1. **State the problem:** We need to find $\left(I - M\right)^{-1} D$ where $$M = \begin{bmatrix}0.4 & 0.1 & 0.0 \\ 0.2 & 0.6 & 0.1 \\ 0.1 & 0.1 & 0.4\end{bmatrix}, \quad D = \begin{bmatrix}21 \\ 37 \\ 12\end{bmatrix}$$ 2. **Calculate $I - M$:** The identity matrix $I$ of size 3 is $$I = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$$ So, $$I - M = \begin{bmatrix}1-0.4 & 0-0.1 & 0-0.0 \\ 0-0.2 & 1-0.6 & 0-0.1 \\ 0-0.1 & 0-0.1 & 1-0.4\end{bmatrix} = \begin{bmatrix}0.6 & -0.1 & 0 \\ -0.2 & 0.4 & -0.1 \\ -0.1 & -0.1 & 0.6\end{bmatrix}$$ 3. **Find the inverse $(I - M)^{-1}$:** Calculate the determinant $\det(I - M)$: $$\det = 0.6(0.4 \times 0.6 - (-0.1)(-0.1)) - (-0.1)(-0.2 \times 0.6 - (-0.1)(-0.1)) + 0(\ldots)$$ $$= 0.6(0.24 - 0.01) - (-0.1)(-0.12 - 0.01) + 0 = 0.6 \times 0.23 - 0.1 \times 0.13 = 0.138 - 0.013 = 0.125$$ Calculate the matrix of cofactors and then transpose to get the adjugate matrix: $$\text{adj}(I - M) = \begin{bmatrix}0.23 & 0.14 & 0.01 \\ 0.13 & 0.36 & 0.06 \\ 0.04 & 0.14 & 0.22\end{bmatrix}$$ Therefore, $$(I - M)^{-1} = \frac{1}{0.125} \times \text{adj}(I - M) = 8 \times \begin{bmatrix}0.23 & 0.14 & 0.01 \\ 0.13 & 0.36 & 0.06 \\ 0.04 & 0.14 & 0.22\end{bmatrix} = \begin{bmatrix}1.84 & 1.12 & 0.08 \\ 1.04 & 2.88 & 0.48 \\ 0.32 & 1.12 & 1.76\end{bmatrix}$$ 4. **Multiply $(I - M)^{-1}$ by $D$:** $$\begin{bmatrix}1.84 & 1.12 & 0.08 \\ 1.04 & 2.88 & 0.48 \\ 0.32 & 1.12 & 1.76\end{bmatrix} \times \begin{bmatrix}21 \\ 37 \\ 12\end{bmatrix} = \begin{bmatrix}1.84 \times 21 + 1.12 \times 37 + 0.08 \times 12 \\ 1.04 \times 21 + 2.88 \times 37 + 0.48 \times 12 \\ 0.32 \times 21 + 1.12 \times 37 + 1.76 \times 12\end{bmatrix}$$ Calculate each entry: $$= \begin{bmatrix}38.64 + 41.44 + 0.96 \\ 21.84 + 106.56 + 5.76 \\ 6.72 + 41.44 + 21.12\end{bmatrix} = \begin{bmatrix}81.04 \\ 134.16 \\ 69.28\end{bmatrix}$$ **Final answer:** $$\boxed{\begin{bmatrix}81.04 \\ 134.16 \\ 69.28\end{bmatrix}}$$