Answer: The technical-coefficient matrix \(A\) has entries \(a_{ij}=\dfrac{z_{ij}}{x_j}\). In matrix form \(A = Z \,\hat{x}^{-1}\), where \(Z\) is the matrix of interindustry flows and \(\hat{x}\) is the diagonal matrix with total outputs \(x_j\) on the diagonal.
Explanation: Each \(a_{ij}\) is the amount of input from sector \(i\) required per unit of output of sector \(j\) (i.e., the fraction of sector \(j\)’s output that goes to sector \(i\)). To get \(x_j\) you usually sum the column \(j\) of \(Z\) and add final demand if your table separates it: \(x_j=\sum_i z_{ij}+f_j\). Use column-wise division of \(Z\) by \(x\).
Steps:
- Compute total output for each sector: \(x_j=\sum_i z_{ij}\) (or \(x_j=\sum_i z_{ij}+f_j\) if final demand \(f\) is given).
- Form the diagonal matrix \(\hat{x}=\mathrm{diag}(x_1,\dots,x_n)\).
- Compute \(A = Z\,\hat{x}^{-1}\) (equivalently, divide each column \(j\) of \(Z\) by \(x_j\)).
Example:
- Let \(Z=\begin{pmatrix}10 & 5\\[4pt]20 & 10\end{pmatrix}\). Column sums: \(x_1=30,\; x_2=15\).
- \(A=\begin{pmatrix}10/30 & 5/15\\[4pt]20/30 & 10/15\end{pmatrix}=\begin{pmatrix}0.3333 & 0.3333\\[4pt]0.6667 & 0.6667\end{pmatrix}.\)
Note: This \(A\) is used in Leontief analysis; the total output response to final demand is \(\mathbf{x}=(I-A)^{-1}\mathbf{f}\) when appropriate.