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Unformatted text preview: ECHELON FORM AND GAUSSJORDAN
ELIMINATION As we noted in the previous section, our method for solving a system of linear equations
will be to pass to the augmented matrix, use elementary row operations to reduce the
augmented matrix, and then solve the simpler but equivalent system represented by the
reduced matrix. This procedure is illustrated in Fig. 1.3. The objective of the GaussJordan reduction process (represented by the middle
block' in Fig. 1.3) is to obtain a system of equations simpliﬁed to the point where we DEFINITION 3 1.2 Echelon Form and GaussJordan Elimination 15 Given ‘ ystem Augmented
 —>
or equations 1 , matrix Figure 1.3 Procedure for solving a system of linear equations can immediately describe the solution. See, for example, Examples 6 and 7 in Section
1.]. We tum now to the question of how to describe this objective in mathematical
terms—that is, how do We know when the system has been simpliﬁed as much as it can
be? The answer is: The system has been simpliﬁed as much as possible when it is in
reduced echelon form. Echelon Form When an augmented matrix is reduced to the form known as echelon form, it is easy to
solve the linear system represented by the reduced matrix. The formal description of
echelon form 18 given in Deﬁnition 3. Then, in Deﬁnition4, We describe an even simpler form known as reduced echelon form An (m x 'n) matrix B is in echelon form if: 1. All rows that consist entirely of zeros are grouped together at the bottom of
the matrix. 2. In every nonzero row, the ﬁrst nonzero entry (counting from left to right) is
a 1. 3. If the (i + l)st row contains nonzero entries, then the ﬁrst nonzero entry is in
a column to the right of the ﬁrst nonzero entry in the ith row. Put informally, a matrix A is in echelon form if the nonzero entries in A form a
staircaselike pattern, such as the four examples shown in Fig. 1.4. (Note: Exercise 46
shows that there are exactly seven different types of echelon form for a (3 x 3) matrix.
Figure [.4 illustrates four of the possible patterns. In Fig. 1.4, the entries marked :1: can be zero or nonzero.) l * * 1 =1: =1: 1 as a: 0 1 * A = 0 l * A = 0 l * A = 0 0 l A = 0 0 l 0 0 l 0 0 0 0 0 0 0 0 0
Figure 1.4 Patterns for four of the seven possible types of (3 x 3)  matrices in echelon fonn. Entries marked =I= can be either zero or nonzero. 16 Chapter 1 Matrices and Systems of Linear Equations Two examples of matrices in echelon form are 1 —1 4 3 O 2 0
0 0 l 8 —4 3 2 O A: O 0 0 l 2 1 2 B: 0 0 1 6 S
O 0 0 0 0 l 3 0
0 O 0 0 O 0 0 We show later that every matrix can be transformed to echelon form with elementary row
operations. It turns out, however, that echelon form is not unique. In order to guarantee
uniqueness, we therefore add one more constraint and deﬁne a form known as reduced
echelon form. As noted in Theorem 2, reduced echelon form is unique. DEleITION 4 l ,l _ , .eﬁ—AsHWW
l
l Figure 1.5 gives four examples (corresponding to the examples in Fig. 1.4) of matrices
in reduced echelon form. 1 O 0 1 O * l * 0 0 1 0
A: 0 l O A: O I * A: 0 0 l A: 0 0 l
0 0 1 O 0 0 0 0 0 O 0 0 Figure 1.5 Patterns for four of the seven possible types of (3 x 3)
matrices in reduced echelon form. Entries marked * can be either zero
or nonzero. Two examples of matrices in reduced echelon form are 1 0 0 2 l 2 O 1 —l
A: O l O —l B: 0 0 l 3 4
0 0 l 3 0 0 0 0 0 As can be seen from these examples and from Figs. 1.4 and 1.5, the feature that distin
guishes reduced echelon form from echelon form is that the leading l in each nonzero
row has only 0’5 above and below it. I I l EXAMPLE 1 For each matrix shown, choose one of the following phrases to describe the matrix. (a) The matrix is not in echelon form. (b) The matrix is in echelon form, but not in reduced echelon form.
(c) The matrix is in reduced echelon form. 1.2 Echelon Form and GaussJordan Elimination 17 l 0 0 l 3 2
A=l:2 l 0 , B: 0 —1 l .
