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(I 1

the reduced row echelon matrix, which corresponds to the system

- p, =0

PI

p2 + 2pl = 0 . 5 .

If we write this system as

PI = Pi

(18)

P2 = 11.5 2p3.

136 SYSTEMS OF LINEAR EqUATloNs [71

we notice that there is no single solution to (I 8): for an? value of p3, system (IX)

determines corresponding values of p, and p2. Since system (IX) has multiple

solutions, so does system (17). For any choice of p.;, (18) determines values of p,

and pz which solve system (17). For example,

(Check that these three are truly solutions of (17).)

As an example of a system with HO solurions, consider an investment model

with state nzturns

R,? = 2

R II = 1 R,z = 3

Rp = 3

Rz, = 3 Rzz = I

Once again by equation (14) in Chapter 6. the corresponding system of equations

for a state price vector (p,, /Q) is

I/вЂ˜, + Ii/I? = I

311, + I/l? = I (19)

- 3111 = I.

2P I

In system ,191. now that the only 11,. ,>2 pair that solves the tirsr T W O equations ib

ox, = 0.25. .x2 = 0.25. Since this pair dots not sarisfy the third equation. (19) ha\

no solution. When we reduce the augmented matrix of (19) ro row echelon form

we ohtain

3 I

0 -8 1 -2

il0 I -0.25 вЂ˜1

0

The last row corresponds to the equation

op + ofi = p.25. (20)

The left-hand side of this equation is always II and thus can never equal -0.25.

So there is no ,I,, ˜1˜ pair which solves this equation. Uote that if we replace the

last equation in (19) hy

2&J + 2pz = I ,

then the new system has the unique solution

p, = 0.25, pz = 0.25,

and the row echelon form of the augmented matrix becomes

which con&ins no contradictions. (Exercise: check all these computations.)

Thcsc cxamplcs raise the following questions about systems of linear equa-

tions.

(1) When does a particular system of linear equations have a solutionвЂ™?

(2) How many solutions does it have? How do we compute them?

(3) What conditions on the coefficient matrix will guarantee the existence of

hiвЂ˜s on the right-hand side of (2)?

nf lctr.st one solution for any choice of

(4) What conditions on the coefficient matrix will guarantee the existence of

b,вЂ˜sвЂ™!

OI mo˜cf one solution for any choice of

138 SYSTEMS OF LINEAR EqtMTloNs 171

For another example, consider an investment model with three assets and four

states. Suppose that shares of the three assets have the following current values:

вЂњ1 = 38, вЂњ2 = 98, вЂњ3 = 153

As in Section 6.2, we write ys; for the value of a share of asset i one year from

now if state s occurs. Suppose the ysiвЂ™s have the following values:

2

y11= 1 y12= y13= 3

Y22 = 12

Yzl= 4 YZ3 = 18

YX = 69

Y31 = 17 Y32 = 46

yd,2 = 10 y41 = 17.

y41= 4

By (13) in Chapter 6, the state prices pl, p2, pi, p4 for this model satisfy the system

Ip1 + 4p2 + 17pz + 4p, = 38

2p, + 12pz + 46˜3 + 10˜4 = 98

3p, + 18˜1 + 69p, + 17˜4 = 153.

Its augmented matrix is

1 4 17 4 1 38

2 12 46 10 I 98 ,

i3 18 69 17 I 153 1

with corresponding echelon form

TOW

1 4 17 4 1 38

0 4 12 2 / 22

0 2 I 61

i0 0

Divide the second row by 4 and the third row by 2 to obtain

I 4 17 4 I 38

01 3 0.5 I 5.5

c0 0 0 I 1 31

Work first with the pivot in the third row to change column 4 from

140 SYSTEMS OF LINEAR EQвЂќATlONS [7]

0

; 1;

00 ; ; i i 1 вЂ˜),

(0 0 0 0 0 0 * w

This matrix is in row echelon form. The corresponding reduced row echelon form

is

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