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Let C and d be as in Exercise 5.

  1. Determine the production level necessary to satisfy a final demand for 1 unit of output from sector 1.
  2. Use an inverse matrix to determine the production level necessary to satisfy a final demand of \(\left[ {\begin{array}{*{20}{c}}{51}\\{30}\end{array}} \right]\).
  3. Use the fact that \(\left[ {\begin{array}{*{20}{c}}{51}\\{30}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{50}\\{30}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]\) to explain how and why the answers to parts (a) and (b) and to Exercise 5 are related.

Short Answer

Expert verified
  1. The production level necessary to satisfy the final demandfor 1 unit of output from sector 1 is \({x_1} = \left[ {\begin{array}{*{20}{c}}{1.6}\\{1.2}\end{array}} \right]\).
  2. The production level necessary to satisfy a final demand of \(\left[ {\begin{array}{*{20}{c}}{51}\\{30}\end{array}} \right]\) is \({x_2} = \left[ {\begin{array}{*{20}{c}}{111.6}\\{121.2}\end{array}} \right]\).
  3. The relation of parts (a) and (b) and Exercise 5 is \({{\mathop{\rm x}\nolimits} _2} = x + {x_1}\).

Step by step solution

01

The value of C and d

Consider the production model \[x = Cx + {\mathop{\rm d}\nolimits} \] for an economy with two sectors, where \(C = \left[ {\begin{array}{*{20}{c}}{.0}&{.5}\\{.6}&{.2}\end{array}} \right],\,{\mathop{\rm d}\nolimits} = \left[ {\begin{array}{*{20}{c}}{50}\\{30}\end{array}} \right]\).

02

Determine the production level

(a)

Theorem 11 states 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 \(x = Cx + {\mathop{\rm d}\nolimits} \).

From Exercise 5, \({\left( {I - C} \right)^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{1.6}&1\\{1.2}&2\end{array}} \right]\).

The production level necessary to satisfy the final demand is shown below:

\[\begin{array}{c}{x_1} = {\left( {I - C} \right)^{ - 1}}{{\mathop{\rm d}\nolimits} _1}\\ = \left[ {\begin{array}{*{20}{c}}{1.6}&1\\{1.2}&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1.6 + 0}\\{1.2 + 0}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1.6}\\{1.2}\end{array}} \right]\end{array}\]

This corresponds to the first column of \({\left( {I - C} \right)^{ - 1}}\).

Thus, the production level necessary to satisfy the final demandfor 1 unit of output from sector 1 is \({x_1} = \left[ {\begin{array}{*{20}{c}}{1.6}\\{1.2}\end{array}} \right]\).

03

Use the inverse matrix to determine the production level

(b)

The production level necessary to satisfy the final demand of \(\left[ {\begin{array}{*{20}{c}}{51}\\{30}\end{array}} \right]\) is shown below:

\[\begin{array}{c}{x_2} = {\left( {I - C} \right)^{ - 1}}{{\mathop{\rm d}\nolimits} _2}\\ = \left[ {\begin{array}{*{20}{c}}{1.6}&1\\{1.2}&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{51}\\{30}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{81.6 + 30}\\{61.2 + 60}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{111.6}\\{121.2}\end{array}} \right]\end{array}\]

Thus, the production level necessary to satisfy the final demand of \(\left[ {\begin{array}{*{20}{c}}{51}\\{30}\end{array}} \right]\) is \({x_2} = \left[ {\begin{array}{*{20}{c}}{111.6}\\{121.2}\end{array}} \right]\).

04

Obtain the relation of parts (a) and (b) and Exercise 5

(c)

From Exercise 5, the production vector x corresponding to

\({\mathop{\rm d}\nolimits} = \left[ {\begin{array}{*{20}{c}}{50}\\{20}\end{array}} \right]\)is \(x = \left[ {\begin{array}{*{20}{c}}{110}\\{120}\end{array}} \right]\).

Use \({{\mathop{\rm d}\nolimits} _2} = d + {d_1}\) to obtain the relation between parts (a), (b), and Exercise 5, as shown below:

\(\begin{array}{c}{{\mathop{\rm x}\nolimits} _2} = {\left( {I - C} \right)^{ - 1}}{{\mathop{\rm d}\nolimits} _2}\\ = {\left( {I - C} \right)^{ - 1}}\left( {{\mathop{\rm d}\nolimits} + {{\mathop{\rm d}\nolimits} _1}} \right)\\ = {\left( {I - C} \right)^{ - 1}}{\mathop{\rm d}\nolimits} + {\left( {I - C} \right)^{ - 1}}{{\mathop{\rm d}\nolimits} _1}\\ = {\mathop{\rm x}\nolimits} + {{\mathop{\rm x}\nolimits} _1}\end{array}\)

Thus, the relation between parts (a), (b) and Exercise 5 is \({{\mathop{\rm x}\nolimits} _2} = x + {x_1}\).

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Most popular questions from this chapter

In Exercises 1 and 2, compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

\(A = \left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right)\), \(C = \left( {\begin{aligned}{*{20}{c}}1&2\\{ - 2}&1\end{aligned}} \right)\), \(D = \left( {\begin{aligned}{*{20}{c}}3&5\\{ - 1}&4\end{aligned}} \right)\) and \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

\(A + 2B\), \(3C - E\), \(CB\), \(EB\).

In Exercises 1โ€“9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5โ€“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.]

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

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.

Use partitioned matrices to prove by induction that for \(n = 2,3,...\), the \(n \times n\) matrices \(A\) shown below is invertible and \(B\) is its inverse.

\[A = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\1&1&0&{}&0\\1&1&1&{}&0\\ \vdots &{}&{}& \ddots &{}\\1&1&1& \ldots &1\end{array}} \right]\]

\[B = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\{ - 1}&1&0&{}&0\\0&{ - 1}&1&{}&0\\ \vdots &{}& \ddots & \ddots &{}\\0&{}& \ldots &{ - 1}&1\end{array}} \right]\]

For the induction step, assume A and Bare \(\left( {k + 1} \right) \times \left( {k + 1} \right)\) matrices, and partition Aand B in a form similar to that displayed in Exercises 23.

Use matrix algebra to show that if A is invertible and D satisfies \(AD = I\) then \(D = {A^{ - {\bf{1}}}}\).

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