3 —4 l 0 0 l
0 l —l 0 12 3 4 S l
C=[0 0 0 I , D: 0 0 12 3 , E: 0 ,
0 0 0 0 0 0 010 0
0
F: 0 , G=[10 0], H=[001]. .— Solution A, B, and F are not in echelon form; D is in echelon form but not in reduced echelon
form; C, E, G, and H are in reduced echelon form. q Solving 3 Linear System Whose Augmented Matrix Is in Reduced
Echelon Form Software packages that can solve systems of equations typically include a command
that produces the reduced echelon form of a matrix. Thus, to solve a linear system on a
machine, we ﬁrst enter the augmented matrix for the system and then apply the machine’s
reduce command. Once we get the machine output (that is, the reduced echelon form
for the original augmented matrix), we have to interpret the output in order to ﬁnd the
solution. The next example illustrates this interpretation process. I I I EXAMPLE .7. Each of the following matrices is in reduced echelon form and is the augmented matrix
for a system of linear equations. In each case, give the system of equations and describe the solution. 1 0 0 3
l 0 —l O
0 1 0 2
B: , C: l 3 0 .,
0 0 l 7
0 l
0 0 0 0
1—3 0 4 2 12 O 5
D: 0 O [—5 l , E: 0 010
0 0 0 0 0 0 0 0 0 Solution Matrix B: Matrix B is the augmented matrix for the following system:
x1 = 3
x; = —2
x3 = 7. Therefore, the system has the unique solution x1 = 3, x2 = —2, and x3 = 7. 18 Chapter 1 Matrices and Systems of Linear Equations Matrix C: Matrix C is the augmented matrix for the following system
xl —x3 = 0
x2 + 3x3 = 0
0x1 + 0X2 + 0X3 =1. Because no values for x., x2, or 3:3 can satisfy the third equation, the system is
inconsistent. Matrix D: Matrix D is the augmented matrix for the following system
x. — 3x2 + 4x4 = 2
X3 — 5x4 = I.
We solve each equation for the leading variable in its row, ﬁnding
xi = 2 + 3x2 — 4x4
163 = 1 + 5x4. In this case, x. and x3 are the dependent (or constrained) variables whereas x2 and x4 are
the independent (or unconstrained) variables. The system has inﬁnitely many solutions,
and particular solutions can be obtained by assigning values to x2 and x4. For example,
setting as; = l and X4 = 2 yields the solution x1 = —3, x2 = 1, x3 = 11, and x4 = 2. Matrix E: The second row of matrix E sometimes leads students to conclude erro
neously that the system of equations is inconsistent. Note the critical difference between
the third row of matrix C (which did represent an inconsistent system) and the second
row of matrix E. In particular, if we write the system corresponding to E, we ﬁnd x1 + 2x2 = 5
X3 = 0.
Thus, the system has inﬁnitely many solutions described by
x] = 5 — 2x2
x3 = 0 where x; is an independent variable. Ill As we noted in Example 2, if an augmented matrix has a row of zeros, we sometimes
jump to the conclusion (an erroneous conclusion) that the corresponding system of
equations is inconsistent (see the discussion of matrix E in Example 2). Similar confusion
can arise when the augmented matrix has a column of zeros. For example, consider the
matrix 1 0 0 —2 0 3
0 0 1 —4 0 1 ,
0 0 0 0 l 2 where F is the augmented matrix for a system of 3 equations in 5 unknowns. Thus,
F represents the system I I , THEOREM 2 1.2 Echelon Form and GaussJordan Elimination 19 x1 — 2X4 = 3
X3 — 4X4 = 2
X5 = 2. The solution of this system is x. = 3 + 2x.;, x3 = l + 4x4, x5 = 2, and x4 is arbitrary.
Note that the equations place no constraint whatsoever on the variable x2. That does not
mean that x2 must be zero; instead, it means that x2 is also arbitrary. Recognizing an Inconsistent System Suppose [A lb] is the augmented matrix for an (m x n) linear system of equations. If
[A lb] is in reduced echelon form, you should be able to tell at a glance whether the
linear system has any solutions. The idea was illustrated by matrix C in Example 2. In particular, we can show that if the last nonzero row of [A  b] has its leading l in
the last column, then the linear system has no solution. To see why this is true, suppose
the last nonzero row of [A  b] has the form [0.0.0. ..,.,0 1].
