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

 

[romsek]

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

 

[topsquark]

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]

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