Contemporary Linear Algebra, Student Solutions Manual

259 Pages·2003·1.04 MB·English

by  Howard Anton

CHAPTER 2
Systems of Linear Equations
EXERCISE SET 2.1
1. (a) and {c) are linear. (b) is not linear due to the x1x3 term. {d) is not' linear due to the x}2 term.
2. (a) a.nd (d) are linear. (b) is not linear because of the xyz term. (c) is not linear because of the
x3/5 term.
1
3. (a) is linear. (b) is linear if k # 0. (c) is linear only if k = 1.
4. (a) is linear. (b) is linear jf m =f: 0. (c) is linear only if m = 1.
5. (a), (d), and (c) are solutions; these sets of values satisfy all three equations. (b) and (c) are not
solutions.
6. (b), (d), and (e) are solutions; these sets of values satisfy all three equations. (a) and (c) are not
solutions.
7. The tluee lines intersect at the point {1,0) (see figure). The values x = 1, y = 0 satisfy all three
equations and this is the unique solution of the system.
3.x-3y"' 3
:c
The augmented matrix of the system is
l ~ ~ ~]·
r231
!
-3 3
Add -2 times row 1 to row 2 and add -3 times row 1 to row 3:
[~ =! !~ ]
Multiply row 2 by - j and add 9 times the new row 2 t.o row 3:
[~ ~ ~]
From the last row we s~ that the system is redundant (reduces to only two equations). From the
= = =
second row we see that y 0 and, from back substitution, it follows that x 1 - 2y 1.
22
Exercise Set 2.1 23
8. The three lines do not intersect in a. common point (see figure). This system has no solution.
)'
The augmented matrix of the system is
and the reduced row echelon form of this matrix (details omitted) is:
The last row corresponds to the equation 0 = 1, so the system is jnconsistent.
=
9. (a) The solution set of the equation 7x - 5y 3 can be described parametrically by (for example)
solving the equation for x in terms of y and then making y into a parameter. This leads to
x == 3'*:,5t, y = t, where -oo < t < oo.
(b) The solution set of 3x1 - 5xz + 4x3 = 7 can be described by solving the equation for x1 in
terms of xz and x3, then making Xz and x3 into parameters. This leads to x1 = I..t5;-<~t,
x:z = s, x3 = t, where - oo < s, t < oo.
(c) The solution set of - 8x1 + 2x2 - 5x3 + 6x4 = 1 can be described by (for example) solving
the equation for x2 in t.erms of x1, x;~, and x4, then making x1, x3, and x4 into parameters.
This leads to :t1 = r, x2 = l±Brt5s- 6t, X3 = s, X4 = t, where -<Xl < r, s, t < oo.
(d) The solution set of 3v-Bw + 2x-y + 4z = 0 can be described by (for example) solving
the equation for y in terms of the other variables, and then making those variables into
parameters. This leads to v = t1, w = t2, x = t3, y = 3tl - 8t2 + Zt3 + 4t4, z = t4, where
- oo < t11 t2, t3, t4 < oo.
10. (a) x = 2- lOt, y = t, where -(X)< t < OO.·
(b) x1 = 3-3s + l2t, x2 = s, x3 = t, where -oo < s, t < oo.
(c) Xt = r, x2 = s, X3 = t, X4 = 20-4r- 2s- 3t, where - oo < r, .s, t. < oo.
(d) v = t1, w = t2, x = -t1 - t2 + 5t3 - 7t4, y = t3, z = t4, where -oo < h, t2, t3, t4 < oo.
11. (a) If the solution set is described by the equations x = 5 + 2t, y = t, then on replacing t by y in
the first equation we have x = 5 + 2y or x- 2y = 5. This is a linear equation with the given
solution set.
(b) The solution set ca.n also be described by solving the equation for y in terms of x. and then
!t-
making x jnto a parameter. This leads to the equations x = t, y = ~-
24 Chapter2
12. (a) If x1 = -3 + t and x2 = 2t, ther. t = Xt + 3 and so x2 = 2x1 + 6 or -2xl + x2 = 6._ This is
a. linear equation with the given solution set.
(b) The solution set can also be described by solving the equation for X2 in terms of x1, and then
making x into a parameter. This leads to the equations x = t, x = 2t + 6.