This row, then, represents the equation
0x1 +0x2 +0x3 + +0x,, =1. Because this equation cannot be satisﬁed, it follows that the linear system represented
by [A  b] is inconsistent. We list this observation formally in the following remark. Remark Let [A I b] be the augmented matrix for an (m x n) linear system of equations,
and let [A l b] be in reduced echelon form. If the last nonzero row of [A  b] has its leading
1 in the last column, then the system of equations has no solution. When you are carrying out the reduction of [A lb] to echelon form by hand, you
might encounter a row that consists entirely of zeros except for a nonzero entry in the last
column. In such a case, there is no reason to continue the reduction process since you
have found an equation in an equivalent system that has no solution; that is, the system
represented by [A l b] is inconsistent. Reduction to Echelon Form . The following theorem guarantees that every matrix can be transformed to one and only
one matrix that is in reduced echelon form.
Let B be an (m x n) matrix. There is a unique (m x n) matrix C such that: (a) C is in reduced echelon form and (b) C is row equivalent to B. q Suppose B is the augmented matrix for an (m x n) system of linear equations. One
important consequence of this theorem is that it shows we can always transform 8 by a 20 Chapter 1 I I l EXAMPLE3 Solution Matrices and Systems of Linear Equations series of elementary row operations into a matrix C which is in reduced echelon form.
Then, because C is in reduced echelon form, it is easy to solve the equivalent linear system represented by C (recall Example 2).
The following steps show how to transform a given matrix B to reduced echelon form. As such, this list of steps constitutes an informal proof of the existence portion
of Theorem 2. We do not prove the uniqueness portion of Theorem 2. The steps listed
assume that B has at least one nonzero entry (because if B has only zero entries, then B is already in reduced row echelon form). Reduction to Reduced Echelon Form for an (m x n) Matrix Step 1. Locate the ﬁrst (leftmost) column that contains a nonzero entry. Step 2. If necessary, interchange the ﬁrst row with another row so that the ﬁrst
nonzero column has a nonzero entry in the ﬁrst row. Step 3. If (1 denotes the leading nonzero entry in row one, multiply each entry
in row one by 1 /a. (Thus, the leading nonzero entry in row one is a 1.) Step 4. Add appropriate multiples of row one to each of the remaining rows so that every entry below the leading l in row one is a 0. Step 5. Temporarily ignore the ﬁrst row of this matrix and repeat Steps 1—4 on
the submatrix that remains. Stop the process when the resulting matrix
is in echelon form. Step 6. Having reached echelon form in Step 5, continue on to reduced echelon
form as follows: Proceeding upward, add multiples of each nonzero
row to the rows above in order to zero all entries above the leading l. The next example illustrates an application of the sixstep process just described.
When doing a small problem by hand, however, it is customary to alter the steps slightly—
instead of going all the way to echelon form (sweeping from left to right) and then going
from echelon to reduced echelon form (sweeping from bottom to top), it is customary
to make a single pass (moving from left to right) introducing 0's above and below the
leading 1. Example 3 demonstrates this singlepass variation. Use elementary row operations to transform the following matrix to reduced echelon
form 0 0 0 2 8 4
0 0 l 3 ll 9
3 —12 —3 —9 —24 —33
—2 8 l 6 l7 2] COCO . The following row operations will transform A to reduced echelon form. 1.2 Echelon Form and GaussJordan Elimination RI <—> R3, (1/3)Rl: Introduce a leading 1 into the ﬁrst row of the ﬁrst
nonzero column. 0 l —4 —1 —3 —8 —ll
0 0 0 l 3 ll 9
O 0 0 0 2 8 4
O —2 8 l 6 17 21 R4 + 2R1: Introduce 0’s below the leading 1 in row 1.
0 l —4 —1 —3 —8 —ll
0 0 0 l 3 ll 9
0 0 0 0 2 8 4
0 0 0 —l O 1 1
R1 + R2, R4 + R2: Introduce 0’s above and below the leading l in row 2. 0 l —4 O 0 3 —2
0 0 0 l 3 11 9
0 0 O O 2 8 4
0 0 0 O 3 12 8
(l/2)R3: Introduce a leading 1 into row 3.
0 l —4 0 0 3 2
0 0 0 l 3 11 9
0 0 0 0 l 4 2 O O 0 O 3 12 8 R2 — 3R3, R4 — 3R3: Introduce 0’s above and below the leading 1 in row 3. 0 l —4 0 0 3 —2
O 0 0 1 0 —l 3
0 O 0 0 l 4 2
O 0 0 0 0 0 2 . (1/2)R4: Introduce a leading 1 into row 4.
