This equals (by expanding either along the row 3 or column 3)
I need some help with the following:
A = the following matrix
(2 -1 0)
(-1 2 0)
(0 0 2)
I need to find the eigenvalues of A
The characteristic equation is det (A-\lambda I) = 0, that is
|2-\lambda -1 0|
|-1 2-\lambda 0| = 0
|0 0 2-\lambda|
We expand this determinant and obtain
(2-\lambda ) |2-\lambda 0| -0 + |-1 2-\lambda|
| 0 2-\lambda| |0 0 |
Is this expansion correct?
Sorry cant get bottonm line to align up. Also how do I get the greek letters and use large brackets?
This gives
Now for your first question the code
\begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & 0 \\ 0 & 0 & 2\end{bmatrix}
creates the first matrix but it must be wrapped in [ TEX] [ /TEX] you can click the go advanced button at the bottom and then the right most TEX tab. Also if you type them manually do not put the spaces [TEX [/TEX.
I am trying to write the following quadric equation in standard form.
2x^2+2y^2+2z^2-2xy+x-y=0
Using the strategy above I have got as far as, the following equation,
3(x')^2+2(y')^2+(z')^2- x'+ z'=0
Is this correct?
I now need to translate the origin by completeing the squares of the equation. Could anyone give me any help?
Personally, I would NOT translate first. This equation can be written in matrix form
Yes, the 3 by 3 matrix has eigenvalues 1, 2, and 3 and the corresponding eigenvectors are , , respectively, which are equivalent to the transition matrix Arron gave above.
More abstractly, we are saying that the original equation is [tex]X^TAX+ BX= 0[/tex[ where A, B, and X are the matrices above. Let P be the transition matrix. We can rewrite that as .
If P is orthogonal, as Arron has it, So and we have
If we let , , the equation becomes
where D is now diagonal with the eigenvalues on the main diagonal. That corresponds to a quadratic form that has no "mixed" quadratic terms and you can now translate.