From Final Demand to Gross Output
A production plan must supply both the goods requested by final users and the goods that sectors use as inputs to produce their own output. The input-output model collects those requirements in a matrix equation. Solving that equation converts a specified final-demand vector into the gross output required from each sector.
Suppose there are \(n\) sectors. Let \(x_j\) be the gross output of sector \(j\), and let \(d_i\) be the final demand for sector \(i\). The technical-coefficient matrix \(\mathbf{A}=(a_{ij})\in\mathbb{R}^{n\times n}\) records input requirements: \(a_{ij}\) is the amount of output from sector \(i\) needed to produce one unit of output from sector \(j\). Write the gross-output and final-demand vectors as columns, \(\mathbf{x}=(x_1,\dots,x_n)^{\top}\) and \(\mathbf{d}=(d_1,\dots,d_n)^{\top}\). Then \(\mathbf{A}\mathbf{x}\) is the vector of intermediate input requirements.
The rearrangement follows by subtracting \(\mathbf{A}\mathbf{x}\) from both sides of the balance equation. The matrix \(\mathbf{I}_n-\mathbf{A}\) is square, so its inverse exists exactly when it is nonsingular. If it is invertible, the Leontief inverse gives a direct formula. If it is not, the system must be examined as a linear system: a specified \(\mathbf{d}\) may give no solution or may fail to determine a unique one.
Proof. Set \(\mathbf{M}=\mathbf{I}_n-\mathbf{A}\). Since \(\mathbf{M}\mathbf{M}^{-1}=\mathbf{I}_n\), the proposed vector satisfies \(\mathbf{M}\mathbf{x}=\mathbf{M}\mathbf{M}^{-1}\mathbf{d}=\mathbf{d}\), so it is a solution. If \(\mathbf{y}\) is any solution, then \(\mathbf{M}\mathbf{y}=\mathbf{d}\). Multiplying both sides by \(\mathbf{M}^{-1}\) gives \(\mathbf{y}=\mathbf{M}^{-1}\mathbf{d}=\mathbf{x}\). Thus a solution exists and every solution equals \(\mathbf{x}\). \(\square\)
Computing with the Leontief Inverse
One way to calculate the inverse is to row-reduce the augmented matrix \([\mathbf{M}\mid\mathbf{I}_n]\), where \(\mathbf{M}=\mathbf{I}_n-\mathbf{A}\). If the left block reduces to \(\mathbf{I}_n\), the right block becomes \(\mathbf{M}^{-1}\). For a particular demand vector, it is often more direct to solve the augmented system \([\mathbf{M}\mid\mathbf{d}]\) and calculate only the needed output vector.
Worked Example: Two Sectors and an Invertible Leontief Matrix
Suppose
First form \(\mathbf{M}=\mathbf{I}_2-\mathbf{A}\), then calculate its inverse by row reduction:
Scale the second row by \(\frac65\), then add half of the second row to the first:
Thus \(\mathbf{M}^{-1}=\begin{bmatrix}\frac65&\frac35\\\frac25&\frac65\end{bmatrix}\). Apply it to the demand vector:
Check in the original balance: \(\mathbf{A}\mathbf{x}=\begin{bmatrix}\frac12(18)\\\frac13(12)\end{bmatrix}=\begin{bmatrix}9\\4\end{bmatrix}\), so \(\mathbf{A}\mathbf{x}+\mathbf{d}=\begin{bmatrix}9\\4\end{bmatrix}+\begin{bmatrix}3\\14\end{bmatrix}=\begin{bmatrix}12\\18\end{bmatrix}=\mathbf{x}\). The calculated vector satisfies the production balance.
Solving the System Directly by Row Reduction
The formula using an inverse is useful when the inverse itself is required, but an individual production plan can also be found by solving \(\mathbf{M}\mathbf{x}=\mathbf{d}\) directly. The augmented matrix \([\mathbf{M}\mid\mathbf{d}]\) encodes the system: each row represents one equation, with the final column holding its right-hand side.
