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Exercise E · Q1
Q.

Solve the following problem using Leontief input-output model.

Sector 1Sector 2Total
Sector 12510
Sector 23420

If the system is viable then discuss the situation for new demand 8 and 12 from sector 1 and sector 2 respectively.

Andaman Nicobar CbseNCERTSubjective· 5mImportance★★★★★est
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✓ Free question

Build the technology matrix, verify viability via the Hawkins–Simon conditions, then solve X=(I−A)−1DX=(I-A)^{-1}D for the new demand.

Technology (input) coefficient aij=xijXja_{ij}=\dfrac{x_{ij}}{X_j} (input from sector ii per unit output of sector jj). Gross output X=(I−A)−1DX=(I-A)^{-1}D, where DD is final demand. Viability (Hawkins–Simon): the leading principal minors of (I−A)(I-A) are all positive.

Given inter-industry flows and totals X1=10,  X2=20X_1=10,\;X_2=20:

From \ ToS1S2Total
S12510
S23420
  1. Coefficients: a11=210=0.2,  a21=310=0.3,  a12=520=0.25,  a22=420=0.2.a_{11}=\dfrac{2}{10}=0.2,\;a_{21}=\dfrac{3}{10}=0.3,\;a_{12}=\dfrac{5}{20}=0.25,\;a_{22}=\dfrac{4}{20}=0.2.

A=[0.20.250.30.2].A=\begin{bmatrix}0.2&0.25\\0.3&0.2\end{bmatrix}.

  1. I−A=[0.8−0.25−0.30.8].I-A=\begin{bmatrix}0.8&-0.25\\-0.3&0.8\end{bmatrix}.
  2. Viability: 0.8>00.8>0 and det⁡(I−A)=0.8(0.8)−(−0.25)(−0.3)=0.64−0.075=0.565>0.\det(I-A)=0.8(0.8)-(-0.25)(-0.3)=0.64-0.075=0.565>0. Both leading minors positive ⇒\Rightarrow viable.
  3. (I−A)−1=10.565[0.80.250.30.8].(I-A)^{-1}=\dfrac{1}{0.565}\begin{bmatrix}0.8&0.25\\0.3&0.8\end{bmatrix}.
  4. New demand D=[812]D=\begin{bmatrix}8\\12\end{bmatrix}:

X1=0.8(8)+0.25(12)0.565=6.4+30.565=9.40.565≈16.64,X_1=\frac{0.8(8)+0.25(12)}{0.565}=\frac{6.4+3}{0.565}=\frac{9.4}{0.565}\approx16.64,

X2=0.3(8)+0.8(12)0.565=2.4+9.60.565=120.565≈21.24.X_2=\frac{0.3(8)+0.8(12)}{0.565}=\frac{2.4+9.6}{0.565}=\frac{12}{0.565}\approx21.24.

  1. Both outputs are positive, so the demand (8,12)(8,12) can be met.
✓Final answer

Viable (det⁡(I−A)=0.565>0\det(I-A)=0.565>0); required gross outputs X1≈16.64X_1\approx16.64 and X2≈21.24X_2\approx21.24 units.

Note

The book's answer key (Exercise E, Q1) writes det⁡(I−A)=1625−340=109200\det(I-A)=\tfrac{16}{25}-\tfrac{3}{40}=\tfrac{109}{200}; the reduction is a typo — 1625−340=128−15200=113200=0.565\tfrac{16}{25}-\tfrac{3}{40}=\tfrac{128-15}{200}=\tfrac{113}{200}=0.565. The verdict (viable) is unaffected.

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