стр. 27 
Because of this connection with Gaussian elimination, it is natural that the first
nonzero entry in each row of a matrix in TOW echelon form he called a pivot.
The row echelon form is the goal in the Gaussian elimination process. In
GaussJordan elimination, one wants to use row operations to reduce the matrix
even further. First. multiply each row of the row echelon form by the reciprocal of
the pivot in that row and create a new matrix all of whose pivots are 1s. Then, use
these new pivots (starting with the 1 in the last row) to turn each nonzero entry
above it (in the same column) into a zew.
I7.21 ELEMENTARY ROW OPERATIONS 133
For example, multiply the second row of (14) by 1/0.X and the third row of
(14) by l/O.7 to achieve the matrix
1 PO.4 0.3 I 130
0 1  0 . 2 5 I 125
(0 0 1 I 300 1
Then, use the pivot in row 3 to turn the entries 0.25 and ˜0.3 above it into
zerosP first by adding 0.25 times row 3 to row 2 and then by adding 0.3 times
row 3 to row 1. The result is
1 PO.4 0 I 220
0 1 0 I 200
0 0 1 I 300
Finally, use the pivot in row 2 to eliminate the nonzem entry above it by adding
0.4 times r o w 2 to TOW 1 to get
the matrix
( 0 0 1 0 0 I 0 0 1 II I300 200 300 ) (1%
Notice that this is the augmented matrix for system (12) and that one can read the
solution right off the last column of this matrix:
x, = 300. x* = 200, xi = 300
We say that matrix (15) is in reduced ˜mvechelonform
Definition A row echelon matrix in which each pivot is a I and in which each
column containing a pivl,tcontainsnoothernon=eroentries is said to beinreduced
row echelon form.
.The matrices in Examples 7.4 and 7.5 above are in reduced row echelon form.
Note that in transforming a matrix to row echelon form we work from top left to
huttom right. To achieve the reduced row echelon form, we continue in the same
way but in the other direction, from bottom right to top left.

EXERCISES
Dcscrihc the row operations involved in going from equations(R) to (10)
1.9
Put the matrices in Examples 7.2 and 7.3 in reduced mw rchclon form.
7.10
134 SYSTElMS OF LINEAR EQвЂќATlвЂќNS 171
7.11 Write the three systems in Exercise 7.3 in matrix form. Then use row operations to
find their corresponding nxv cchclon and rcduccd row echelon forms and to find the
solution.
7.12 Use GaussJordan elimination in matrix form to solw the system
W+ x+ 3y2r=0
2w+3x+ 7y2z=Y
3w+sx+13yYz=l
˜2IVf x  z=n.
7.3 SYSTEMS WITH MANY OR NO SOLUTIONS
As we will study in more detail lalcr, the locus of all points (xl, x1) which satisfy
the linear equation ulixl + alznz = h, is a straight line in the plane. Therefore,
the solution (xt, x1) of the two lincu equations in two unknowns
h,
(16)
= II2
is a point which lies on both lines of (16) in the Cartesian plane. Solving system
(16) is equivalent to finding where the two lines given by (16) cross. In general.
two lines in the plane will be nonparallel and will cross in exactly one point.
However, the lines given by (16) can be parallel to each other. In thih cast. they
will either coincide (II they will ncvcr cross. lfthey coincide; every point on either
line is a wlution to (I 6); and (16) has irr/inire/˜ many solutions. An example is the
system
*, f 2x2 = 3
21, TV 4x2 = 6.
In the cast whcrc the two parallel lines do not CTOSS, the corresponding system has
no solution. as the example
illustrates. Thcrcforc. it follows from geometric considerations that two linear
equations in two unknowns can have one solution, no solution, or infinitely many
solutions. We will see later in this chapter that this principle holds for cvcry system
of m liwor equations in II unknowns.
I7.31 SYSTEMS WITH MANY OR NO SOLUTIONS 135
So far we have worked with examples of systems in which there are exactly
as many equations as there are unknowns. As we saw in the inputoutput model
and the Markov model in Chapter 6, systems in which the number of equations
differs from the number of unknowns arise naturally.
For example, let us look for a state price system for the investment model in
Example 6.1. Substitution of the state returns R,i from Example 6.1 into equations
(14) in Chapter 6 for the state prices leads to the system
whose augmented matrix is
(3 1 22 31 11 11 )
Adding 3 times the first row to the second yields the row echelon matrix
( 1 2 31
0 4 8 ( 2
7
To obtain the reduced IDW echelon form, multiply the last equation hy l/4:
Then, add 2 times the new last row to the first row to eliminate the 2 in the first
row above the pivot. The result is
стр. 27 