Let C be an \(n \times n\) consumption matrix whose column sums are less than 1. Let x be the production vector that satisfies a final demand d, and let \[\Delta {\mathop{\rm x}\nolimits} \] be a production vector that satisfies a different final demand \[\Delta {\mathop{\rm d}\nolimits} \].

a. Show that if the final demand changes from d to \[{\mathop{\rm d}\nolimits} + \Delta {\mathop{\rm d}\nolimits} \], then the new production level must be \[{\mathop{\rm x}\nolimits} + \Delta {\mathop{\rm x}\nolimits} \]. Thus \[\Delta {\mathop{\rm x}\nolimits} \] gives the amounts by which production must change in order to accommodate the change \[\Delta {\mathop{\rm d}\nolimits} \] in demand.

b. Let \[\Delta {\mathop{\rm d}\nolimits} \] be the vector in \({\mathbb{R}^n}\) with as the first entry and 0’s elsewhere. Explain why the corresponding production \[\Delta {\mathop{\rm x}\nolimits} \] is the first column of \({\left( {I - C} \right)^{ - 1}}\). This shows that the first column of \({\left( {I - C} \right)^{ - 1}}\) gives the amounts the various sectors must produce to satisfy an increase of 1 unit in the final demand for output from sector 1.

Let x be a production vector that satisfies the final demand d, and \[\Delta {\mathop{\rm x}\nolimits} \] be a production vector that satisfies a different final demand \(\Delta {\mathop{\rm d}\nolimits} \).

It is given that \(\left( {I - C} \right){\mathop{\rm x}\nolimits} = {\mathop{\rm d}\nolimits} \) and \(\left( {I - C} \right)\Delta {\mathop{\rm x}\nolimits} = \Delta {\mathop{\rm d}\nolimits} \).

The new production level of \({\mathop{\rm x}\nolimits} + \Delta {\mathop{\rm x}\nolimits} \) is shown below:

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

Thus, it is proved that \({\mathop{\rm x}\nolimits} + \Delta {\mathop{\rm x}\nolimits} \) is the new production level that corresponds to a final demand of \({\mathop{\rm d}\nolimits} + \Delta {\mathop{\rm d}\nolimits} \).

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

Let \[\Delta {\mathop{\rm d}\nolimits} \] be the vector in \({\mathbb{R}^n}\) as the first entry and as zero elsewhere.

\[\Delta {\mathop{\rm d}\nolimits} \]is the first column of I. So, the production vector \[\Delta {\mathop{\rm x}\nolimits} \] is the first column of \({\left( {I - C} \right)^{ - 1}}\) because \(\Delta {\mathop{\rm x}\nolimits} = {\left( {I - C} \right)^{ - 1}}\Delta {\mathop{\rm d}\nolimits} \).

Thus, it is proved that the production vector \[\Delta {\mathop{\rm x}\nolimits} \] is the first column of \({\left( {I - C} \right)^{ - 1}}\).