Answer to Question #303982 in Linear Algebra for Daniella

Question

A company produces three products which are interdependent. These are A, B and C. The flow of inputs and outputs between the products is represented in the table below:

Inputs (in thousands of units)

A B C Final demand

Outputs

(in thousands of units) A

B

C 40

60

80 65

130

65 75

75

25 20

60

80

Required:

i) Derive the technical coefficients matrix (3 marks)

ii) Determine the Leontief’s inverse matrix (12 marks)

iii) Compute the output level for each product if the final demand for product A increased by 2000 units, that of product C decreased by 1,000 units and the final demand for product B remained unchanged. (5 marks)

Answer 1

i) The technical coefficients matrix

The technical coefficient matrix refers to the matrix of B.

B= $\begin{bmatrix} b11 & b12 & b13 \\ b21 & b22 & b23 \\ b31 & b32 & b33 \\ \end{bmatrix}$

where,

b11 = 40/200 = 0.2

b12= 65/325= 0.2

b13= 75/250 = 0.3

b21=60/200 = 0.3

b22=130/325 = 0.4

b23=75/250 = 0.3

b31=80/200= 0.4

b32=65/325 = 0.2

b33= 25/250 = 0.1

hence,

B= $\begin{bmatrix} 0.2 & 0.2 & 0.3 \\ 0.3 & 0.4 & 0.3 \\ 0.4 & 0.2 & 0.1 \\ \end{bmatrix}$ [Answer]

ii) The Leontief’s inverse matrix

Leontief's inverse matrix is given by (I-B)^-1

I-B = $\begin{bmatrix} 1 &0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 &1 \\ \end{bmatrix} - \begin{bmatrix} 0.2 & 0.2 & 0.3 \\ 0.3 & 0.4 & 0.3 \\ 0.4 & 0.2 & 0.1 \\ \end{bmatrix}=\begin{bmatrix} 0.8 & -0.2 & -0.3 \\ -0.3 & 0.6 & -0.3 \\ -0.4 & -0.2 & 0.9 \\ \end{bmatrix}$

(I-B)^-1

$\begin{bmatrix} 0.8 & -0.2 & -0.3 \\ -0.3 & 0.6 & -0.3 \\ -0.4 & -0.2 & 0.9 \\ \end{bmatrix}|\begin{bmatrix} 1 &0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 &1 \\ \end{bmatrix}$ Turn the left matrix into the right matrix

R1 ▶ ( $\frac{5}{4}$ )R1

$\begin{bmatrix} 1 & \frac{-1}{4} & \frac{-3}{8} \\ \frac{-3}{10} & \frac{3}{5} & \frac{-3}{10} \\ \frac{-2}{5} &\frac{-1}{5} & \frac{9}{10} \\ \end{bmatrix}|\begin{bmatrix} \frac{5}{4} &0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 &1 \\ \end{bmatrix}$

R2 ▶R2- ( $\frac{-3}{10}$ )R1

$\begin{bmatrix} 1 & \frac{-1}{4} & \frac{-3}{8} \\ 0 & \frac{21}{40} & \frac{-33}{80} \\ \frac{-2}{5} &\frac{-1}{5} & \frac{9}{10} \\ \end{bmatrix}|\begin{bmatrix} \frac{5}{4} &0 & 0 \\ \frac{3}{8} & 1 & 0 \\ 0 & 0 &1 \\ \end{bmatrix}$

R3 ▶ R3 -$(\displaystyle - \frac{2}{5} −​5 ​ ​2 ​​)R1$

$\begin{bmatrix} 1 & \frac{-1}{4} & \frac{-3}{8} \\ 0 & \frac{21}{40} & \frac{-33}{80} \\ 0 &\frac{-3}{10} & \frac{3}{4} \\ \end{bmatrix}|\begin{bmatrix} \frac{5}{4} &0 & 0 \\ \frac{3}{8} & 1 & 0 \\ \frac{1}{2} & 0 &1 \\ \end{bmatrix}$

R2 ▶ $(\displaystyle 1.90476)R2$

$\begin{bmatrix} 1 & \frac{-1}{4} & \frac{-3}{8} \\ 0 & 1 & \frac{-11}{14} \\ 0 &\frac{-3}{10} & \frac{3}{4} \\ \end{bmatrix}|\begin{bmatrix} \frac{5}{4} &0 & 0 \\ \frac{3}{8} & \frac{40}{21} & 0 \\ \frac{1}{2} & 0 &1 \\ \end{bmatrix}$