0 1 —4 O 0 3 ~2
0 0 0 l 0 — 1 3
0 O 0 0 I 4 2
O 0 0 0 0 0 1 R1 + 2R4, R2 — 3R4, R3  2R4: Introduce 0’s above the leading I in row 4. 0 1 —4 0 0 3 O
0 0 0 l 0 —1 0
O 0 O O l 4 0
O 0 0 0 0 0 l 21 22 Chapter 1 l l I EXAMPLE 4 Solution Matrices and Systems of Linear Equations Having provided this example of how to transform a matrix to reduced echelon form,
we can be more speciﬁc about the procedure for solving a system of equations that is diagrammed in Fig. 1.3. Solving a System of Equations Given a system of equations:
Step 1. Create the augmented matrix for the system. Step 2. Transform the matrix in Step 1 to reduced echelon form. Step 3. Decode the reduced matrix found in Step 2 to obtain its associated
system of equations. (This system is equivalent to the original system.) Step 4. By examining the reduced system in Step 3, describe the solution set
for the original system. The next example illustrates the complete process. Solve the following system of equations: 2xl~4x2+ 3x3—4x4—llx5= 28
—x1+2x2— x3+2X4+ 5x5=—13  3X3+ x4+ 6x5=—10
3xl6x2lle3—8X4—28x5: 61. We ﬁrst create the augmented matrix and then transform it to reduced echelon form. The
augmented matrix is
2 —4 3 —4 — 11 28
—l 2 ——1 2 5  l3
0 0 —3 l 6 —10
3 —6 10 —8 —28 61 . The ﬁrst step is to introduce a leading 1 into row 1. We can introduce the leading 1
if we multiply row 1 by 1/2, but that would create fractions that are undesirable for hand
work. As an alternative, we can add row 2 to row I and avoid fractions.
R1 + R2= 1 —2 2 —2 —6 15
——l 2 —l 2 5 — 13
0 0 —3 l 6 — IO
3 —6 10 —8 —28 61 1.2 Echelon Form and GaussJordan Elimination R2 + R1, R4 — 3R1: Introduce 0’s below the leading 1 in row 1. I 2 2 —2 6 l5
0 0 l 0 l 2
0 0 3 l 6 —10
0 0 4 —2 —10 16 R1 — 2R2, R3 + 3R2, R4 — 4R2: Introduce 0’s above and below the leading l in row 2.
1 —2 0 —2 —4 11
0 O l 0 —l 2
0 O 0 1 3 —4
0 0 0 —2 —6 8 R1 + 2R3, R4 + 2R3: Introduce 0’s above and below the leading l in row 3. l —2 0 0 2 3
0 0 l 0 —1 2
0 O 0 l 3 —4
O O 0 O O 0 The matrix above represents the system of equations x;  2x2 + 2x5 = 3
X3 — X5 = 2
X4 + 3X5 = —4. Solving the preceding system, we ﬁnd: X] = 3 + ZXZ  2x5
X3: 2 + x5
' x4 = —4 —— 3x5 23 (1)
In In Eq. (1) we have a nice description of all of the inﬁnitely many solutions to the
original system—it is called the general solution for the system. For this example,
x2 and x5 are viewed as independent (or unconstrained) variables and can be assigned
values arbitrarily. The variables x1. x3, and x4 are dependent (or constrained) variables,
and their values are determined by the values assigned to x; and x5. For example, in
Eq. (1), setting x2 = l and X5 = —1 yields a particular solution given by x] = 7, x2 = 1,133 = 1,164 = —1,andx5= —1. N3 0368: H 733:8 HS: 3:83: a. 5:3... Magma—Hm > U U 20 _ Z‘Imom a 3:503:38: 3HHnHoHd Ham HH 3:" 05:8 :Hmoosuda :5 33:5: H + m + u + . . . + a N no: + CE 23:: H8 Sam 0:; 8: «a»; GE. .3 08:3 :39 HHHm 8:39. $an :8
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H: 98: 288.8. «52$ :8 85:0: man 9. was H3: :8 mu. a. l NH: H w mvaHmH: Hm 88358: N: I Ah: I H 1.2 Echelon Form and GaussJordan Elimination 27 24. xl—x2+ X3: 3
2xl+x24X3=—3 25. X+ XZ=2
3x+3x2=6 26. x1— xz+ X3=4
2X] —2.X2+3X3 == 2 27. x1+xz x3: 2
~3x1—3xz+3X3=—6 28. 2X1+ 3x2—4x3= 3
x1— szQX3=—2
—X1+16X2+2x;;= 16 29. x1+xz— x3: I
