(a) Technology matrix from $a_{ij}=x_{ij}/X_j$; solve $(I-B)X=D$ with $D=(23,3)$: $X_1\approx34.16$, $X_2\approx17.31$. (b) $\dfrac{Eq}{Ep_1}=\dfrac{4}{3}$, $\dfrac{Eq}{Ep_2}=8$.

Part (a) — Input–output (Leontief) analysis

Step 1 — form the technology matrix $B$ using $a_{ij}=\dfrac{x_{ij}}{X_j}$ (each input divided by the consuming sector's gross output $X_1=20,\ X_2=12$):

$a_{11}=\frac{4}{20}=0.2,\quad a_{12}=\frac{3}{12}=0.25,\quad a_{21}=\frac{5}{20}=0.25,\quad a_{22}=\frac{4}{12}=\frac13\approx0.333.$

$B=\begin{bmatrix}0.2 & 0.25\\0.25 & 0.333\end{bmatrix}.$

(Check: $BX+D$ reproduces the original outputs, e.g. sector 1: $0.2(20)+0.25(12)+13=4+3+13=20$ ✓.)

Step 2 — set up $X=(I-B)^{-1}D$ with new demand $D=\begin{bmatrix}23\\3\end{bmatrix}$.

$I-B=\begin{bmatrix}\tfrac45 & -\tfrac14\\[2pt]-\tfrac14 & \tfrac23\end{bmatrix}.$

Step 3 — solve $(I-B)X=D$. Writing the equations and clearing fractions:

$16X_1-5X_2=460,\qquad -3X_1+8X_2=36.$

Eliminating $X_2$: $128X_1-40X_2=3680$ and $-15X_1+40X_2=180$; adding,

$113X_1=3860\;\Rightarrow\;X_1=\frac{3860}{113}\approx34.16.$

Then $8X_2=36+3X_1=36+102.48=138.48\Rightarrow X_2=\dfrac{1956}{113}\approx17.31.$

Part (a) result: to meet a final demand of $23$ (sector 1) and $3$ (sector 2), gross outputs are $X_1\approx34.16$ units and $X_2\approx17.31$ units.

Part (b) — Partial elasticities of demand

Given: $q=13-2p_1-3p_2^2$.

Step 1 — partial derivatives.

$\frac{\partial q}{\partial p_1}=-2,\qquad \frac{\partial q}{\partial p_2}=-6p_2.$

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