Chapter 2, Problem 22RP

Explanation of Solution

a.

Determining A:

Let $a_{ij}$ denote the $i{j}^{th}$ element of matrix A.

Assume that rows for the matrix refer to the inputs and column for the matrix refer to the output.

Define steel as product 1, car as product 2, and machine as product 3.

Thus, $a_{11}$ is the dollar value of steel required to produce $1 of steel, $a_{21}$ is the dollar value of cars required to produce $1 of steel, and $a_{31}$ is the dollar value of machines required to produce $1 of steel.

Therefore,

$\begin{array}{l}{a}_{11}=0.30\\ a_{21}=0.15\\ a_{31}=0.40\end{array}$

Similarly, $a_{12}$ is the dollar value of steel required to produce $1 of cars, $a_{22}$ is the dollar value of cars required to produce $1 of cars, and $a_{32}$ is the dollar value of machines required to produce $1 of cars.

Therefore,

$\begin{array}{l}{a}_{12}=0\end{array}$

Explanation of Solution

b.

Proof:

It is given that s, c, and m represents the dollar value of steel, cars, and machines, produced in the economy respectively, and $d_s$, $d_c$, and $d_m$ represents the computation of steel, cars, and machines respectively.

Note that a product has two uses in the economy, one to produce other goods and the other one for the consumption. So, the total production of a product must be equal to the amount of that product used to produce other goods and the amount consumed, that is supply is equal to the demand.

For instance, for steel, its production must be equal to the dollar value of steel used in the steel’s production plus the dollar value of the steel used in the car’s production plus the dollar value of steel used in the machines production plus the dollar value of the steel’s consumption.

$s=a_{11}s+a_{12}c+a_{13}m+d_s\mathrm{......}(1)$

Similarly,

$c=a_{21}s+a_{22}c+a_{23}m+d_c\mathrm{}$

Explanation of Solution

c.

Proof:

Let,

$\text{X=}\left[\begin{array}{l}\text{s}\\ \text{c}\\ \text{m}\end{array}\right]andB=\left[\begin{array}{l}{d}_s\\ d_c\\ d_m\end{array}\right]$

Then, the equations (1), (2) and (3) can be written as,

$X=AX+B\mathrm{......}(4)$

Since a matrix multiplied by the identity matrix is the matrix itself, so the equation (4) can be further written as,

$IX=AX+B$

Subtract from both sides of the equation

Explanation of Solution

d.

Consider the equation derived in part (c.)

$(I-A)\left[\begin{array}{l}\text{s}\\ \text{c}\\ \text{m}\end{array}\right]=\left[\begin{array}{l}{d}_s\\ d_c\\ d_m\end{array}\right]$

Pre-multiply both sides by ${(I-A)}^{-1}$.

${(I-A)}^{-1}(I-A)\left[\begin{array}{l}\text{s}\\ \text{c}\\ \text{m}\end{array}\right]={(I-A)}^{-1}\left[\begin{array}{l}{d}_s\\ d_c\\ d_m\end{array}\right]\mathrm{}$

Explanation of Solution

e.

It is given that the demand for steel goes up by $1.

Denote the new production levels by $s\text{'},c\text{'},m\text{'}$.

Then,

$\left[\begin{array}{l}\text{s'}\\ \text{c'}\\ \text{m'}\end{array}\right]={(I-A)}^{-1}\left[\begin{array}{l}{d}_s+1\\ d_c\\ d_m\end{array}\right]\mathrm{......}(6)$

Thus, the change in production levels can be derived by subtracting equation (5) from the above equation (6).

$\left[\begin{array}{l}\text{s'}\\ \text{c'}\\ \text{m'}\end{array}\right]-\left[\begin{array}{l}\text{s}\\ \text{c}\\ \text{m}\end{array}\right]={(I-A)}^{-1}\left[\begin{array}{l}{d}_s+1\\ d_c\\ d_m\end{array}\right]-{(I-A)}^{-1}[\begin{array}{l}{d}_s\end{array}$