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The Leontief production equation \({\bf{x}} = C{\bf{x}} + {\bf{d}}\), is usually accompanied by a dual price equation,

\({\bf{p}} = {C^T}{\bf{p}} + {\bf{v}}\)

Where \({\bf{p}}\) is a price vector whose entries list the price per unit for each sector’s output, and \({\bf{v}}\) is a value added vector whose entries list the value added per unit of output. (Value added includes wages, profit, depreciation, etc.). An important fact in economics is that the gross domestic product (GDP) can be expressed in two ways:

{gross domestic product} \( = {{\bf{p}}^T}{\bf{d}} = {{\bf{v}}^T}{\bf{x}}\)

Verify the second equality. [Hint: Compute \({{\bf{p}}^T}{\bf{x}}\)in two ways.]

Short Answer

Expert verified

\({{\bf{v}}^T}{\bf{x}} = {{\bf{p}}^T}{\bf{d}}\)

Step by step solution

01

Solve the price equation

The given equation is \({\bf{p}} = {C^T}{\bf{p}} + {\bf{v}}\),

\(\begin{array}{c}{{\bf{p}}^T} = {\left( {{C^T}{\bf{p}} + {\bf{v}}} \right)^T}\\ = {\left( {{C^T}{\bf{p}}} \right)^T} + {{\bf{v}}^T}\\ = {{\bf{p}}^T}C + {{\bf{v}}^T}\end{array}\)

Multiply the above equation by \({\bf{x}}\) on both sides.

\(\begin{array}{c}{{\bf{p}}^T}{\bf{x}} = \left( {{{\bf{p}}^T}C + {{\bf{v}}^T}} \right){\bf{x}}\\ = {{\bf{p}}^T}C{\bf{x}} + {{\bf{v}}^T}{\bf{x}}\end{array}\)

02

Find the value of \({{\bf{p}}^T}{\bf{x}}\)

Multiply the equation \({{\bf{p}}^T} = {{\bf{p}}^T}C + {{\bf{v}}^T}\) by \({\bf{x}}\) on both sides.

\(\begin{array}{c}{{\bf{p}}^T}{\bf{x}} = \left( {{{\bf{p}}^T}C + {{\bf{v}}^T}} \right){\bf{x}}\\ = {{\bf{p}}^T}C{\bf{x}} + {{\bf{v}}^T}{\bf{x}}\end{array}\)

03

Find \({{\bf{p}}^T}{\bf{x}}\) using the production equation

Find the value of \({{\bf{p}}^T}{\bf{x}}\) using the equation \({\bf{x}} = C{\bf{x}} + {\bf{d}}\).

\(\begin{array}{c}{{\bf{p}}^T}{\bf{x}} = {{\bf{p}}^T}\left( {C{\bf{x}} + {\bf{d}}} \right)\\ = {{\bf{p}}^T}C{\bf{x}} + {{\bf{p}}^T}{\bf{d}}\end{array}\)

04

Compare the two equations of \({{\bf{p}}^T}{\bf{x}}\)

\(\begin{array}{c}{{\bf{p}}^T}C{\bf{x}} + {{\bf{v}}^T}{\bf{x}} = {{\bf{p}}^T}C{\bf{x}} + {{\bf{p}}^T}{\bf{d}}\\{{\bf{v}}^T}{\bf{x}} = {{\bf{p}}^T}{\bf{d}}\end{array}\)

So, the equation\({{\bf{v}}^T}{\bf{x}} = {{\bf{p}}^T}{\bf{d}}\)is true.

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Most popular questions from this chapter

Let Abe an invertible \(n \times n\) matrix, and let B be an \(n \times p\) matrix. Show that the equation \(AX = B\) has a unique solution \({A^{ - 1}}B\).

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.

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

If Ais an \(n \times n\) matrix and the equation \(A{\bf{x}} = {\bf{b}}\) has more than one solution for some b, then the transformation \({\bf{x}}| \to A{\bf{x}}\) is not one-to-one. What else can you say about this transformation? Justify your answer.

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