# 27.2 solve the system Y'={}y

#### 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:
$$\displaystyle \left[ \begin{array}{ccc} - \lambda+1&2&-3\\ 0&-\lambda+1&0\\ 6&3&-\lambda-8 \end{array} \right] =-18\lambda +\left(-\lambda-8\right) \left(- \lambda + 1\right)^{2} + 18$$

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#### romsek

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

#### topsquark

Forum Staff
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:
$$\displaystyle \left[ \begin{array}{ccc} - \lambda+1&2&-3\\ 0&-\lambda+1&0\\ 6&3&-\lambda-8 \end{array} \right] =-18\lambda +\left(-\lambda-8\right) \left(- \lambda + 1\right)^{2} + 18$$

View attachment 39339
Okay, so what are the eigenvalues and eigenvectors?

-Dan

#### HallsofIvy

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