Solve the initial valued problem $x' = \begin{pmatrix} 4 & -3 \\ 8 & -6 \end{pmatrix} x$, $x(0) = \begin{pmatrix} -1 \\ 1 \end{pmatrix}$.
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The characteristic equation is given by: $\text{det}(A - \lambda I) = \begin{vmatrix} 4-\lambda & -3 \\ 8 & -6-\lambda \end{vmatrix} = (4-\lambda)(-6-\lambda) - (-3)(8) = \lambda^2 + 2\lambda - 24 + 24 = \lambda^2 + 2\lambda = \lambda(\lambda+2) = 0$ The Show more…
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