Let C and d be as in Exercise 5.

1. Determine the production level necessary to satisfy a final demand for 1 unit of output from sector 1.
2. Use an inverse matrix to determine the production level necessary to satisfy a final demand of \(\left[ {\begin{array}{*{20}{c}}{51}\\{30}\end{array}} \right]\).
3. Use the fact that \(\left[ {\begin{array}{*{20}{c}}{51}\\{30}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{50}\\{30}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]\) to explain how and why the answers to parts (a) and (b) and to Exercise 5 are related.

Consider the production model \[x = Cx + {\mathop{\rm d}\nolimits} \] for an economy with two sectors, where \(C = \left[ {\begin{array}{*{20}{c}}{.0}&{.5}\\{.6}&{.2}\end{array}} \right],\,{\mathop{\rm d}\nolimits} = \left[ {\begin{array}{*{20}{c}}{50}\\{30}\end{array}} \right]\).

(a)

Theorem 11 states 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 \(x = Cx + {\mathop{\rm d}\nolimits} \).

From Exercise 5, \({\left( {I - C} \right)^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{1.6}&1\\{1.2}&2\end{array}} \right]\).

The production level necessary to satisfy the final demand is shown below:

\[\begin{array}{c}{x_1} = {\left( {I - C} \right)^{ - 1}}{{\mathop{\rm d}\nolimits} _1}\\ = \left[ {\begin{array}{*{20}{c}}{1.6}&1\\{1.2}&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1.6 + 0}\\{1.2 + 0}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1.6}\\{1.2}\end{array}} \right]\end{array}\]

This corresponds to the first column of \({\left( {I - C} \right)^{ - 1}}\).

Thus, the production level necessary to satisfy the final demandfor 1 unit of output from sector 1 is \({x_1} = \left[ {\begin{array}{*{20}{c}}{1.6}\\{1.2}\end{array}} \right]\).

(b)

The production level necessary to satisfy the final demand of \(\left[ {\begin{array}{*{20}{c}}{51}\\{30}\end{array}} \right]\) is shown below:

\[\begin{array}{c}{x_2} = {\left( {I - C} \right)^{ - 1}}{{\mathop{\rm d}\nolimits} _2}\\ = \left[ {\begin{array}{*{20}{c}}{1.6}&1\\{1.2}&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{51}\\{30}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{81.6 + 30}\\{61.2 + 60}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{111.6}\\{121.2}\end{array}} \right]\end{array}\]

Thus, the production level necessary to satisfy the final demand of \(\left[ {\begin{array}{*{20}{c}}{51}\\{30}\end{array}} \right]\) is \({x_2} = \left[ {\begin{array}{*{20}{c}}{111.6}\\{121.2}\end{array}} \right]\).

(c)

From Exercise 5, the production vector x corresponding to

\({\mathop{\rm d}\nolimits} = \left[ {\begin{array}{*{20}{c}}{50}\\{20}\end{array}} \right]\)is \(x = \left[ {\begin{array}{*{20}{c}}{110}\\{120}\end{array}} \right]\).

Use \({{\mathop{\rm d}\nolimits} _2} = d + {d_1}\) to obtain the relation between parts (a), (b), and Exercise 5, as shown below:

\(\begin{array}{c}{{\mathop{\rm x}\nolimits} _2} = {\left( {I - C} \right)^{ - 1}}{{\mathop{\rm d}\nolimits} _2}\\ = {\left( {I - C} \right)^{ - 1}}\left( {{\mathop{\rm d}\nolimits} + {{\mathop{\rm d}\nolimits} _1}} \right)\\ = {\left( {I - C} \right)^{ - 1}}{\mathop{\rm d}\nolimits} + {\left( {I - C} \right)^{ - 1}}{{\mathop{\rm d}\nolimits} _1}\\ = {\mathop{\rm x}\nolimits} + {{\mathop{\rm x}\nolimits} _1}\end{array}\)

Thus, the relation between parts (a), (b) and Exercise 5 is \({{\mathop{\rm x}\nolimits} _2} = x + {x_1}\).