2x1—x2+7x3= 8
—x; +X2 —5.\’3 = 5 30.II+X2 — xs=l
Iz+ZX3+X4+3JC5=1 X] — X3+X4+ X5=0
31. x. + x3+x42x5=l 2xl+X2+3X3X4+ X5=0
3X1—x2+4X3+X4+ 175:] 32. x+X2=l 33. xl+x2=l
xl—x2=3 xl—X2=3
2X1+X2=3 2X]+X2=2 34. x1+2x2= 1 35.xI—x2— X3=1
21144121,: 2 x] + X3=2
X[2x2=l XQ+ZX3=3 1n Exercises 36—40, ﬁnd all values a for which the sys—
tem has no solution. 36. xl+ZX2=3 37. x1+3x2=4 ax1—2x2= 5 2x1+6x2 =a
38. 2x1 + 4):; = a 39. 3xl + cur; = 3 3x+6x2=5 ax+3x2=5
40. x1 +. axz = 6 ax1+2ax2=4 In Exercises 41 and 42. ﬁnd all values at and )3 where
05a527rand05/3527r. 41.2cosa+4sin,8= 3
3cosa—Ssinﬁ=—1
42. 2cosza— sin2ﬁ= 1
lZcosZa + 8sin2ﬂ :13 43. Describe the solution set of the following system in terms 0fX3: x1 + x2+x3 = 3 x1 + 2x; 2'5.
For x], x2,x3 in the solution set: a) Find the maximum value of A73 such that
x1 anndxz 20. b) Find the maximum value of
y = 2x1  4x2 + x3 subject to x. a 0 and
X2 _>_ 0. c) Find the minimum value of
y = (x, — 1)2 + (x2 + 3)2 + (x3 + 1)2 with no
restriction on x] or x2. [Hint: Regard y as a
function of x3 and set the derivative equal to 0;
then apply the secondderivative test to verify
that you have found a minimum] 44. Let A and I be as follows: 4:2]. 33]. Prove that if b — cd 75 0, then A is row equivalent
to I . 45. As in Fig. 1.4, display all the possible conﬁgurations
for a (2 x 3) matrix that is in echelon form. [Hint
There are seven such conﬁgurations. Consider the
various positions that can be occupied by one, two,
or none of the symbols] 46. Repeat Exercise 45 for a (3 x 2) matrix, for a (3 x 3)
matrix, and for a (3 x 4) matrix. 47. Consider the matrices B and C: hm. 4:2]. By Exercise 44, B and C are both row equivalent to
matrix I in Exercise 44. Determine elementary row
operations that demonstrate that B is row equivalent
to C. 48. Repeat Exercise 47 for the matrices an} em 49. A certain threedigit number N equals ﬁfteen times
the sum of its digits. If its digits are reversed, the
resulting number exceeds N by 396. The one’s digit
is one larger than the sum of the other two. Give
a linear system of three equations whose three un
knowns are the digits of N. Solve the system and
ﬁnd N. 50. Find the equation of the parabola, y = ax2+bx +c,
that passes through the points (—1, 6), (1, 4), and
(2, 9). [Hint For each point, give a linear equation
in a, b, and c.] 51. Three people play a game in which there are al
ways two winners and one loser. They have the 28 52. 53. Chapter 1 Matrices and Systems of Linear Equations understanding that the loser gives each winner an
amount equal to what the winner already has. After
three games, each has lost just once and each has
$24. With how much money did each begin? Find three numbers whose sum is 34 when the sum
of the ﬁrst and second is 7, and the sum of the second
and third is 22. A zoo charges $6 for adults, $3 for students, and
$.50 for children. One morning 79 people enter and
pay a total of $207. Determine the possible numbers
of adults, students, and children. 54. Findacubic polynomial, p(x) = a+bx+cx2+dx3.
such that p(l) = 5, p’(l) = 5,p(2) = 17, and
p'(2) = 21. In Exercises 55—58, use Eq. (2) to ﬁnd the formula for
the sum. If available, use linear algebra software for
Exercises 57 and 58. 55.l+2+3++n 56.12+22+32++n2
57.1“+24+34++n4
53.15+25+35++n5 ...
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