$\Phi(t) = e^{A t}$

The matrix $A$ has eigenvalues $-\tfrac{7}{10}$ , $-\tfrac{3}{10}$ and $0$ , with corresponding eigenvectors $\left(\begin{array}{c}1 \\ 1 \\ -3 \end{array}\right) \hspace{1.5cm} \left(\begin{array}{c} 0 \\ 1 \\ 0 \end{array}\right) \hspace{1.5cm} \left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right)$ Thus $e^{At}\left(\begin{array}{c}1 \\ 1 \\ -3 \end{array}\right) = e^{-\frac{7t}{10}}\left(\begin{array}{c}1 \\ 1 \\ -3 \end{array}\right) \hspace{1cm} e^{At} \left(\begin{array}{c} 0 \\ 1 \\ 0 \end{array}\right) = e^{-\frac{3t}{10}}\left(\begin{array}{c} 0 \\ 1 \\ 0 \end{array}\right) \hspace{1cm} e^{At}\left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right) = \left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right)$ Thus, if $\mathbf{v}(0) \; = \; u\left(\begin{array}{c}1 \\ 1 \\ -3 \end{array}\right) + v \left(\begin{array}{c} 0 \\ 1 \\ 0 \end{array}\right) + w\left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right)$ then $\mathbf{v}(t) \;=\; e^{At}\mathbf{v}(0) \; = \; ue^{-\frac{7t}{10}}\left(\begin{array}{c}1 \\ 1 \\ -3 \end{array}\right) + v e^{-\frac{3t}{10}}\left(\begin{array}{c} 0 \\ 1 \\ 0 \end{array}\right) + w\left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right)$ and hence $\lim_{t \to \infty}\mathbf{v}(t) \; = \; w\left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right)$ so that $\lim_{t \to \infty}x(t) \; = \; w$ In this case we have $u=v=-\tfrac34$ and $w = \tfrac74$ , so the desired answer is $\boxed{\tfrac74}$ .

Writing the equations in a non-matrix form gives:

$\dot{x} = A_1 x + B_1y + C_1 z$ $\dot{y} = A_2 x + B_2y + C_2 z$ $\dot{z} = A_3x + B_3y + C_3 z$

Taking Laplace transform on both sides:

$sX - x(0) = A_1 X + B_1Y + C_1 Z$ $sY - y(0) = A_2 X + B_2Y + C_2 Z$ $sZ - z(0) = A_3 X + B_3Y + C_3 Z$

Solving for X (Using Wolfram-Alpha):

$X = \frac{40s+49}{40s^2 + 28s}$

Now, Applying the final value theorem:

$\lim_{s \to 0} sX(s) = \lim_{t \to \infty} x(t)$ $\implies \lim_{t \to \infty} x(t) = \lim_{s \to 0} \frac{40s+49}{40s + 28}$ $\implies \boxed{ \lim_{t \to \infty} x(t) = \frac{7}{4}}$