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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

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

In Exercise 10 mark each statement True or False. Justify each answer.

10. a. A product of invertible \(n \times n\) matrices is invertible, and the inverse of the product of their inverses in the same order.

b. If A is invertible, then the inverse of \({A^{ - {\bf{1}}}}\) is A itself.

c. If \(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\) and \(ad = bc\), then A is not invertible.

d. If A can be row reduced to the identity matrix, then A must be invertible.

e. If A is invertible, then elementary row operations that reduce A to the identity \({I_n}\) also reduce \({A^{ - {\bf{1}}}}\) to \({I_n}\).

Use partitioned matrices to prove by induction that the product of two lower triangular matrices is also lower triangular. [Hint: \(A\left( {k + 1} \right) \times \left( {k + 1} \right)\) matrix \({A_1}\) can be written in the form below, where \[a\] is a scalar, v is in \({\mathbb{R}^k}\), and Ais a \(k \times k\) lower triangular matrix. See the study guide for help with induction.]

\({A_1} = \left[ {\begin{array}{*{20}{c}}a&{{0^T}}\\0&A\end{array}} \right]\).

3. Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{5}}\\{ - {\bf{7}}}&{ - {\bf{5}}}\end{aligned}} \right)\).

Prove the Theorem 3(d) i.e., \({\left( {AB} \right)^T} = {B^T}{A^T}\).

Let \(A = \left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\), and \(D = \left( {\begin{aligned}{*{20}{c}}2&0&0\\0&3&0\\0&0&5\end{aligned}} \right)\). Compute \(AD\) and \(DA\). Explain how the columns or rows of A change when A is multiplied by D on the right or on the left. Find a \(3 \times 3\) matrix B, not the identity matrix or the zero matrix, such that \(AB = BA\).
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