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Chapter 3: Problem 13

In the notation of the open model of Leontief, suppose that $$ A=\left(\begin{array}{cc} \frac{1}{2} & \frac{1}{5} \\ \frac{1}{3} & \frac{1}{5} \end{array}\right) \quad \text { and } \quad d=\left(\begin{array}{l} 2 \\ 5 \end{array}\right) $$ are the input-output matrix and the demand vector, respectively. How much of each commodity must be produced to satisfy this demand?

Short Answer

Expert verified
To satisfy the given demand, \(13\) units of the first commodity and \(\frac{40}{3}\) units of the second commodity must be produced.

Step by step solution

01

Identify the given matrices

: We are given the input-output matrix A and the demand vector d as follows: A=\(\left(\begin{array}{cc} \frac{1}{2} & \frac{1}{5} \\\ \frac{1}{3} & \frac{1}{5} \end{array}\right)\) d=\(\left(\begin{array}{l} 2 \\\ 5 \end{array}\right)\)
02

Find the matrix I - A

: First, we need to calculate the matrix I - A, where I is the identity matrix. The identity matrix is a square matrix with ones on the diagonal and zeros elsewhere. So, in this case: I=\(\left(\begin{array}{cc} 1 & 0 \\\ 0 & 1 \end{array}\right)\) Now, find I - A by subtracting corresponding elements: I - A = \(\left(\begin{array}{cc} 1 - \frac{1}{2} & 0 - \frac{1}{5} \\\ 0 - \frac{1}{3} & 1 - \frac{1}{5} \end{array}\right)\) I - A = \(\left(\begin{array}{cc} \frac{1}{2} & -\frac{1}{5} \\\ -\frac{1}{3} & \frac{4}{5} \end{array}\right)\)
03

Find the inverse matrix (I - A)^{-1}

: To find the inverse of a 2x2 matrix, we can follow these steps: 1. Swap the elements in positions (1, 1) and (2, 2); 2. Change the signs of the elements in positions (1, 2) and (2, 1); 3. Divide each element by the determinant of the original matrix. So first, we need to find the determinant of (I - A): |I - A| = \(\frac{1}{2}*\frac{4}{5} - (-\frac{1}{5})*(-\frac{1}{3})\) |I - A| = \(\frac{2}{5} - \frac{1}{15}\) = \(\frac{3}{15}\) Now, find the inverse matrix (I - A)^{-1}: \((I - A)^{-1}\) = \( \frac{1}{\frac{3}{15}} * \left(\begin{array}{cc} \frac{4}{5} & \frac{1}{5} \\\ \frac{1}{3} & \frac{1}{2} \end{array}\right)\) \((I - A)^{-1}\) = \( 5 * \left(\begin{array}{cc} \frac{4}{5} & \frac{1}{5} \\\ \frac{1}{3} & \frac{1}{2} \end{array}\right)\) \((I - A)^{-1}\) = \(\left(\begin{array}{cc} 4 & 1 \\\ \frac{5}{3} & \frac{5}{2} \end{array}\right)\)
04

Multiply (I - A)^{-1} with the demand vector d

: Now, we can find the production quantities required to satisfy the demand by multiplying the matrix (I - A)^{-1} with the demand vector d: x = \((I - A)^{-1}*d\)= \(\left(\begin{array}{cc} 4 & 1 \\\ \frac{5}{3} & \frac{5}{2} \end{array}\right)\) * \(\left(\begin{array}{l} 2 \\\ 5 \end{array}\right)\) x = \(\left(\begin{array}{l} 4*2 + 1*5 \\\ \frac{5}{3}*2 + \frac{5}{2}*5 \end{array}\right)\) x = \(\left(\begin{array}{l} 13 \\\ \frac{40}{3} \end{array}\right)\) So, to satisfy the given demand, \(13\) units of the first commodity and \(\frac{40}{3}\) units of the second commodity must be produced.

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Most popular questions from this chapter

In the closed model of Leontief with food, clothing, and housing as the basic industries, suppose that the input-output matrix is $$ A=\left(\begin{array}{rrr} \frac{7}{16} & \frac{1}{2} & \frac{3}{16} \\ \frac{5}{16} & \frac{1}{6} & \frac{5}{16} \\ \frac{1}{4} & \frac{1}{3} & \frac{1}{2} \end{array}\right) \text {. } $$ At what ratio must the farmer, tailor, and carpenter produce in order for equilibrium to be attained?

A system \(\mathscr{L}\) of linear equations is said to be consistent if no linear combination of its equations is a degenerate equation \(L\) with a nonzero constant. Show that \(\mathscr{L}\) is consistent if and only if \(\mathscr{L}\) is reducible to echelon form.

Solve each of the following systems using its augmented matrix \(M:\) a. \(\begin{aligned} x+2 y-z &=3 \\ x+3 y+z &=5 \\ 3 x+8 y+4 z &=17 \end{aligned}\) b. \(\begin{aligned} x-2 y+4 z &=2 \\ 2 x-3 y+5 z &=3 \\ 3 x-4 y+6 z &=7 \end{aligned}\) c. \(x+y+3 z=1$$2 x+3 y-z=3$$5 x+7 y+z=7\)

Let \(V\) denote the set of all solutions to the system of linear equations $$ \begin{array}{r} x_{1}-x_{2}+2 x_{4}-3 x_{5}+x_{6}=0 \\ 2 x_{1}-x_{2}-x_{3}+3 x_{4}-4 x_{5}+4 x_{6}=0 . \end{array} $$ (a) Show that \(S=\\{(0,-1,0,1,1,0),(1,0,1,1,1,0)\\}\) is a linearly independent subset of V. (b) Extend \(S\) to a basis for \(\mathrm{V}\).

Suppose that the augmented matrix of a system \(A x=b\) is transformed into a matrix \(\left(A^{\prime} \mid b^{\prime}\right)\) in reduced row echelon form by a finite sequence of elementary row operations. (a) Prove that rank $\left(A^{\prime}\right) \neq \operatorname{rank}\left(A^{\prime} \mid b^{\prime}\right)$ if and only if \(\left(A^{\prime} \mid b^{\prime}\right)\) contains a row in which the only nonzero entry lies in the last column. (b) Deduce that \(A x=b\) is consistent if and only if $\left(A^{\prime} \mid b^{\prime}\right)$ contains no row in which the only nonzero entry lies in the last column.
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