Answer
See solution
Work Step by Step
First get inverse of $(I-C):$
\[
(I-C)^{-1}=\left[\begin{array}{cc}
1 & -.5 \\
-.6 & .8
\end{array}\right]^{-1}=\frac{1}{.3-.8}\left[\begin{array}{cc}
.8 & .5 \\
.6 & 1
\end{array}\right]=\left[\begin{array}{cc}
1.6 & 1 \\
1.2 & 2
\end{array}\right]
\]
Production model: $x=(I-C)^{-1} d$.
We need $d_{1}=\left[\begin{array}{l}1 \\ 0\end{array}\right],$ so $x=\left[\begin{array}{cc}1.6 & 1 \\ 1.2 & 2\end{array}\right] \cdot\left[\begin{array}{l}1 \\ 0\end{array}\right]=\left[\begin{array}{c}1.6 \\ 1.2\end{array}\right]$
$b)$
\[
d_{2}=\left[\begin{array}{l}
51 \\
30
\end{array}\right]
\]
Use same production model as in part $a$ ):
\[
x=\left[\begin{array}{ll}
1.6 & 1 \\
1.2 & 2
\end{array}\right] \cdot\left[\begin{array}{l}
51 \\
30
\end{array}\right]=\left[\begin{array}{l}
111.6 \\
121.2
\end{array}\right]
\]
c)
From exercise $5,$ we know that the production corresponding to demand
\[
d=\left[\begin{array}{l}
50 \\
30
\end{array}\right] \text { is } x=\left[\begin{array}{l}
110 \\
120
\end{array}\right]
\]
Also, note that $d_{2}=d+d_{1}$.
Using the production model for $x_{2}$, we obtain:
\[
\begin{array}{l}
x_{2}=(I-C)^{-1} \cdot d_{2} \\
=(I-C)^{-1}\left(d+d_{1}\right)=\text { , distributive property of matrix multiplication } \\
=(I-C)^{-1} d+(I-C)^{-1} d_{1}= \\
=x+x_{1}
\end{array}
\]
