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Exercise E · Q2
Q.

Solve the following problem using Leontief input-output model.

FIAITotal
Food industry201040
Agricultural industry302060

If the system is viable then discuss the situation for new demand 80 and 120 from FI and AI respectively.

Arunachal CbseNCERTSubjective· 5mImportance★★★★★est
73% · 58/80 Questions
✓ Free question

Form the technology matrix, confirm viability, then compute X=(I−A)−1DX=(I-A)^{-1}D for the new final demand (80,120)(80,120).

aij=xijXja_{ij}=\dfrac{x_{ij}}{X_j}; gross output X=(I−A)−1DX=(I-A)^{-1}D; viable when the leading principal minors of (I−A)(I-A) are positive.

Totals XFI=40,  XAI=60X_{FI}=40,\;X_{AI}=60:

From \ ToFIAITotal
Food (FI)201040
Agri (AI)302060
  1. Coefficients: a11=2040=12,  a21=3040=34,  a12=1060=16,  a22=2060=13.a_{11}=\dfrac{20}{40}=\dfrac12,\;a_{21}=\dfrac{30}{40}=\dfrac34,\;a_{12}=\dfrac{10}{60}=\dfrac16,\;a_{22}=\dfrac{20}{60}=\dfrac13.

A=[1/21/63/41/3].A=\begin{bmatrix}1/2&1/6\\3/4&1/3\end{bmatrix}.

  1. I−A=[1/2−1/6−3/42/3].I-A=\begin{bmatrix}1/2&-1/6\\-3/4&2/3\end{bmatrix}.
  2. Viability: 12>0\tfrac12>0 and det⁡(I−A)=12⋅23−16⋅34=13−18=8−324=524>0⇒\det(I-A)=\dfrac12\cdot\dfrac23-\dfrac16\cdot\dfrac34=\dfrac13-\dfrac18=\dfrac{8-3}{24}=\dfrac{5}{24}>0\Rightarrow viable.
  3. (I−A)−1=15/24[2/31/63/41/2]=245[2/31/63/41/2].(I-A)^{-1}=\dfrac{1}{5/24}\begin{bmatrix}2/3&1/6\\3/4&1/2\end{bmatrix}=\dfrac{24}{5}\begin{bmatrix}2/3&1/6\\3/4&1/2\end{bmatrix}.
  4. D=[80120]D=\begin{bmatrix}80\\120\end{bmatrix}:

XFI=245(23⋅80+16⋅120)=245(1603+20)=245⋅2203=528015=352,X_{FI}=\frac{24}{5}\left(\tfrac23\cdot80+\tfrac16\cdot120\right)=\frac{24}{5}\left(\tfrac{160}{3}+20\right)=\frac{24}{5}\cdot\frac{220}{3}=\frac{5280}{15}=352,

XAI=245(34⋅80+12⋅120)=245(60+60)=245⋅120=576.X_{AI}=\frac{24}{5}\left(\tfrac34\cdot80+\tfrac12\cdot120\right)=\frac{24}{5}(60+60)=\frac{24}{5}\cdot120=576.

  1. Both positive ⇒\Rightarrow demand (80,120)(80,120) is feasible.
✓Final answer

Viable (det⁡(I−A)=5/24>0\det(I-A)=5/24>0); required gross outputs XFI=352X_{FI}=352 and XAI=576X_{AI}=576 units.

Note

The official CBSE book's answer key (Exercise E, Q2) prints this system as not viable, reporting det⁡(I−A)=−312<0\det(I-A)=-\tfrac{3}{12}<0. That key computes a21=3060a_{21}=\tfrac{30}{60} (dividing the agricultural-industry flow 3030 by the wrong total, 6060) instead of the correct a21=3040=34a_{21}=\tfrac{30}{40}=\tfrac34 (the flow 3030 is consumed by the food industry, whose output is 4040). Using the standard column convention — the same one that makes the neighbouring parts Q3 (det⁡=−160\det=-\tfrac1{60}) and Q4 (det⁡=1990\det=\tfrac{19}{90}) match the book's key exactly — gives det⁡(I−A)=524>0\det(I-A)=\tfrac{5}{24}>0, so the system is genuinely viable. Our outputs XFI=352, XAI=576X_{FI}=352,\ X_{AI}=576 satisfy AX+D=XAX+D=X exactly, confirming this.

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