Proof. A row swap only changes the order of the equations, so it does not change which vectors satisfy all of them. Multiplying one equation by a nonzero scalar preserves its solutions because the equation can be recovered by dividing by that scalar. Replacing one equation by itself plus a scalar multiple of another also preserves the solutions: any vector satisfying the original equations satisfies the new equation, and the original equation is recovered by subtracting the same multiple of the unchanged equation. Each elementary row operation therefore has a reversible effect on the equations, so a sequence of such operations leaves the solution set unchanged. \(\square\)
Worked Example: Three Sectors by Row Reduction
Consider the technical-coefficient matrix and final-demand vector
Subtract \(\mathbf{A}\) from \(\mathbf{I}_3\) and augment by \(\mathbf{d}\). Scale the first, second and third rows by \(\frac43\), \(\frac32\) and \(2\), respectively:
Now use the first row to clear the entry in row 2, column 1, and use the updated second row to clear the entry in row 3, column 2:
The solution is \(\mathbf{x}=\begin{bmatrix}8&12&20\end{bmatrix}^{\top}\). Check: multiplying by the original technical-coefficient matrix gives \(\mathbf{A}\mathbf{x}=\begin{bmatrix}2\\2+4\\4+10\end{bmatrix}=\begin{bmatrix}2\\6\\14\end{bmatrix}\). Therefore \(\mathbf{A}\mathbf{x}+\mathbf{d}=\begin{bmatrix}2\\6\\14\end{bmatrix}+\begin{bmatrix}6\\6\\6\end{bmatrix}=\begin{bmatrix}8\\12\\20\end{bmatrix}=\mathbf{x}\).
What to Check When No Inverse Exists
A row reduction that fails to produce a pivot in every column does not give an inverse. It is not valid to apply an inverse formula in that case. The augmented system still tells us whether the specified demand has a solution: a row that asserts \(0\) equals a nonzero number is a contradiction. Even when a solution exists, a free variable means that the equations do not determine a unique output vector.
Worked Example: A Singular System with No Production Plan
Let
Then \(\mathbf{M}=\mathbf{I}_2-\mathbf{A}=\begin{bmatrix}\frac12&-\frac12\\-\frac12&\frac12\end{bmatrix}\). Its rows are negatives of each other, so it is not invertible. Row-reducing the augmented system gives
The last row represents \(0=1\), so there is no solution to \(\mathbf{M}\mathbf{x}=\mathbf{d}\). Consequently, there is no vector satisfying the original production balance for this specified demand. This example shows why the existence of a Leontief inverse must be checked rather than assumed.
Common Pitfall: Transposing the Input Coefficients
The convention \(a_{ij}=\) input from sector \(i\) per unit of sector \(j\)'s output places the supplying sector in row \(i\) and the producing sector in column \(j\). Thus the intermediate requirements are \(\mathbf{A}\mathbf{x}\), not \(\mathbf{A}^{\top}\mathbf{x}\). A mistaken transpose can produce a vector that solves a different system.
For the two-sector example, the correct output was \(\mathbf{x}=\begin{bmatrix}12&18\end{bmatrix}^{\top}\). Using the transpose instead gives \(\mathbf{A}^{\top}\mathbf{x}=\begin{bmatrix}6\\6\end{bmatrix}\), and then \(\mathbf{A}^{\top}\mathbf{x}+\mathbf{d}=\begin{bmatrix}9\\20\end{bmatrix}\), which is not \(\mathbf{x}\). The correction is to use the stated column convention: compute \(\mathbf{A}\mathbf{x}=\begin{bmatrix}9\\4\end{bmatrix}\), and check that this plus \(\mathbf{d}\) equals \(\mathbf{x}\).
A final check should always return to the original balance, not just the rearranged system used in the calculation. For a proposed output vector, calculate \(\mathbf{A}\mathbf{x}\), add \(\mathbf{d}\), and compare the result with \(\mathbf{x}\). This checks the model equation directly and can reveal a sign error, a transposed matrix, or an arithmetic mistake.
Check Your Understanding
For each problem, state the equation you would solve and use the original production balance to check any computed output.
- If \(\mathbf{A}=\begin{bmatrix}0&\frac12\\\frac13&0\end{bmatrix}\) and \(\mathbf{d}=\begin{bmatrix}6\\10\end{bmatrix}\), find \(\mathbf{x}\) using the Leontief inverse.
- For the same \(\mathbf{A}\), form \(\mathbf{I}_2-\mathbf{A}\) and explain why its inverse, if computed, must be applied to a column demand vector on the left.
- What does a row \(\begin{bmatrix}0&0&|&5\end{bmatrix}\) in a reduced augmented system tell you about the production plan?
- Explain why an invertible \(\mathbf{I}_n-\mathbf{A}\) guarantees exactly one solution for every specified \(\mathbf{d}\).
- For the three-sector example, calculate \(\mathbf{A}\mathbf{x}+\mathbf{d}\) once more and identify what each component represents in the balance.
