6. \text{SYSTEMET PÅ MATHUSFORM (M0049M)}\begin{array}{l} \left[ \begin{array}{l} x \\ y \end{array} \right] ^ {\prime} = \left[ \begin{array}{l l} - 4 & 6 \\ - 3 & 5 \end{array} \right] \left[ \begin{array}{l} x \\ y \end{array} \right], \quad \left[ \begin{array}{l} x (0) \\ y (0) \end{array} \right] = \left[ \begin{array}{l} 0 \\ 1 \end{array} \right] \\ \text{LÄT } A = \left[ \begin{array}{l l} - 4 & 6 \\ - 3 & 5 \end{array} \right]. \end{array}

BESTÄM EGENVÄNDEN/VECTOREN FÖR A.
EGENVÄNDEN: KAR. EKV.: $\det(A - \lambda I) = 0$

\begin{array}{l} ( \Rightarrow ) \left| \begin{array}{cc} - 4 - \lambda & 6 \\ - 3 & 5 - \lambda \end{array} \right| = 0 \quad ( \Rightarrow ) (- 4 - \lambda) (5 - \lambda) + 18 = 0 \\ ( \Rightarrow ) - 20 - \lambda + \lambda^2 + 18 = 0 \quad ( \Rightarrow ) \lambda^2 - \lambda - 2 = 0 \\ ( \Rightarrow ) \lambda = \frac{1}{2} \pm \sqrt{\frac{1}{4} + \frac{8}{4}} = \frac{1}{2} \pm \frac{3}{2} = -1 \text{ j } 2 \\ \end{array} \begin{array}{l} \text{Koll: } -1 + 2 = 1, \text{ tr } A = -4 + 5 = 1 \\ \quad (-1) 2 = -2, \text{ det } A = -20 + 18 = -2 \quad \text{ok} \end{array}

EGENVECTOR:

\begin{array}{l} \lambda = -1: A \vec{v} = -1 \vec{v} \quad ( \Rightarrow ) (A + I) \vec{v} = \vec{0} \\ \left[ \begin{array}{ll} - 3 & 6 \\ - 3 & 6 \end{array} \right] \sim \left[ \begin{array}{ll} 1 & 2 \\ 0 & 0 \end{array} \right] \quad \begin{array}{l} x_1 \text{ BUNDEN } x_2 \text{ FRI} \\ x_2 = t, x_1 = 2t \\ \therefore \vec{v} = t \left[ \begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array} \right] \text{ VÄLJ} \quad \left[ \begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array} \right] \end{array} \\ \end{array} \begin{array}{l} \lambda = 2: A \vec{v} = 2 \vec{v} \quad ( \Rightarrow ) (A - 2 I) \vec{v} = \vec{0} \\ \left[ \begin{array}{ll} - 6 & 6 \\ - 3 & 3 \end{array} \right] \sim \left[ \begin{array}{ll} 1 & 1 \\ 0 & 0 \end{array} \right] \quad \begin{array}{l} x_1 \text{ BUNDEN } x_2 \text{ FRI} \\ x_2 = t, x_1 = t \end{array} \\ \therefore \vec{v} = t \left[ \begin{array}{ll} 1 & 1 \\ 1 & \end{array} \right] \text{ VÄLJ} \quad \left[ \begin{array}{ll} 1 & 1 \\ 1 \end{array} \right] / \\ \end{array}

SAMTLIGA LÖSNINGAR (PÅ VECTORFORM)

\begin{array}{l} \text{Än } A \left[ \begin{array}{l} 2 \\ 1 \end{array} \right] e^{-t} + B \left[ \begin{array}{l} 1 \\ 1 \end{array} \right] e^{2t} \\ \text{MEN, } \left[ \begin{array}{l} x (0) \\ y (0) \end{array} \right] = \left[ \begin{array}{l} 0 \\ 1 \end{array} \right], \text{ och detta gär} \\ \left[ \begin{array}{l} 0 \\ 1 \end{array} \right] = \left[ \begin{array}{ll} 2 & 1 \\ 1 & 1 \end{array} \right] \left[ \begin{array}{l} A \\ B \end{array} \right] \quad ( \Rightarrow ) \left\{ 2A + B = 0 \quad A + B = 1 \right. \\ ( \Rightarrow ) A = -1, B = 2 \\ \therefore \quad - \left[ \begin{array}{l} 2 \\ 1 \end{array} \right] e^{-t} + 2 \left[ \begin{array}{l} 1 \\ 1 \end{array} \right] e^{2t}, \quad \text{på komponentform} \\ \left\{ \begin{array}{l} x = -2 e^{-t} + 2 e^{2t} \\ y = -e^{-t} + 2 e^{2t} \end{array} \right. \\ \end{array}