Exercise · ZW-LINEAR-SYSTEM-AND-EIGENVALUES-006 · ★★☆ Intermediate
If \(A=\begin{pmatrix}-1&0\\0&-3\end{pmatrix}\), what happens to the solutions of \(\mathbf x'=A\mathbf x\) as \(t\to\infty\)?
Choose one option.
A Todas tienden al origen B Todas crecen sin cota C Unas tienden al origen y otras crecen D Todas oscilan sin amortiguarse Show answer A. Todas tienden al origen← Back to the question
Explanation The eigenvalues \(-1\) and \(-3\) are negative, so \(e^{-t}\) and \(e^{-3t}\) tend to \(0\), and so does every combination .
Homogeneous system with a constant \(n\times n\) matrix \(A\), whose elementary solutions are built from an eigenvalue \(\lambda\) and an eigenvector \(\mathbf v\) of \(A\).
\[\mathbf x'=A\mathbf x,\qquad \mathbf x(t)=e^{\lambda t}\mathbf v,\quad A\mathbf v=\lambda\mathbf v\]
More practice problems: linear system and eigenvalues