Pulse TF matrix (G = [3 1; −2 1], D = [1; 0])

The pulse transfer function matrix is $\mathbf{F}(z) = \mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} + \mathbf{D}$. There is one input and two outputs, so $\mathbf{F}(z)$ is 2×1.

Given

$$\mathbf{G} = \begin{bmatrix} 3 & 1 \\ -2 & 1 \end{bmatrix},\quad \mathbf{H} = \begin{bmatrix} 1 \\ 1 \end{bmatrix},\quad \mathbf{C} = \begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix},\quad \mathbf{D} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$

Step 1: Inverse of $z\mathbf{I}-\mathbf{G}$

$$z\mathbf{I}-\mathbf{G} = \begin{bmatrix} z-3 & -1 \\ 2 & z-1 \end{bmatrix},\qquad |z\mathbf{I}-\mathbf{G}| = (z-3)(z-1) + 2 = z^2 - 4z + 5$$ $$(z\mathbf{I}-\mathbf{G})^{-1} = \frac{1}{z^2-4z+5}\begin{bmatrix} z-1 & 1 \\ -2 & z-3 \end{bmatrix}$$

Step 2: Multiply by H

$$(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \frac{1}{z^2-4z+5}\begin{bmatrix} z \\ z-5 \end{bmatrix}$$

Step 3: Multiply by C

$$\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \frac{1}{z^2-4z+5}\begin{bmatrix} z + (z-5) \\ 2z + (z-5) \end{bmatrix} = \frac{1}{z^2-4z+5}\begin{bmatrix} 2z-5 \\ 3z-5 \end{bmatrix}$$

Step 4: Add D

$$\begin{aligned} F_1(z) &= \frac{2z-5}{z^2-4z+5} + 1 = \frac{z^2-2z}{z^2-4z+5}\\ F_2(z) &= \frac{3z-5}{z^2-4z+5} \end{aligned}$$

Result

$$\mathbf{F}(z) = \begin{bmatrix} \dfrac{z(z-2)}{z^2-4z+5} \\[3mm] \dfrac{3z-5}{z^2-4z+5} \end{bmatrix}$$

The poles are $z = 2 \pm j1$ ($|z| = \sqrt5 > 1$), so the system is unstable.

Answer: $Y_1(z)/U(z) = \dfrac{z^2-2z}{z^2-4z+5}$, $Y_2(z)/U(z) = \dfrac{3z-5}{z^2-4z+5}$