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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.

Dnh Dd 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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