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Chapter 2: Q2.6-5Q (page 93)

Consider the production model \[{\mathop{\rm x}\nolimits} = C{\mathop{\rm x}\nolimits} + d\] for an economy with two sectors, where

\(C = \left[ {\begin{array}{*{20}{c}}{.0}&{.5}\\{.6}&{.2}\end{array}} \right],\,{\mathop{\rm d}\nolimits} = \left[ {\begin{array}{*{20}{c}}{50}\\{30}\end{array}} \right]\)

Use an inverse matrix to determine the production level necessary to satisfy the final demand.

Short Answer

Expert verified

The production level necessary to satisfy the final demand is \(x = \left[ {\begin{array}{*{20}{c}}{110}\\{120}\end{array}} \right]\).

Step by step solution

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01

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

\(\begin{array}{c}{\mathop{\rm x}\nolimits} = {\left( {I - C} \right)^{ - 1}}{\mathop{\rm d}\nolimits} \\ = {\left( {\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{.0}&{.5}\\{.6}&{.2}\end{array}} \right]} \right)^{ - 1}}\left[ {\begin{array}{*{20}{c}}{50}\\{20}\end{array}} \right]\\ = {\left[ {\begin{array}{*{20}{c}}1&{ - .5}\\{ - .6}&{.8}\end{array}} \right]^{ - 1}}\left[ {\begin{array}{*{20}{c}}{50}\\{20}\end{array}} \right]\\ = \frac{1}{{0.5}}\left[ {\begin{array}{*{20}{c}}{.8}&{.5}\\{.6}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{50}\\{20}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1.6}&1\\{1.2}&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{50}\\{20}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{110}\\{120}\end{array}} \right]\end{array}\)

Thus, the production level necessary to satisfy the final demand is \(x = \left[ {\begin{array}{*{20}{c}}{110}\\{120}\end{array}} \right]\)

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Most popular questions from this chapter

Let \(A = \left[ {\begin{array}{*{20}{c}}B&{\bf{0}}\\{\bf{0}}&C\end{array}} \right]\), where \(B\) and \(C\) are square. Show that \(A\)is invertible if an only if both \(B\) and \(C\) are invertible.

In Exercises 1 and 2, compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

\(A = \left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right)\), \(C = \left( {\begin{aligned}{*{20}{c}}1&2\\{ - 2}&1\end{aligned}} \right)\), \(D = \left( {\begin{aligned}{*{20}{c}}3&5\\{ - 1}&4\end{aligned}} \right)\) and \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

\(A + 2B\), \(3C - E\), \(CB\), \(EB\).

In exercise 5 and 6, compute the product \(AB\) in two ways: (a) by the definition, where \(A{b_{\bf{1}}}\) and \(A{b_{\bf{2}}}\) are computed separately, and (b) by the row-column rule for computing \(AB\).

\(A = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{2}}}\\{ - {\bf{3}}}&{\bf{0}}\\{\bf{3}}&{\bf{5}}\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{3}}\\{\bf{2}}&{ - {\bf{1}}}\end{aligned}} \right)\)

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{ - {\bf{3}}}\\{ - {\bf{4}}}&{\bf{6}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{4}}\\{\bf{5}}&{\bf{5}}\end{aligned}} \right)\) and \(C = \left( {\begin{aligned}{*{20}{c}}{\bf{5}}&{ - {\bf{2}}}\\{\bf{3}}&{\bf{1}}\end{aligned}} \right)\). Verfiy that \(AB = AC\) and yet \(B \ne C\).

Exercises 15 and 16 concern arbitrary matrices A, B, and Cfor which the indicated sums and products are defined. Mark each statement True or False. Justify each answer.

15. a. If A and B are \({\bf{2}} \times {\bf{2}}\) with columns \({{\bf{a}}_1},{{\bf{a}}_2}\) and \({{\bf{b}}_1},{{\bf{b}}_2}\) respectively, then \(AB = \left( {\begin{aligned}{*{20}{c}}{{{\bf{a}}_1}{{\bf{b}}_1}}&{{{\bf{a}}_2}{{\bf{b}}_2}}\end{aligned}} \right)\).

b. Each column of ABis a linear combination of the columns of Busing weights from the corresponding column of A.

c. \(AB + AC = A\left( {B + C} \right)\)

d. \({A^T} + {B^T} = {\left( {A + B} \right)^T}\)

e. The transpose of a product of matrices equals the product of their transposes in the same order.
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