Thread: 27.2 solve the system Y'={}y

1. 27.2 solve the system Y'={}y

Solve the system $\displaystyle Y'=\begin{bmatrix} 1 & 3 & -3 \\ 0 & 1 & 0 \\ 6 & 3 & -8 \end{bmatrix}Y$

not real sure but W|A returned this but no steps
so assume first thing we so is Eigenvalues

subtracting $\lambda$ from the diagonal entries of the given matrix and Find the determinant of the obtained matrix:
$\displaystyle \left[ \begin{array}{ccc} - \lambda+1&2&-3\\ 0&-\lambda+1&0\\ 6&3&-\lambda-8 \end{array} \right] =-18\lambda +\left(-\lambda-8\right) \left(- \lambda + 1\right)^{2} + 18$ 2. Re: 27.2 solve the system Y'={}y

it's solved just like the 2 dimensional case but now there are 3 terms instead of 2

3. Re: 27.2 solve the system Y'={}y Originally Posted by bigwave Solve the system $\displaystyle Y'=\begin{bmatrix} 1 & 3 & -3 \\ 0 & 1 & 0 \\ 6 & 3 & -8 \end{bmatrix}Y$

not real sure but W|A returned this but no steps
so assume first thing we so is Eigenvalues

subtracting $\lambda$ from the diagonal entries of the given matrix and Find the determinant of the obtained matrix:
$\displaystyle \left[ \begin{array}{ccc} - \lambda+1&2&-3\\ 0&-\lambda+1&0\\ 6&3&-\lambda-8 \end{array} \right] =-18\lambda +\left(-\lambda-8\right) \left(- \lambda + 1\right)^{2} + 18$ Okay, so what are the eigenvalues and eigenvectors?

-Dan

4. Re: 27.2 solve the system Y'={}y

Yes, you have the equation for the eigenvalues and it is fairly simple to solve it: the eigenvalues are 1, 1, and 8.