and this will give youR1/-2 and R2/2
I don't know what you mean by this. Dividing one row by another is not a "row-operation".R2/R1
If I ignore the previous line and subtract R1- R2, I get and I can stop here. An eigenvector is .R2-R1
First, that is not a vector, it is a matrix. Also you should have realized that there was something wrong. 9 was an eigenvalue because det(A- 9I)= 0 which means that A is not invertible and cannot be row reduced to the identity matrix as you have.R2/2
until i got the vector
|1 0 0|
|0 1 0|
[quot]I cant really find out what to do now. And then i have to find a matrix, P, that diagonalizes the matrix A.[/quote]
A more fundamental way of doing this: from the definition of "eigevalue" there exist a non-zero vector v such that .
Multiplying that out gives the three equations 7x+ 2y= 9x, 12x- 3y= 9y or 2x+ 2y= 0 and 3x- 3y= 0 both of which are satified by any x,y such that y= x. Again, that says that is an eigenvector.
You don't show how you got that equation but that's not what I get.In the other example i had the vector
A= |-1 -2 -2|
|1 2 1|
|-1 -1 0|
and i solved for the eigenvalues and got
But if i solve it with my texas instrument, i get λ=-0,295598 wich doesn't sound so right. So how do i solve it?