Exercise · ZW-LINEAR-SYSTEM-AND-EIGENVALUES-004 · ★☆☆ Basic
Solve \(\mathbf x'=\begin{pmatrix}2&0\\0&-5\end{pmatrix}\mathbf x\) with \(\mathbf x(0)=(1,1)\).
Choose one option.
Show answer
- B. \((e^{2t},\,e^{-5t})\)
Explanation
The matrix is diagonal, its eigenvalues are \(2\) and \(-5\), with vectors \((1,0)\) and \((0,1)\). With the initial conditions both coefficients equal \(1\).
Review the concept: Linear system and eigenvalues
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
- ZW-LINEAR-SYSTEM-AND-EIGENVALUES-001 What are the eigenvalues of A=[1 2; 2 1]?
- ZW-LINEAR-SYSTEM-AND-EIGENVALUES-002 For A=[1 2; 2 1], which eigenvector corresponds to λ=3?
- ZW-LINEAR-SYSTEM-AND-EIGENVALUES-003 The general solution of mathbf x'=[1 2; 2 1]mathbf x is:
- ZW-LINEAR-SYSTEM-AND-EIGENVALUES-005 For A=[2 1; 1 2], which is a solution of mathbf x'=Amathbf x?
- ZW-LINEAR-SYSTEM-AND-EIGENVALUES-006 If A=[-1 0; 0 -3], what happens to the solutions of mathbf x'=Amathbf x as t→∞?
- ZW-LINEAR-SYSTEM-AND-EIGENVALUES-007 For A=[0 1; -2 -3], the solution of mathbf x'=Amathbf x with mathbf x(0)=(1,0) has first component:
- ZW-LINEAR-SYSTEM-AND-EIGENVALUES-008 If A=[0 1; -4 0], its eigenvalues are pm2i. How do the solutions behave?
- ZW-LINEAR-SYSTEM-AND-EIGENVALUES-009 A 2×2 matrix has eigenvalues 2 and -3. What are its trace and determinant?
← ZW-LINEAR-SYSTEM-AND-EIGENVALUES-003 ZW-LINEAR-SYSTEM-AND-EIGENVALUES-005 →