1. An economy has three sectors producing products Product 1, Product 2, and Prod- uct 3. • To produce 1 unit of Product 1 requires 0.20 units of Product 1, 0.15 units of Product 2, and 0.10 units of Product 3. • To product 1 unit of Product 2 it takes 0.14 units of Product 1, 0.05 units of Product 2, and 0.12 units of Product 3. • To produce 1 unit of Product 3 it takes 0.14 units of Product 1, and 0.08 units of Product 2. (a) What should the total production be set at in order to satisfy an external de- mand of 100 units of Product 1, 120 units of Product 2, and 150 units of Product 3? (b) Individually interpret each entry in the third column of (I − M)−1.
Answers: 1
![\left[\begin{array}{c}P_1\\P_2\\P_3\end{array}\right] =\left[\begin{array}{ccc}0.02&.0.15&0.10\\0.14&0.05&0.12\\0.14&0.08&0\end{array}\right] \left[\begin{array}{c}P_1\\P_2\\P_3\end{array}\right] +\left[\begin{array}{c}P_1\\P_2\\P_3\end{array}\right] _{external}](/tpl/images/0512/0177/6d45c.png)
![\left[\begin{array}{c}P_1\\P_2\\P_3\end{array}\right] _{external}=\left[\begin{array}{c}100\\120\\150\end{array}\right]](/tpl/images/0512/0177/b2b35.png)
![\left[\begin{array}{c}P_1\\P_2\\P_3\end{array}\right] =\left[\begin{array}{ccc}0.02&.0.15&0.10\\0.14&0.05&0.12\\0.14&0.08&0\end{array}\right] \left[\begin{array}{c}P_1\\P_2\\P_3\end{array}\right] +\left[\begin{array}{c}100\\120\\150\end{array}\right]](/tpl/images/0512/0177/c888a.png)
![P=\left[\begin{array}{c}P_1\\P_2\\P_3\end{array}\right] \\M=\left[\begin{array}{ccc}0.02&.0.15&0.10\\0.14&0.05&0.12\\0.14&0.08&0\end{array}\right] \\\Delta=\left[\begin{array}{c}100\\120\\150\end{array}\right]](/tpl/images/0512/0177/b7a9b.png)
![\\I-M=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] -\left[\begin{array}{ccc}0.02&0.15&0.10\\0.14&0.05&0.12\\0.14&0.08&0\end{array}\right] \\ I-M=\left[\begin{array}{ccc}1-0.02&0-0.15&0-0.10\\0-0.14&1-0.05&0-0.12\\0-0.14&0-0.08&1-0\end{array}\right] \\I-M=\left[\begin{array}{ccc}0.98&-0.15&-0.10\\-0.14&0.95&-0.12\\-0.14&-0.08&1\end{array}\right]](/tpl/images/0512/0177/61845.png)
![adj(I-M)=adj(\left[\begin{array}{ccc}0.98&-0.15&-0.10\\-0.14&0.95&-0.12\\-0.14&-0.08&1\end{array}\right])\\adj(I-M)=\left[\begin{array}{ccc}0.9404&0.1580 & 0.1130\\0.1568&0.9660& 0.1316\\0.1442 &0.0994 & 0.9100\end{array}\right]\\](/tpl/images/0512/0177/cfc6b.png)
![(I-M)^{-1}=\frac{1}{0.8837}\left[\begin{array}{ccc}0.9404&0.1580 & 0.1130\\0.1568&0.9660& 0.1316\\0.1442 &0.0994 & 0.9100\end{array}\right]\\\\(I-M)^{-1}=\left[\begin{array}{ccc}1.0642 & 0.1788 & 0.1279\\ 0.1774 &1.0932 & 0.1489\\ 0.1632 & 0.1125 & 1.0298\end{array}\right]\\](/tpl/images/0512/0177/17675.png)
![\left[\begin{array}{c}P_1\\P_2\\P_3\end{array}\right]=\left[\begin{array}{ccc}1.0642 & 0.1788 & 0.1279\\ 0.1774 &1.0932 & 0.1489\\ 0.1632 & 0.1125 & 1.0298\end{array}\right]\left[\begin{array}{c}100\\120\\150\end{array}\right]\\\\\left[\begin{array}{c}P_1\\P_2\\P_3\end{array}\right]=\left[\begin{array}{c}147.06\\ 171.27 \\184.29\end{array}\right]\\](/tpl/images/0512/0177/d6af0.png)
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