1 1 2
13. We can find parametric equations fN the line of intersection by (for example) solving the given
equations for x and y in terms of z, then making z into a parameter:
3+z} 4-3 z}
=
x+y= 2x+y
2x + y = 4- 3.z =} -x -y = -3- z =} x = l-4z
From the above it follows that y = 3 + z - x = 3 + x - 1 + 4z = 2 + Sz, and this leads to the
parametric equations
X = 1 - 4t, y = 2 + 5t, Z = t
for the line of intersection. The corresponding vector equation is
(x,y,z) = (1,2,0) + t(-4,5, 1)
14. We can find parametric equations for the line of intersection by (for example) solving the given
equations for x and y in terms of z, then making z into a parameter:
1-3z}
x + 2y = :::;. 4x = ( 1 - 3z) + (2 - z) = 3 - 4z ::::} X=3--4-z
3x-2y = 2-z 4
From the above it follows hat y = 1t,8z _ This leads to the parametric equations
i -
X :::: t, y = ~ - t, Z = t
and the corresponding vector equation is
(x, y, z) = (~, l· 0) + t( -1, -1, 1}
15. If k :fi 6, then the equations x- y = 3, 2x - 2y = k represent nonintersecting parallel lines and so
the system of equations has no solution. If k = 6, the two lines coincide and so there are infinitely
many solutions: x = 3 + t, y = t, where -oo < t < oo.
=
16. No solutions if k-:/: 3; infinitely many solutions if k 3.
3 -2 -1]
!
[
17. The augmented matrix of the system is 4 5 l 3 .
7 3: 2
[!
0 2
18. The augmented matrix is -1 4
1 -1
Exercise Set 2.1 25
2 0 - 1 1
[~ ~]
19. The augmented matrix o! the system is 3 2 0 - 1
3 1 7 0
11
~ 0 0 :
12 .
20. The augmented matrix is [ 1 0
!
J
0 1 3
21. A system of equations corresponding to the given augmented matrix is: .
=
2xt 0
3xl-4x2 = 0
X2 = 1
22. A system of equations corresponding to the given augmented matrix is:
=
3x1 - 2x3 5
7xl + x2 + 4x3 = - 3
- 2x2 + X3 = 7
23. A system of equations corresponding to the given augmented ma.tri.x is:
?x1 + 2x2 + x3 - 3x-t = 5
+ +
x 1 2x2 4x3 = 1
24. X}= 7, Xz = -2, X3 = 3, :J:4 = 4
25. (a) B is obtained from A by adding 2 times the first row to the second row. A is obtained from
B by adding -2 times the first row to the second row.
(b) B is obtained from A by multiplying the first row by!· A is obtained from B by multiplying
the first row by 2.
26, (a) B is obtained from A by interchanging the first and third rows. A is obtained from B by
interchanging the first and third rows.
(b) B is obtained from A by multiplying the third row by 5. A is obtained from B by multiplying
the third row by }.
27. 2x + 3y + z = 7 28. 2x + 3y + 12z = 4
2x + y + 3z = 9 8x + 9y + 6z= 8
4x + 2y + 5z = 16 6x + 6y + 12z = 7
29. X -J- y + Z = 12 30. X + y + Z = 3
2x + y + 2z = 5 y+z=lO
X - z= - 1 -y + z = 6
31. (a) Jc1 + c2 + 2c3 - C4 = 5 (b) 3cl + cz + 2c3 - C4 = 8
cz + 3c3 + 2c4 = 6 Cz + 3c3 + 2c4 = 3
-c1 + c2 + 5c4 = 5 - c1 + c2 + 5c4 = -2
2ct + c2 + 2c3 """ 5 2c1 + cz + 2c3 = 6
26 Chapter 2
(c) 3cl + c~ + 2c3 - c4:;;;; 4
c2 + 3c3 + 2c, := 4
-cl + Cz + Sc4 = 6
2ct + c2 + 2c3 = 2
32. (a) Ct + C2 + 2c3 = 2 (b) Ct + c2 + 2c3 = 5
2c1 - 2c3 + 5C4 = -2 2Ct - 2c3 + 6c4 = -3
- 4cl + 2cz- c3 + 4c4 = -8 -4ct + 2cz- C3 + 4c4 = -9
+ 2Cl + CJ- C4 = 0 + 2cz + CJ- c.&= 4
5ct - Cz + 3c3 + C4 = 12 5Ct - Cz + 3c3 + C4 = 11
(c) c1 + cz + 2c3 = 4
2c1 - 2c3 + Sc4 = - 4
- 4cl + 2c2 - c3 + 4c• = 2
+ 2c2 + c3 - C4 = 0
5c1 - c2 + 3c3 + c4 = 24
DISCUSSION AND DISCOVERY
Dl. (a) There is no common intersection point.
(b) There is exactly one common point of intersection.
(c) The three lines coincide.
D2. A consistent system has at least one solution; moreover, it either has exactly one solution or it
has infinitely many solutions.
If the system has exactly one solution, then there are two possibilities. If the three lines are
all distinct but have a. common point of intersECtion, then any one of the three equations can be
discarded without altering the solution set. On the other hand, if two of the lines coincide, then
one of the corresponding equations can be discarded without altering the solution set.