R3 ▶ R3 - $(\displaystyle - \frac{3}{10}​​)R2$

$\begin{bmatrix} 1 & \frac{-1}{4} & \frac{-3}{8} \\ 0 & 1 & \frac{-11}{14} \\ 0 &0 & \frac{18}{35} \\ \end{bmatrix}|\begin{bmatrix} \frac{5}{4} &0 & 0 \\ \frac{3}{8} & \frac{40}{21} & 0 \\ \frac{5}{7} & \frac{4}{7} &1 \\ \end{bmatrix}$

R3 ▶ $(\displaystyle 1.94444)R3$

$\begin{bmatrix} 1 & \frac{-1}{4} & \frac{-3}{8} \\ 0 & 1 & \frac{-11}{14} \\ 0 &0 & 1 \\ \end{bmatrix}|\begin{bmatrix} \frac{5}{4} &0 & 0 \\ \frac{3}{8} & \frac{40}{21} & 0 \\ \frac{25}{18} & \frac{10}{9} &\frac{35}{18} \\ \end{bmatrix}$

R2 ▶ R2 - $(\displaystyle -0.785714)R3$

$\begin{bmatrix} 1 & \frac{-1}{4} & \frac{-3}{8} \\ 0 & 1 & 0\\ 0 &0 & 1 \\ \end{bmatrix}|\begin{bmatrix} \frac{5}{4} &0 & 0 \\ \frac{65}{36} & \frac{25}{9} & \frac{55}{36} \\ \frac{25}{18} & \frac{10}{9} &\frac{35}{18} \\ \end{bmatrix}$

R1 ▶ R1 - $(\displaystyle - \frac{3}{8}​​)R3$

$\begin{bmatrix} 1 & \frac{-1}{4} & 0 \\ 0 & 1 & 0\\ 0 &0 & 1 \\ \end{bmatrix}|\begin{bmatrix} \frac{85}{48} &\frac{5}{12} & \frac{35}{48} \\ \frac{65}{36} & \frac{25}{9} & \frac{55}{36} \\ \frac{25}{18} & \frac{10}{9} &\frac{35}{18} \\ \end{bmatrix}$

R1 ▶ R1 - (\displaystyle - \frac{1}{4}​​)R2

$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 &0 & 1 \\ \end{bmatrix}|\begin{bmatrix} \frac{20}{9} &\frac{10}{9} & \frac{10}{9} \\ \frac{65}{36} & \frac{25}{9} & \frac{55}{36} \\ \frac{25}{18} & \frac{10}{9} &\frac{35}{18} \\ \end{bmatrix}$

Therefore,

[I-B]^-1= $\begin{bmatrix} \frac{20}{9} &\frac{10}{9} & \frac{10}{9} \\ \frac{65}{36} & \frac{25}{9} & \frac{55}{36} \\ \frac{25}{18} & \frac{10}{9} &\frac{35}{18} \\ \end{bmatrix}$ [Answer]

iii) Compute the output level for each product if the final demand for product A increased by 2000 units, that of product C decreased by 1,000 units and the final demand for product B remained unchanged.

X= [I-B]^-1D

D=$\begin{bmatrix} 20+2 \\ 60\\ 80-1\\ \end{bmatrix} = \begin{bmatrix} 22 \\ 60\\ 79\\ \end{bmatrix}$

X=$\begin{bmatrix} \frac{20}{9} &\frac{10}{9} & \frac{10}{9} \\ \frac{65}{36} & \frac{25}{9} & \frac{55}{36} \\ \frac{25}{18} & \frac{10}{9} &\frac{35}{18} \\ \end{bmatrix} \begin{bmatrix} 22 \\ 60\\ 89\\ \end{bmatrix}$

Multiply the rows of Matrix A with the columns of Martrix B

X= $\begin{bmatrix} \frac{20.22}{9} +\frac{10.60}{9} + \frac{10.79}{9} \\ \frac{22.65}{36} + \frac{55.79}{36} + \frac{25.60}{9} \\ \frac{25.25}{18} + \frac{10.60}{9} +\frac{35.79}{18} \\ \end{bmatrix}$

=$\begin{bmatrix} \frac{610}{3} \\ \frac{3925}{12} \\ \frac{1505}{6} \\ \end{bmatrix}$ (In thousands of units) [Answer]