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Substitution is the method usually taught in beginning algebra classes. To use this

method, solve one equation of system (2) for one variable, say I,,, in terms of the

other variables in that equation. Substitute this expression for x,, into the other

n, I e q u a t i o n s . T h e r e s u l t i s a n e w s y s t e m o f m I e q u a t i o n s i n t h e n I

u n k n o w n s x,, , .I,, , Continue this process by solving one equation in the new

system for I,, , and substituting this expression into the other m 2 equations ro

ohlain a sys,em of m - 2 equations in the n 2 variables x,, I,, 2. Proceed

until you reach a system with just a single equalion, a situation which is easily

solved. Finally, use the earlier expressions of one variable in terms of the others

t o find all the x;вЂ˜s.

This sounds complicated but it really is straightforward. We used whstilulion

6.2. Let us see how it works on a

to solve the input-output system in Section

ihrvc-good input-oulput model.

7.1

L\uI@ The production process for a three-good economy is summarized

I hv the input-outpul table:

0 0.4 0.3

˜

0.2 0.12 0.14 Table

0.2 il.05 7.1

0.5

Recall from the last chapter that the cntrics in the second column uf Tahlc 7.1

declare that it takes 0.4 unit of good I 0. I2 unit of good 2, and 0.2 unit of good

3 t o produce 1 unh U C good 2. We ignore the labor component in this example.

Supposc that thcrc is an cxogcnous demand for 130 units of good I. 74 units of

24 SYSTEMS OF LINEAR EqlJATloNs [71

good 2, and 95 units of good 3. How much will the economy have to produce

to meet this demand?

Let xi denote the amount of good i produced. As we described last chap-

ter, вЂњsupply equals demandвЂќ leads to the following system of equations:

x, = 0 x, + 0.4 x* + 0.3 x1 + 130

x2 = 0.72, + 0.12x2 + 0.14x1 + 74

A-3 = 0.5x, + 0.2 x2 + 0.05x3 + 95,

which can be rewritten as the system

Xl - 0.4 x* - 0.3 X) = 130

-0.2x1 + 0.88x2 - 0.14x3 = 7 4 (3)

-0.5321 0 . 2 x2 + 0.95x2 = 95.

We write (3a), (3b) and (3˜) for the three equations in system (3) in the order

given, and similarly for following systems. Solving equation (3a) forx, in terms

of x2 and x2 yields

x1 = 0.4x* + 0.3x1 + 130. (4)

Substitute (4) into equations (3b) and (3˜):

-0.2(0.4x2 + 0.3˜˜ + 130) +0.88x2 0.14˜˜ = 74

-0.5(0.4x2 + 0.31, + 130) - 0.2 x* + 0.95x1 = 95,

which simplifies to

0.8nz 0.2x3 = 100

(5)

-0.4x? +o.xx> = IhO

Now, use substitution to solve subsystem (5) by solving the first equation (Sa)

for xL in terms of x3:

x2 = 125 + 0.25X& (6)

and plugging this expression into the second equation (5b):

-0.4(125 + 0.25x?) + 0.8˜1 = 160,

OT x3 = 300.

12.5

r7.11 GAUSSIAN AND GAUSS-JORDAN ELIMINATION

Substitute x1 = 300 into (6) to compute that x2 = 200. (Check.) Finally,

substitute rz = 200 and x2 = 300 into (4) to compute that

x1 = 0.4˜200 + 0.3 300 + 130 = 300.

Therefore, this economy needs to produce 300 units of good one, 200 units of

good two, and 300 units of good three to meet the exogenous demands.

As this example shows, the substitution method is straightforward, but it can

be cumbersome. Furthermore, it does not provide much insight into the nature of

the general solution to systems like (3). It is not a method around which one can

build a general theory of linear systems. However, it is the most direct method for

solving certain systems with a special, very simple form. As such, it will play a

role in the general solution technique we now develop.

Elimination of Variables

The method which is most conducive to theoretical analysis is elimination of

variables, another technique that should be familiar from high school algebra.

First, consider the simple system

x, 2YZ =8

(7)

3x1 + x2 = 3.

We can вЂњeliminateвЂќ the variable XI from this system by multiplying equation (7a)

hy ˜3 to obtain -3x, + 6x2 = ˜24 and adding this new equation to (7b). The

result is

7x2 = -21, or x2 = -3.

TV find x,, we substitute x2 = ˜3 hack into (7b) or (7a) to compute that x1 = 2.

We chose to multiply equation (7a) by -3 precisely so that when we added the

new equation to equation (7b), we would вЂњeliminateвЂќ x1 from the system.

To solve a general system of m equations by elimination of variables, use

the coefficient of x, in the first equation tu eliminate the x, term from all the

equations below it. To do this, add proper multiples of the first equation to each

of the succeeding equations. Now disregard the first equation and eliminate the

next variable-usually x2 ˜ from the last m 1 equations just as before, that

is. by adding proper multiples of the second equation to each of the succeeding

equations. If the second equation does not contain an xz but a lower equation does,

you will have to interchange the order of these two equations before proceeding.

Continue eliminating variables until you reach the last equation. The resulting

simplified system can then easily bc solved by substitution.

126 SYSTEMS OF LINEAR EQUATIONS [7]

Let us try this method on the system (3) arising from the three-good input-

output Table 7.1:

n, - 0.4 x2 0.3 x3 = 131)

˜0.2x, + 0.88x2 - 0.14x1 = 74 (8)

-0sxj ˜ 0 . 2 x* + O.Y5x3 = 9s

We first try to eliminate x1 from the last t w o equations by adding to each of these

equations a proper multiple of the first equation. To eliminate the ˜lJ.2x,-term in

(8h), we multiply @a) by 0.2 and add this new equation to (Xh). The result is the

following calculation:

0.2x, -0.08x, - 0.06˜3 = 26

+ -0.2x, +0.8xX2 - 0.14x3 = 74

+ 0.8x2 - 0.2x3 = 100.

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