Question

Repeat Exercise 5 with

\(C = \left[ {\begin{array}{*{20}{c}}{.1}&{.6}\\{.5}&{.2}\end{array}} \right],\,{\mathop{\rm d}\nolimits} = \left[ {\begin{array}{*{20}{c}}{18}\\{11}\end{array}} \right]\)

Short answer

The production level necessary to satisfy the final demand is \(x = \left[ {\begin{array}{*{20}{c}}{50}\\{45}\end{array}} \right]\).

Step-by-step solution

Use an inverse matrix to determine the production level

Theorem 11states that C is the consumption matrix for an economy. Let d be the final demand. If C and d have non-negative entries and each column sum of C is less than 1, then \({\left( {I - C} \right)^{ - 1}}\) exists. Also, the production vector \(x = {\left( {I - C} \right)^{ - 1}}{\mathop{\rm d}\nolimits} \) has non-negative entries and is the unique solution of \(x = Cx + {\mathop{\rm d}\nolimits} \).

\[\begin{array}{c}{\mathop{\rm x}\nolimits} = {\left( {I - C} \right)^{ - 1}}{\mathop{\rm d}\nolimits} \\ = {\left( {\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{.1}&{.6}\\{.5}&{.2}\end{array}} \right]} \right)^{ - 1}}\left[ {\begin{array}{*{20}{c}}{18}\\{11}\end{array}} \right]\\ = {\left[ {\begin{array}{*{20}{c}}{.9}&{ - .6}\\{ - .5}&{.8}\end{array}} \right]^{ - 1}}\left[ {\begin{array}{*{20}{c}}{18}\\{11}\end{array}} \right]\\ = \frac{1}{{0.42}}\left[ {\begin{array}{*{20}{c}}{.8}&{.6}\\{.5}&{.9}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{18}\\{11}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1.9}&{1.42}\\{1.19}&{2.14}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{18}\\{11}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{50}\\{45}\end{array}} \right]\end{array}\]

Thus, the production level necessary to satisfy the final demand is \(x = \left[ {\begin{array}{*{20}{c}}{50}\\{45}\end{array}} \right]\).