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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that if the columns of Bare linearly dependent, then so are the columns of AB.

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 \(AB\).

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

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 \({I_n}\).

In Exercises 33 and 34, Tis a linear transformation from \({\mathbb{R}^2}\) into \({\mathbb{R}^2}\). Show that T is invertible and find a formula for \({T^{ - 1}}\).

33. \(T\left( {{x_1},{x_2}} \right) = \left( { - 5{x_1} + 9{x_2},4{x_1} - 7{x_2}} \right)\)

Suppose \(\left( {B - C} \right)D = 0\), where Band Care \(m \times n\) matrices and \(D\) is invertible. Show that B = C.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free