Question

For some background on this exercise, see exercise 1.1.24. Consider an economy with three industries, I₁, I₂, I₃. What outputs x₁, x₂, x₃ should they posses to satisfy both consumer demand and interindustry demand? The demand put on the three industries are shown in the accompanying figure.

Short answer

Thus, to satisfy both the consumer and interindustry demand the output should be ${\mathrm{x}}_1=500$,${\mathrm{x}}_2=300$ and ${\mathrm{x}}_3=400$.

Step-by-step solution

Concept introduction

Two vectors are perpendicular to each other if their dot product is zero,

Let us suppose that $\overrightarrow{\mathbf{x}}$and $\overrightarrow{\mathbf{y}}$are two perpendicular then these two vectors are said to be perpendicular, if$\overrightarrow{\mathbf{x}}\mathbf{.}\overrightarrow{\mathbf{y}}\mathbf{=}\mathbf{0}$

Find the output so that they satisfy both consumer demand and interindustry demand.

The system of the equation from the given diagram is,

$$\begin{gathered} {\mathrm{x}}_1=0.2{\mathrm{x}}_2+0.3{\mathrm{x}}_3+320 \\ {\mathrm{x}}_2=0.1{\mathrm{x}}_1+0.4{\mathrm{x}}_3+90 \\ {\mathrm{x}}_3=0.2{\mathrm{x}}_1+0.5_2+150 \end{gathered}$$

We can write above system of equation in matrix form,

$\left|\begin{array}{c}{\mathrm{x}}_1-0.2{\mathrm{x}}_2-0.3{\mathrm{x}}_3=320\\ -0.1{\mathrm{x}}_1+{\mathrm{x}}_2-0.4{\mathrm{x}}_3=90\\ -0.2{\mathrm{x}}_1-0.5{\mathrm{x}}_2+{\mathrm{x}}_3=150\end{array}\right|$

Augmented matrix is,

$\left[\begin{array}{cccc}1& -0.2& -0.3& |320\\ -0.1& 1& -0.4& |90\\ -0.2& -0.5& 1& |150\end{array}\right]$

Use operation${\mathrm{R}}_2\to 10{\mathrm{R}}_2$ and${\mathrm{R}}_3\to 10{\mathrm{R}}_3$ in the above equation,

$\left[\begin{array}{cccc}1& -0.2& -0.3& |320\\ -1& 10& -4& |900\\ 2& -5& 10& |1500\end{array}\right]$

Use operation${\mathrm{R}}_2\to {\mathrm{R}}_2+{\mathrm{R}}_1$ and${\mathrm{R}}_3\to {\mathrm{R}}_3+2{\mathrm{R}}_1$

$\left[\begin{array}{cccc}1& -0.2& -0.3& |320\\ 0& 9.8& -4.3& |1220\\ 0& -5.4& 9.4& |2140\end{array}\right]$

Use operation,${\mathrm{R}}_3\to 9.8{\mathrm{R}}_2+5.4{\mathrm{R}}_2$

$\left[\begin{array}{cccc}1& -0.2& -0.3& |320\\ 0& 9.8& -4.3& |1220\\ 0& 0& 68.9& |27560\end{array}\right]$

Now write above matrix in terms of the equation,

$\begin{array}{rcl}{\mathrm{x}}_1-0.2{\mathrm{x}}_2-0.3{\mathrm{x}}_3& =& 320\\ 9.8{\mathrm{x}}_2-4.3{\mathrm{x}}_3& =& 1220\\ 68.9{\mathrm{x}}_3& =& 27560\end{array}$

On solving the above three equation,

${\mathrm{x}}_1=500,{\mathrm{x}}_2=300$, and${\mathrm{x}}_3=400$

Therefore, to satisfy both the consumer and interindustry demand the output should be ${\mathrm{x}}_1=500,{\mathrm{x}}_2=300$, and ${\mathrm{x}}_3=400$.