If the system has infinitely many solutions, then the three lines coincide. In this case any one
(in fact any two) of the equations can be discarded without altering the solution set.
D3. Yes. If B can be obtained from A by multiplying a row by a. nonzero constant, then A can be
obtained from B by multiplying the same row by the reciprocal of that constant. If B can be
obtained from A by interchanging two rows, then A can be obtained from B by interchanging the
same two rows. Finally, if B can be obtained from A by adding a multiple of a row to another
row, then A can be obtained from B by subtracting the same multiple of that row from the other
row.
= = = =
D4. If k = l m 0, then x y 0 is a solution of all three equations and so the system is consistent.
If the system has exactly one solution then the three lines intersect at the origin.
05. The parabolay = ax2 + bx + c will pass through the points {1, 1), (2,4), and (-1, 1) if and only if
a+ b+c=l
4a + 2b + c = 4
a - b+c=I
Since there is a unique parabola passing through any three non-collinear points, one would expect
this system to ba.ve exactly one·s olution.
Discussion and Discovery 27
D6. The parabola y == ax2 + bx + c passes through the points (x1, Yt), (xz, Y2), a.nd (x3, Ys) if and only
if
ax~ + b~1 + c = Yl
4ax~ + Zbx2 + c = Y2
+ =
ax~ - bxs c Y3
i.e. if and only if a, b, and c satisfy the linear sy5tem whose augmented matrix is
Y1]
1
1 Y2
1 Y3
D7. To say that the equations have the same solution set is the same thing as to say that they represent
the same line. From the first equation the x -intercept of the line is x1 = c, and from the second
1
=
equation the x1-intercept is x1 = d; thus c d. If the line ist , vertical then k = l = 0. If the line
is not vertical then from the first equation the slope is m = and from the second equation the
slope ism= f; thus k = l. In summary, we conclude that c = d and k = .l; thus the two equations
are identicaL
08. (a) True. If there are n ~ 2 columns, then the first n-1 columns correspond to the coefficients
of the variables that appear in the equations ar1d the last column corresponds to the constants
that appear on the right-hand side of the equal sign.
{b) False. Referring to Example 6: The sequen_c~ of linear systems appearing in the left-hand
column all have the same solution set, but the corresponding augmented matrices appearing
in the right-hand column are all different. ·
(c) False. Multiplying a row of the augmented matrix by zero corresponds to multiplying··both
sides of the corresponciing equation by zero. But this is equivalent to discarding one of the
equations!
(d) True. If the system is consistent, one can solve for two of the variables in terms of the third
or (if further redundancy is present) for one of the variables in terms of the other two. ln any
case, there is at lca.c:;t one "free" variable that ca.n be made into a parameter in describing
the solution set of the system. Thus if the system is consistent, it will have infinitely many
solutions.
D9. (a) True. A plane in 3-space corresponds to a linear equation in three variables. Thus a set of
four planes corresponds to a system of four linear equations in three variables. If there is
enough redundancy in the equations so that the system reduces to a system of two indepen
dent equations, then the solution set will be a line. For example, four vertical planes each
containing the z-axis and intersecting the xy-plane in four distinct lines.
(b) False. Interchanging the first two columns corresponds to interchanging the coefficients of
the first two variables. This results in a different system with a differeut solution set. [It
is oka.y to interchange rows since this corresponds to interchanging equations and therefore
does not alter the solution set.]
(c) False. If there Is enough redundancy so that the system reduces to a system of only two
(or fewer) equations, and if these equations are consistent, then the original system will be
consistent.
= = =
(d) True. Such a system will always have the trivial solution x1 x2 = · · · Xn 0.
28 Chapter2
EXERCISE SET 2.2
1. The matrice1' (a), (c), and (d) are in reduced row echelon form. The matrix (b) does not satisfy
property 4 of the definition, and the matrix (e) does not satisfy property 2.
2. Tbe mo.trices (c), {d), an(l {e) are in reduced row echelon form. The matrix (a) does not satisfy
property 3 of the definition, and the matrix {b} does not satisfy property 4.
3. The matrices (a) and (b) are in row echelon form. The matrix (c) does not satisfy property 1 or
property 3 of the definition.
4. The matrices (a) and {b) are in row echelon form. The matrix (c) does not satisfy property 2.
5. The matrices (a) and (c) are in reduced row echelon form. The matrix (b) does not satisfy property
3 and thus ls not in row echelon form or reduced row echelon form.
6. The matrix (c) is in reduced row echelon form. The matrix (a) is in row echelon frem but does
not satisfy property 4. The matrix (b) does not satisfy property 3 and thus is not in row echelon
form or reduced row echelon form.
[8 g). [8 A]. [6 [A o]
7. The pos:;ible 2 by 2 reduced row echelon forms are ~), and with any real
number substituted for the *.
