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Chapter 2: Q2.6-11Q (page 93)

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鈥檚 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

2. Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{3}}&{\bf{2}}\\{\bf{7}}&{\bf{4}}\end{aligned}} \right)\).

Let \(X\) be \(m \times n\) data matrix such that \({X^T}X\) is invertible., and let \(M = {I_m} - X{\left( {{X^T}X} \right)^{ - {\bf{1}}}}{X^T}\). Add a column \({x_{\bf{0}}}\) to the data and form

\(W = \left[ {\begin{array}{*{20}{c}}X&{{x_{\bf{0}}}}\end{array}} \right]\)

Compute \({W^T}W\). The \(\left( {{\bf{1}},{\bf{1}}} \right)\) entry is \({X^T}X\). Show that the Schur complement (Exercise 15) of \({X^T}X\) can be written in the form \({\bf{x}}_{\bf{0}}^TM{{\bf{x}}_{\bf{0}}}\). It can be shown that the quantity \({\left( {{\bf{x}}_{\bf{0}}^TM{{\bf{x}}_{\bf{0}}}} \right)^{ - {\bf{1}}}}\) is the \(\left( {{\bf{2}},{\bf{2}}} \right)\)-entry in \({\left( {{W^T}W} \right)^{ - {\bf{1}}}}\). This entry has a useful statistical interpretation, under appropriate hypotheses.

In the study of engineering control of physical systems, a standard set of differential equations is transformed by Laplace transforms into the following system of linear equations:

\(\left[ {\begin{array}{*{20}{c}}{A - s{I_n}}&B\\C&{{I_m}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\bf{x}}\\{\bf{u}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{y}}\end{array}} \right]\)

Where \(A\) is \(n \times n\), \(B\) is \(n \times m\), \(C\) is \(m \times n\), and \(s\) is a variable. The vector \({\bf{u}}\) in \({\mathbb{R}^m}\) is the 鈥渋nput鈥� to the system, \({\bf{y}}\) in \({\mathbb{R}^m}\) is the 鈥渙utput鈥� and \({\bf{x}}\) in \({\mathbb{R}^n}\) is the 鈥渟tate鈥� vector. (Actually, the vectors \({\bf{x}}\), \({\bf{u}}\) and \({\bf{v}}\) are functions of \(s\), but we suppress this fact because it does not affect the algebraic calculations in Exercises 19 and 20.)

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 \(B \ne C\).

Use the inverse found in Exercise 3 to solve the system

\(\begin{aligned}{l}{\bf{8}}{{\bf{x}}_{\bf{1}}} + {\bf{5}}{{\bf{x}}_{\bf{2}}} = - {\bf{9}}\\ - {\bf{7}}{{\bf{x}}_{\bf{1}}} - {\bf{5}}{{\bf{x}}_{\bf{2}}} = {\bf{11}}\end{aligned}\)

Suppose \(AD = {I_m}\) (the \(m \times m\) identity matrix). Show that for any b in \({\mathbb{R}^m}\), the equation \(A{\mathop{\rm x}\nolimits} = {\mathop{\rm b}\nolimits} \) has a solution. (Hint: Think about the equation \(AD{\mathop{\rm b}\nolimits} = {\mathop{\rm b}\nolimits} \).) Explain why Acannot have more rows than columns.
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