$$\text{Consider the differential equation}\overrightarrow{\mathrm{ w}}=A\overrightarrow{w},\text{with}\overrightarrow{\mathrm{ w}}(t=0)=\left[\begin{array}{c}1\\ 1\end{array}\right].\text{If}\overrightarrow{\mathrm{ w}}\left(t\right)=e^t{\overrightarrow{u}}_x+e^{-2t}{\overrightarrow{u}}_y\text{ be the solution to the equation}$$

\text{where}{\overrightarrow{u}}_x\text{and}{\overrightarrow{u}}_{\mathrm{y }}{\mathrm{are\; unit\; vectors\; along\; the\; positive\; x\; and\; y\; axes\; respectively,\; then\; which\; of\; the\; following\; options}}_{}

{\mathrm{is\; the\; correct\; matrix\; representing\; A\; ?}}_{}

The correct option representing the matrix $A$ is:

$$\left[\begin{array}{cc}1& 0\\ 0& -2\end{array}\right]$$

Step-by-Step Derivation:

We are given the system of linear differential equations:
$$\frac{d\overrightarrow{w}}{dt}=A\stackrel{���}{w}$$ with the solution vector:
$$\overrightarrow{w}\left(t\right)=e^t{\overrightarrow{u}}_x+e^{-2t}{\overrightarrow{u}}_y$$ where ${\overrightarrow{u}}_x$ and ${\overrightarrow{u}}_y$ are unit vectors along the positive x and y axes, respectively:

$${\overrightarrow{u}}_x=\left[\begin{array}{c}1\\ 0\end{array}\right],{\overrightarrow{u}}_y=\left[\begin{array}{c}0\\ 1\end{array}\right]$$

Substituting these unit vectors, we can write $\overrightarrow{w}\left(t\right)$ in column vector form as:

$$\overrightarrow{w}\left(t\right)=\left[\begin{array}{c}{e}^t\\ e^{-2t}\end{array}\right]$$

Now, we calculate the time derivative of $\overrightarrow{w}\left(t\right)$:

$$\frac{d\overrightarrow{w}}{dt}=\left[\begin{array}{c}\frac{d}{dt}\left(e^t\right)\\ \frac{d}{dt}\left(e^{-2t}\right)\end{array}\right]=\left[\begin{array}{c}{e}^t\\ -2e^{-2t}\end{array}\right]$$

Let the 2 × 2 matrix $A$ be represented by:
$$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$$ We substitute $\frac{d\overrightarrow{w}}{dt}$, $A$, and $\overrightarrow{w}$ into the differential equation $\frac{d\overrightarrow{w}}{dt}=A\overrightarrow{w}$:

$$\left[\begin{array}{c}{e}^t\\ -2e^{-2t}\end{array}\right]=\left[\begin{array}{c}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}{e}^t\\ e^{-2t}\end{array}\right]$$

This matrix multiplication yields the system of equations:
1) $e^t=a{e}^t+b{e}^{-2t}$
2) $-2e^{-2t}=c{e}^t+d{e}^{-2t}$

Since the exponential functions $e^t$ and $e^{-2t}$ are linearly independent, the coefficients of matching functions on both sides of each equation must be equal. Therefore, we obtain:
From equation (1): $a=1$ and $b=0$
From equation (2): $c=0$ and $d=-2$

Substituting these values back into matrix $A$ gives:

$$A=\left[\begin{array}{cc}1& 0\\ 0& -2\end{array}\right]$$