Question

Consider the differential equation \(\dot{\vec{w}}=A\vec{w}\), with \(\vec{w}(t=0)=\begin{bmatrix}1\\1\end{bmatrix}\). If \(\vec{w}(t)=e^{t}\vec{u}_x+e^{-2t}\vec{u}_y\) is the solution to the equation, where \(\vec{u}_x\) and \(\vec{u}_y\) are unit vectors along the positive \(x\) and \(y\) axes respectively, then which of the following options is the correct matrix representing \(A\)?

• A. \(\begin{bmatrix}1 & 0\\0 & -2\end{bmatrix}\)
• B. \(\begin{bmatrix}-1 & 0\\0 & 2\end{bmatrix}\)
• C. \(\begin{bmatrix}0 & -2\\1 & 0\end{bmatrix}\)
• D. \(\begin{bmatrix}0 & 2\\-1 & 0\end{bmatrix}\)

Hint:

Since u_x and u_y are the coordinate unit vectors, write w(t) as the column vector [e^t, e^{-2t}] and differentiate it directly.

Answer 1

The Correct Option is A

Step 1: Recognize the exponential form as an eigen-solution.
A linear system $\dot{\vec w}=A\vec w$ has solutions built from $e^{\lambda t}\vec v$, where $\lambda$ is an eigenvalue of $A$ and $\vec v$ is its eigenvector. The given solution is a sum of two such terms, $e^{t}\vec u_x$ and $e^{-2t}\vec u_y$.

Step 2: Read off the eigenvalues and eigenvectors.
The term $e^{t}\vec u_x$ gives eigenvalue $\lambda_1=1$ with eigenvector $\vec u_x=[1,0]^T$. The term $e^{-2t}\vec u_y$ gives eigenvalue $\lambda_2=-2$ with eigenvector $\vec u_y=[0,1]^T$.

Step 3: Build A from its eigenvalues and eigenvectors.
Since the eigenvectors are exactly the standard basis vectors, $A$ is already diagonal in this basis, with the eigenvalues placed on the diagonal in the same order as their eigenvectors:
\[ A=\begin{bmatrix}\lambda_1 & 0\\0 & \lambda_2\end{bmatrix}=\begin{bmatrix}1 & 0\\0 & -2\end{bmatrix} \]

Step 4: Confirm the other matrices do not share these eigenpairs.
The matrix in (B) has eigenvalues $-1$ and $2$, the wrong sign for both growth rates. The matrices in (C) and (D) are not diagonal in the standard basis, so $\vec u_x$ and $\vec u_y$ would not even be their eigenvectors.
\[ \boxed{\begin{bmatrix}1 & 0\\0 & -2\end{bmatrix}} \]