To solve this system of linear differential equations, we proceed step by step:

Part (a): General Solution

The given system is: $\mathbf{x}' = \begin{pmatrix} 3 & -2 \\ 10 & -6 \end{pmatrix} \mathbf{x}.$ This is a first-order linear system. To find the general solution, we:

1. Find the eigenvalues of the coefficient matrix $A = \begin{pmatrix} 3 & -2 \\ 10 & -6 \end{pmatrix}$: $\det(A - \lambda I) = \det\begin{pmatrix} 3 - \lambda & -2 \\ 10 & -6 - \lambda \end{pmatrix} = 0.$ Expanding the determinant: $(3-\lambda)(-6-\lambda) - (-2)(10) = \lambda^2 + 3\lambda - 8 = 0.$ Solve the quadratic equation: $\lambda = \frac{-3 \pm \sqrt{3^2 - 4(1)(-8)}}{2(1)} = \frac{-3 \pm \sqrt{41}}{2}.$ So, the eigenvalues are: $\lambda_1 = \frac{-3 + \sqrt{41}}{2}, \quad \lambda_2 = \frac{-3 - \sqrt{41}}{2}.$

2. Find the eigenvectors for each eigenvalue**: For each eigenvalue $\lambda$, solve $(A - \lambda I)\mathbf{v} = 0$.

   • For $\lambda_1 = \frac{-3 + \sqrt{41}}{2}$: Subtract $\lambda_1$ from the diagonal elements of $A$, and find the null space of the resulting matrix.

   • Repeat this for $\lambda_2 = \frac{-3 - \sqrt{41}}{2}$.

   These calculations will give two eigenvectors.

3. Write the solution: The general solution is a linear combination of eigenvector solutions: $\mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2.$

Part (b): Behavior as $t \to \infty$

Here, we examine the eigenvalues:

• If $\lambda_1 > 0$, the corresponding term $e^{\lambda_1 t}$ will dominate as $t \to \infty$, leading to growth.
• If $\lambda_2 < 0$, the term $e^{\lambda_2 t}$ will decay as $t \to \infty$.

Since one eigenvalue is larger than the other, the solution's behavior depends on the relative magnitudes of $c_1$ and $c_2$.

Part (c): Direction Field

Direction fields represent the slope of the system's trajectories. Based on the eigenvalues, the system's phase portrait will show:

• Stable/unstable node or saddle point depending on the signs of eigenvalues.
• If both eigenvalues have the same sign, it’s either a node or spiral.
• A saddle point arises if one eigenvalue is positive and the other is negative.

Let me know if you would like me to explicitly solve for eigenvectors or analyze further.