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Chapter 2: Q2.6-1Q (page 93)

Exercises 1-4 refer to an economy that is divided into three sectors - manufacturing, agriculture, and services. For each unit of output, manufacturing requires .10 unit from other companies in that sector, .30 unit from services. For each unit of output, agriculture uses .20 unit of its own output, .60 unit from manufacturing, and .10 unit from services. For each unit of output, the services sector consumes .10 unit from services, .60 unit from manufacturing, but no agricultural products.

1. Construct the consumption matrix for this economy, and determine what intermediate demands are created if agriculture plans to produce 100 units.

Short Answer

Expert verified

The intermediate demand is \(Cx = \left[ {\begin{array}{*{20}{c}}{60}\\{20}\\{10}\end{array}} \right]\).

Step by step solution

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01

Construct the consumption matrix for the economy

The important point is that each column represents the unit consumption vector for the corresponding vector. When you order sector manufacturing, agriculture, and services, the consumption matrix is

\(C = \left[ {\begin{array}{*{20}{c}}{.10}&{.60}&{.60}\\{.30}&{.20}&0\\{.30}&{.10}&{.10}\end{array}} \right]\).

02

Determine the intermediate demand created

The total intermediate demand from all three sectors is

\(\begin{array}{c}\left\{ {intermediate demand} \right\} = {x_1}{c_1} + {x_2}{c_2} + {x_3}{c_3}\\ = Cx.\end{array}\)

Here, C is theconsumption matrix\(\left[ {\begin{array}{*{20}{c}}{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right]\).

\(Cx\)represents the intermediate demands produced by the production vector x. If agriculture plans to produce 100 units, then the intermediate vector is

\(\begin{array}{c}Cx = \left[ {\begin{array}{*{20}{c}}{.10}&{.60}&{.60}\\{.30}&{.20}&0\\{.30}&{.10}&{.10}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}0\\{100}\\0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{0.1 \times 0 + 0.6 \times 100 + 0.6 \times 0}\\{0.3 \times 0 + 0.2 \times 100 + 0 \times 0}\\{0.3 \times 0 + 0.1 \times 100 + 0.1 \times 0}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{0 + 60 + 0}\\{0 + 20 + 0}\\{0 + 10 + 0}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{60}\\{20}\\{10}\end{array}} \right].\end{array}\)

Thus, the intermediate demand is \(Cx = \left[ {\begin{array}{*{20}{c}}{60}\\{20}\\{10}\end{array}} \right]\).

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Most popular questions from this chapter

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.]

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

Suppose the second column of Bis all zeros. What can you

say about the second column of AB?

A useful way to test new ideas in matrix algebra, or to make conjectures, is to make calculations with matrices selected at random. Checking a property for a few matrices does not prove that the property holds in general, but it makes the property more believable. Also, if the property is actually false, you may discover this when you make a few calculations.

37. Construct a random \({\bf{4}} \times {\bf{4}}\) matrix Aand test whether \(\left( {A + I} \right)\left( {A - I} \right) = {A^2} - I\). The best way to do this is to compute \(\left( {A + I} \right)\left( {A - I} \right) - \left( {{A^2} - I} \right)\) and verify that this difference is the zero matrix. Do this for three random matrices. Then test \(\left( {A + B} \right)\left( {A - B} \right) = {A^2} - {B^{\bf{2}}}\) the same way for three pairs of random \({\bf{4}} \times {\bf{4}}\) matrices. Report your conclusions.

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]

6. \[\left[ {\begin{array}{*{20}{c}}X&{\bf{0}}\\Y&Z\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&{\bf{0}}\\B&C\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\{\bf{0}}&I\end{array}} \right]\]

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 “input” to the system, \({\bf{y}}\) in \({\mathbb{R}^m}\) is the “output” and \({\bf{x}}\) in \({\mathbb{R}^n}\) is the “state” 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.)
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