "This gives:", you need to check your math. I see $$(1-\lambda)^2(-1-\lambda)-4(1-\lambda)-4(1-\lambda)$$

To find eigenvectors, see if this resource helps: http://www.sosmath.com/matrix/eigen2/eigen2.html

Essentially, you will set up $$A \left(\begin{matrix} x \\ y \\ z \end{matrix}\right) = \lambda \left(\begin{matrix} x \\ y \\ z \end{matrix}\right)$$ where you substitute in each $\lambda$ found earlier. Then you'll have 3 equations and 3 unknowns for each $\lambda$.

5. Mar 26, 2012

### Office_Shredder

Staff Emeritus
What does it mean for there to be an eigenvalue of -1?
If you write down the equation that the eigenvector satisfies, you should get three equations and three unknowns - note in general the solution for the eigenvector will be non-unique since you can scale it to get a new eigenvector, but you should be able to get a one-dimensional set of vectors that works

6. Mar 26, 2012

### rohan03

ok I had anotherlook at my simplification and I have spotted some error while finding eigen values using 3x3 matrix

what I get is :
( i am using y instaed of lambda as I am typing here so apologies)

(1-y){(1-y)(-1-y)-(-2x-2)} -2{(0x-2) -(-2)(1-y)}
=(1-y){-1-y+y+y^2-4} -2 { 0-(-2)(1-y)}
=(1-y){-1+y^2-4-4(1-y)}
=(1-y)(y^2-9)
=(1-y) (y-3) (y+3)
hence giving eigen values of y=1, y=-3 and y=3

is this correct?

7. Mar 26, 2012

### alfredska

That looks correct.

8. Mar 26, 2012

### rohan03

thank you so much for all your help.

9. Mar 26, 2012

### Ray Vickson

I still do not see an attachment.

RGV