\[ \mathrm{e}^{\boldsymbol{A}t} = \mathscr{L}^{-1}\left[(s\boldsymbol{I}-\boldsymbol{A})^{-1}\right] = \mathscr{L}^{-1}\begin{bmatrix} \dfrac{1}{s+a} & 0 \\ 0 & \dfrac{1}{s+b} \end{bmatrix} = \begin{bmatrix} \mathrm{e}^{-at} & 0 \\ 0 & \mathrm{e}^{-bt} \end{bmatrix} \]

\[ \boldsymbol{x}_{\delta}(t) = \int_{0}^{t}\begin{bmatrix} \mathrm{e}^{-a\tau} & 0 \\ 0 & \mathrm{e}^{-b\tau} \end{bmatrix}\begin{bmatrix} \dfrac{1}{b-a} \\ \dfrac{1}{a-b} \end{bmatrix}\delta(t-\tau)\mathrm{d}\tau = \begin{bmatrix} \dfrac{1}{b-a}\mathrm{e}^{-at} \\ \dfrac{1}{a-b}\mathrm{e}^{-bt} \end{bmatrix} \]

\[ \boldsymbol{x}_{u}(t) = \int_{0}^{t}\begin{bmatrix} \mathrm{e}^{-a\tau} & 0 \\ 0 & \mathrm{e}^{-b\tau} \end{bmatrix}\begin{bmatrix} \dfrac{1}{b-a} \\ \dfrac{1}{a-b} \end{bmatrix}\cdot 1(1-\tau)\mathrm{d}\tau = \int_{0}^{t}\begin{bmatrix} \dfrac{1}{b-a}\mathrm{e}^{-a\tau} \\ \dfrac{1}{a-b}\mathrm{e}^{-b\tau} \end{bmatrix}\mathrm{d}\tau = \begin{bmatrix} \dfrac{1}{a(b-a)}(1-\mathrm{e}^{-at}) \\ \dfrac{1}{b(a-b)}(1-\mathrm{e}^{-bt}) \end{bmatrix} \]

\[ \boldsymbol{x}_{r}(t) = \int_{0}^{t}\begin{bmatrix} \mathrm{e}^{-a\tau} & 0 \\ 0 & \mathrm{e}^{-b\tau} \end{bmatrix}\begin{bmatrix} \dfrac{1}{b-a} \\ \dfrac{1}{a-b} \end{bmatrix}(t-\tau)\mathrm{d}\tau = \begin{bmatrix} \dfrac{1}{a(b-a)}t - \dfrac{1}{a^{2}(b-a)}(1-\mathrm{e}^{-at}) \\ \dfrac{1}{b(a-b)}t - \dfrac{1}{b^{2}(a-b)}(1-\mathrm{e}^{-bt}) \end{bmatrix} \]

\[ \mathrm{e}^{\boldsymbol{A}t} = \mathscr{L}^{-1}\left[(s\boldsymbol{I}-\boldsymbol{A})^{-1}\right] = \mathscr{L}^{-1}\left[\begin{matrix} \left(\dfrac{b}{b-a}\cdot\dfrac{1}{s+a} + \dfrac{a}{a-b}\cdot\dfrac{1}{s+b}\right) & \left(\dfrac{1}{b-a}\cdot\dfrac{1}{s+a} + \dfrac{1}{a-b}\cdot\dfrac{1}{s+b}\right) \\ \left(\dfrac{-ab}{b-a}\cdot\dfrac{1}{s+a} + \dfrac{-ab}{a-b}\cdot\dfrac{1}{s+b}\right) & \left(\dfrac{a}{a-b}\cdot\dfrac{1}{s+a} + \dfrac{-b}{a-b}\cdot\dfrac{1}{s+b}\right) \end{matrix}\right] \]

\[ = \begin{bmatrix} \dfrac{b}{b-a}\mathrm{e}^{-at} + \dfrac{a}{a-b}\mathrm{e}^{-bt} & \dfrac{1}{b-a}\mathrm{e}^{-at} + \dfrac{1}{a-b}\mathrm{e}^{-bt} \\ \dfrac{-ab}{b-a}\mathrm{e}^{-at} + \dfrac{-ab}{a-b}\mathrm{e}^{-bt} & \dfrac{a}{a-b}\mathrm{e}^{-at} + \dfrac{-b}{a-b}\mathrm{e}^{-bt} \end{bmatrix} \]

\[ \boldsymbol{x}_{\delta}(t) = \int_{0}^{t}\mathrm{e}^{\boldsymbol{A}\tau}\begin{bmatrix} 0 \\ 1 \end{bmatrix}\delta(t-\tau)\mathrm{d}\tau = \begin{bmatrix} \dfrac{1}{b-a}(\mathrm{e}^{-at} - \mathrm{e}^{-bt}) \\ \dfrac{1}{a-b}(a\mathrm{e}^{-at} - b\mathrm{e}^{-bt}) \end{bmatrix} \]

\[ \boldsymbol{x}_{u}(t) = \int_{0}^{t}\mathrm{e}^{\boldsymbol{A}\tau}\begin{bmatrix} 0 \\ 1 \end{bmatrix} 1(t-\tau)\mathrm{d}\tau = \int_{0}^{t}\begin{bmatrix} \dfrac{1}{b-a}(\mathrm{e}^{-a\tau} - \mathrm{e}^{-b\tau}) \\ \dfrac{1}{a-b}(a\mathrm{e}^{-a\tau} - b\mathrm{e}^{-b\tau}) \end{bmatrix}\mathrm{d}\tau \]

\[ = \begin{bmatrix} \dfrac{1}{b(a-b)}(1-\mathrm{e}^{-bt}) - \dfrac{1}{a(b-a)}(1-\mathrm{e}^{at}) \\ \dfrac{1}{a-b}(\mathrm{e}^{-bt} - \mathrm{e}^{-at}) \end{bmatrix} \]