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WC

Our system (8) has been transformed to the simpler system

In transforming system (8) to system (9). we used only one operation: w e

added a multiple of one equation t u another. This operation is reversible. For

example, we can recover (8) from (9) by adding -0.2 times (Ya) to (Yh) t o obtain

(8h) and then by adding -0.5 times (9a) to (SC) to obtain (Xc). (We continue to

write (Yaj to denote the first equation in system (Y).) There are two other rcversibls

operations one often uses to transform a system of equations: 1) multiplying both

sides of an equation hy a nonzao scalar and 2) interchanging t w o equations.

These three operations are called the elementary equation operations. Since

equals arc always added to or subtracted from equals or multiplied hy the same

scalar, the set ofxiвЂ™swhich solve the original system will also solve the transformed

system. In fact, since these three operations are reversible, any solution of the

transformed system will also be a solution of the original system. C:onsequently.

both systems will have the exact same set of solutions. We call two systems of

linear equations equivalent if any solution of one system is also a solution of

the other.

L7.1 I GAL˜SSIAN ANвЂќ CAвЂќSS˜IORDAN ELlMlNATlON 127

If one system of linear equations is derived from another by elementary

Fact

equation operations, then both systems hevc the same solutions; that is, the systems

are equivalent.

Let us return to our elimination procedure and continue working on system

(9). Having eliminated XI from the last two equations, we now want to eliminate

x2 from the last equation. We apply the elimination process to the system of two

equations (9b) and (9˜) in two unknowns. Multiply (9b) by I/2 and add this new

equation to (9˜) to oh&in the new system:

Ix, - o.4xz ˜ 0.3X˜ = 130

+ 08x1 0.2x1 = 100 (10)

+ 0.7x1 = 210.

Since each equation in (10) has one fewer variable than the previous one, this

system is particularly amenable to solution by substitution. Thus, xi = 300 from

(10˜). Substitutingxi = 300 into (lOb)givesxz = 200. Finally, substituting these

two values into (lOa) yields x, = 300. The method used in this paragraph is

usually called back substitution.

This method of reducing a given system of equations by adding a multiple of

one equation to another or by interchanging equations until one reaches a system

of the form (10) and then solving (10) via back substitution is called Gaussian

The important characteristic of system (10) is that each equation

elimination.

contains fewer variables than the previous equation.

At each stage of the Gaussian elimination process. we want to change some

coefficient of our linear system to 0 by adding a multiple of an cadim equation

to the given one. For cxample, if you want to use the cocfficicnt aji, in the third

equation tu climinare the coefficient oix in the fifth equation, we add (--oin/aix)

times the third equation to the fifth equation, to get a new fifth equation whose

kth coefficient is The is then called and we sa!; that

0. coefficient ulii a pivot.

wc вЂњpivot on uik to eliminate ujx.вЂќ At each stage of the elimination pnxcdurc;

we use a pivot to climinatc all coefticients directly belou: it. For example, in

transforming system (8) to system (9). the cocfficicnt I in equation (8вЂќ) is the

pivot: in tranforming system (9) to system (lo), the cocficient 0.8 in equation

(вЂ˜Jh) is the pivot.

iVote that 0 can never be a pivot iti this process. If you want to eliminate xi from

a subsystem of equations and if the coefficient ofsh˜ is zcw in the first equation of

this suhsystcm and nonze˜ in a suhscqocnt equation, you will have to IZYLвЂ˜ISC the

wdcr of these twc> equations before pmcccding.

We did not use the operation of transforming an equation by simply multiplying

it h! a non˜rm scalar. Thcrc is B wriant of Gaussian elimination, called Gauss-

which uses all three elementary equation operations. This

Jordan eliminations

method starts like Gaussian elimination. c-g,. by transforming (8) to (IO). After

reaching system (10). multiply each equation in (10) hy a scalar so that the first

128 SYSTEMS OF LINEAR EQUATIONS [7]

nonzero coefficient is 1:

x, - 0.4x2 - 0.3 xj = 130

x* - 0.25x2 = 125 (11)

x3 = 300

Now, instead of using back substitution, use Gaussian elimination methods

from the bottom equation to the top to eliminate all but the first term on the left-

hand side in each equation in (I I). For example, add 0.25 times equation (1 Ic)

to equation (1 lb) to eliminate the coefficient of x3 in (1 lb) and obtain x2 = 200.

Then, add 0.3 times (11˜) to (lla) and 0.4 times (llb) to (1 la) to obtain the new

system:

= 300

XI

= 200

x2 (12)

13 = 300,

which needs no further work to see the solution. Gauss-Jordan elimination is

particularly useful in developing the theoryoflinearsyst˜ms; Gaussian elimination

is usually more efficient in solving actual linear systems.

Earlier we mentioned a third method for solving linear systems, namely matrix

methods. We will study these methods in the next two chapters, when we discuss

matrix inversion and CramerвЂ™s rule. For now, it suffices to note that all the intuition

behind these more advanced methods derives from Gaussian elimination. The

understanding of this technique will provide a solid base on which to build your

knowledge of linear algebra.

EXERCISES

7.1 Which of the follawing equations are linear?

0) 3s, - 4x2 + 5.9 = 6; h) X,.X˜lj = - 2 : c) .?+6?;= I:

d) (I + ?.)(I - z) = -7: e) IT + 3вЂќвЂ˜Z = 4; f) * + 3;вЂњвЂќ = -4

7.2 Solve the following systems by substitution, Gaussian elimination, and GauwJardan

elimination:

7.3 Solve the following systems by Gauss-Jordan elimination. Note that the third system

requires an equation interchange.

b) 4x + 2y 32 = I

a) 3x + 3y = 4 c)2x+2y- z= 2

x y = 10; 6x + 3y 52 = 0 Xf y + z=-2

x+ y+2z=9; 2.x 4y + 32 = 0,

7.4 Formalize the three elementary equation operations using the abstract notation of

system (2), and for each operation, write out the operation which reverses its effect.

7.5 Solve the IS-LM system in Exercise 6.7 by substitution.

7.6 Consider the general IS-LM model with no liscal policy in Chapter 6. Suppose that

M, = MвЂќ; that is, the intercept of the LM-curve is 0.

a) Use substitution to solve this system for Y and I in terms of the other parameters.

b) How does the equilibrium GNP depend on the marginal propensity to save?

c:) How does the equilibrium interest late depend on the marginal propensity to save?

7.7 Use Gaussian elimination to solve

3x + 3y = 4

{ --x- y=lO

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