91Ó°ÊÓ

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, and the production vector \(x = {\left( {I - C} \right)^{ - 1}}{\mathop{\rm d}\nolimits} \) has non-negative entries and is the unique solution of \(x = Cx + {\mathop{\rm d}\nolimits} \).

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

The production level needed to satisfy a final demand of 18 units for agriculture and no demand for the other sectors.

For d, solve the equation \(x = Cx + {\mathop{\rm d}\nolimits} \).

\[\begin{array}{c}{\mathop{\rm d}\nolimits} = x - Cx\\\left[ {\begin{array}{*{20}{c}}0\\{18}\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{.10}&{.60}&{.60}\\{.30}&{.20}&0\\{.30}&{.10}&{.10}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}0\\{18}\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{.10{x_1}}&{.60{x_2}}&{.60{x_3}}\\{.30{x_1}}&{.20{x_2}}&0\\{.30{x_1}}&{.10{x_2}}&{.10{x_3}}\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}0\\{18}\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{.9{x_1}}&{ - .60{x_2}}&{ - .60{x_3}}\\{ - .30{x_1}}&{.8{x_2}}&0\\{ - .30{x_1}}&{.1{x_2}}&{.9{x_3}}\end{array}} \right]\end{array}\]

Write the matrix in the system of equations, as shown below:

\(\begin{array}{c}.9{x_1} - .6{x_2} - .6{x_3} = 0\\ - .3{x_1} - .8{x_2} = 18\\ - .3{x_1} - .1{x_2} + .9{x_3} = 0\end{array}\)

The augmented matrix of the system of the equation is

\(\left[ {\begin{array}{*{20}{c}}{.90}&{ - .60}&{ - .60}&0\\{ - .30}&{.80}&{.00}&{18}\\{ - .30}&{ - .10}&{.90}&0\end{array}} \right]\).

At row one, multiply row one by \( - \frac{1}{{0.90}}\).

\( \sim \left[ {\begin{array}{*{20}{c}}1&{ - .666}&{ - .666}&0\\{ - .30}&{.80}&{.00}&{18}\\{ - .30}&{ - .10}&{.90}&0\end{array}} \right]\)

At row two, multiply row one by 0.3 and add it to row two. At row three, multiply row one by 0.30 and add it to row three.

\( \sim \left[ {\begin{array}{*{20}{c}}1&{ - .6666}&{ - .6666}&0\\0&{0.6}&{ - 0.20}&{18}\\0&{ - 0.3}&{0.7}&0\end{array}} \right]\)

At row two, multiply row two by \(\frac{1}{{0.6}}\).

\( \sim \left[ {\begin{array}{*{20}{c}}1&{ - .666}&{ - .666}&0\\0&1&{ - 0.333}&{30}\\0&{ - 0.3}&{0.7}&0\end{array}} \right]\)

At row one, multiply row two by 0.666 and add it to row one. At row three, multiply row two by 0.3 and add it to row three.

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&{ - 0.888}&{20}\\0&1&{ - 0.33}&{30}\\0&0&{0.6}&9\end{array}} \right]\)

At row two, multiply row three by 0.333 and add it to row two.

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&0&{33.33}\\0&1&0&{35}\\0&0&1&{15}\end{array}} \right]\)

Thus, \(x = \left( {33.33,35,15} \right)\).