# Toronto Math Forum

## MAT244-2013S => MAT244 Math--Tests => Quiz 4 => Topic started by: Victor Ivrii on March 22, 2013, 04:18:05 AM

Title: Q4--day section
Post by: Victor Ivrii on March 22, 2013, 04:18:05 AM
Please post the problem and its solution (in contrast to a popular opinion I am not omniscient :D and unless someone advises me I am not sure which problem it was).
Title: Q4--day section--problem 1
Post by: Changyu Li on March 22, 2013, 10:22:22 PM
I think the first problem was from the textbook
$\mathbf x' = \left(\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 0 & -1 & 1 \\ \end{array}\right) \mathbf x \\$
The characteristic polynomial is
$-4 + 3k^2 - k^3 = 0 \\ k = 2, 2, -1$
The eigenvectors are
$\lambda = -1, \left(\begin{array}{c} -3 \\ 4 \\ 2 \end{array}\right), \lambda = 2, \left(\begin{array}{c} -0 \\ -1 \\ 1 \end{array}\right)$
Jordan decomposition yields the similarity transform of
$\mathbf T = \left(\begin{array}{ccc} -3 & 0 & -1 \\ 4 & -1 & -1 \\ 2 & 1 & 0 \end{array}\right)$
Thus the solution is
$\mathbf x = c_1 \left(\begin{array}{c} -3 \\ 4 \\ 2 \end{array}\right) e^{-t} + c_2 \left(\begin{array}{c} 0 \\ -1 \\ 1 \end{array}\right) e^{2t} + c_3 \left( \left(\begin{array}{c} 0 \\ -1 \\ 1 \end{array}\right) t e^{2t} + \left(\begin{array}{c} -1 \\ -1 \\ 0 \end{array}\right) e^{2t}\right)$