Solve the Leontief production equation for an economy with three sectors, given that

\(C = \left[ {\begin{array}{*{20}{c}}{.2}&{.2}&{.0}\\{.3}&{.1}&{.3}\\{.1}&{.0}&{.2}\end{array}} \right]\)and \({\mathop{\rm d}\nolimits} = \left[ {\begin{array}{*{20}{c}}{40}\\{60}\\{80}\end{array}} \right]\).

The Leontief input-output model or production equationis \(x = Cx + {\mathop{\rm d}\nolimits} \). The equation \(x = Cx + {\mathop{\rm d}\nolimits} \) can be written as \(I{\mathop{\rm x}\nolimits} - C{\mathop{\rm x}\nolimits} = {\mathop{\rm d}\nolimits} \) or \(\left( {I - C} \right){\mathop{\rm x}\nolimits} = {\mathop{\rm d}\nolimits} \).

\(\begin{array}{c}I - C = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{.2}&{.2}&{.0}\\{.3}&{.1}&{.3}\\{.1}&{.0}&{.2}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{.8}&{ - .2}&{.0}\\{ - .3}&{.9}&{ - .3}\\{ - .1}&{.0}&{.8}\end{array}} \right]\end{array}\)

It is given that \({\mathop{\rm d}\nolimits} = \left[ {\begin{array}{*{20}{c}}{40}\\{60}\\{80}\end{array}} \right]\).

The augmented matrix \(\left[ {\begin{array}{*{20}{c}}{I - C}&{\mathop{\rm d}\nolimits} \end{array}} \right]\) is shown below:

\(\left[ {\begin{array}{*{20}{c}}{.8}&{ - .2}&{.0}&{40.0}\\{ - .3}&{.9}&{ - .3}&{60.0}\\{ - .1}&{.0}&{.8}&{80.0}\end{array}} \right]\)

At row one, multiply row one by \(\frac{1}{{0.8}}\).

\( \sim \left[ {\begin{array}{*{20}{c}}1&{ - 0.25}&0&{50}\\{ - .3}&{.9}&{ - .3}&{60.0}\\{ - .1}&{.0}&{.8}&{80.0}\end{array}} \right]\)

At row two, multiply row one by 0.3 and add it to row two. At row three, multiply row one by 0.1 and add it to row three.

\( \sim \left[ {\begin{array}{*{20}{c}}1&{ - 0.25}&0&{50}\\0&{0.825}&{ - .3}&{75}\\0&{ - .025}&{.8}&{85}\end{array}} \right]\)

At row two, multiply row two by \(\frac{1}{{0.825}}\).

\( \sim \left[ {\begin{array}{*{20}{c}}1&{ - 0.25}&0&{50}\\0&1&{ - .3636}&{90.90}\\0&{ - .025}&{.8}&{85}\end{array}} \right]\)

At row one, multiply row two by 0.25 and add it to row one. At row three, multiply row two by 0.025 and add it to row three.

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&{ - 0.0909}&{72.727}\\0&1&{ - 0.3636}&{90.90}\\0&0&{0.7909}&{87.272}\end{array}} \right]\)

At row three, multiply row three by 0.79090.

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&{ - 0.0909}&{72.727}\\0&1&{ - 0.3636}&{90.90}\\0&0&1&{110.3448}\end{array}} \right]\)

At row one, multiply row three by 0.09090 and add it to row one.

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&0&{82.7586}\\0&1&{ - 0.3636}&{90.90}\\0&0&1&{110.3448}\end{array}} \right]\)

At row two, multiply row three by 0.3636 and add it to row two.

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&0&{82.7586}\\0&1&0&{131.03}\\0&0&1&{110.3448}\end{array}} \right]\)

Thus, \({\mathop{\rm x}\nolimits} = \left( {82.8,131.0,110.3} \right)\).