8. The possible 3 by 3 reduced row echelon forms are
. !] . !
[~ ~ ~]. [~ ~ ~] [~ ~ [~ ~]
and
:]
[~ ~ ~]·[~ ~ ~] [~ ~ ~]·[~ ~
.o
0 0 0 0 0 0 0 0 0 0 0
with any real numbers substituted for the *'s.
9. The given matrix corresponds to the system
X1 = -3
Xz 0
X3 = 7
which clearly has the unique solution x1 = -3, x2 = 0, X3 = 7.
10. The given matrix corresponds to the system
x1 + 2x2 + Zx4 = - 1
X3 + 3x4 = 4
Solving these equations for the leading variables (x and x3) in terms of the free variables (x2 and
1
= =
x4) results in x1 -1- 2x2 - 2x4 and x3 4- 3x4. Thus, by assigning arbitrary values to x2
and x4, the solution set of the system can be represented by the parametric equa.tious
Xt = -1- 28-2t, X2 = S, X3 = 4- 3t, X4 = t
where - oo < s, t < S>O· The corresponding vector form is
(x~. x2, X3, X4) = (-1, 0, 4, 0} + s( -2, 1, 0, G)+ t( -2,0, -3, 1)
Exercl&e Set 2.2 29
11. The given matrix corresponds to the system
+ 3xs = -2
+ 4xs = 7
X4 + 5xs = 8
where the equation corresponding to the zero row has been omitted. Solving these equations
for the leading variables (x1, xs, and x4) in terms of the free variables (x2 and xs) results in
Xt = -2 + 6x2-3xs, X3 = 7-4xs and X4 = 8-5xs. Thus, assigning arbitrary values to x2 and
xs, the solution set can be represented by the parametric equations
X] = -2 + 6s-3t, X2 = S, X3 = 7-4t, X4 = 8-5t, X5 = t
where -oo < s, t < oo. The correspondjng vector form is
12. The given matrix corresponds to the system
x1 - 3x2 = 0
X3 = 0
0=1
which is clearly inconsistent since the last equation is not satisfied for any values of x1, x2, and x3.
13. The given matrix corresponds to the system
- 7x4 = 8
+ =
3x4 2
X3 + X4 = -5
Solving these equations for the leading variables in terms of the free variable results in x1 == 8 +· 7 x4,
x2 =o 2- 3x1, and x3 = -5 - x4. Thus, making x4 into a parameter, the solution set of the system
can be represented by the parametric equations
X1 = 8 + 7t, Xz == 2- 3t, X3 = -5- t, X4 = t
where -oo < t < oo. The corresponding vector form is
14. The given matrix corresponds to the single equation Xt + 2xz + 2x.t - x5 = 3 in which x3 does n0t
appear. Solving for x1 in terms of the other variables results in x1 = 3-2xz- 2x4 + X5. Thus,
making x2, x3, x4, and Xs into parameters, the solution set of the equation is given by
Xt ::::: 3 - 2s - 2u + V, X2 = S, X3 = t, X4 = u, X5 = V
where -oo < s, t, u, v < :x>. The corresponding (column) vector form is
r
XI 3-2s-2u+v 3 -2 0 l-~
r~
X2 s 0 1 0 !
X3 = 0 +s 0 +t 1 +.
X4 u 0 0 0 +u
l~
xs v 0 0 0
3C Chapter2
15. The system of equations corresponding to the given matrix is
x1 - 3xz + 4x3 = 7
x2 + 2x3 = 2
X3 = 5
Starting with the last equation and W'orking up, it follows that Z3 == 5, Xz = 2 - 2x3 = 2 - 10 = - 8,
and x1 = 7 + 3x2- 4x3 = 7-24-20 = -37.
Alternate solution via Gauss-Jordan (starting from the original matrix and reducing further):
1 -3 4
0 1 2
[
0 0 1
Add - 2 times row 3 to row 2. Add -4 times row 3 to row 1.
0 -13]
0 --8
l 5
Add 3 times row 2 to row 1. 0 -37]
0
1 0 - 8
0 1 5
=
From this we conclude (as before) that x1 -37, x2 = - 8, and x3 = 5.
16. The system of equations corresponding to the given matrix is
x + 8x3 - S:c4 = 6
1
x2 + 4x3 - 9x.: = 3
X3 + X4 = 2
Starting with the last equation and working up, we have .1:3 = 2 - x4, x2 = 3 - 4x3 + 9x4 =
3-4(2- .r4) + 9x4 = -5 + 13x4, and x1 = 6-8xa + 5x4 = 6-8(2- x4) + 5x4 = .;:::10 + 13x4·
Finally, assigning an arbitrary value to x4, the solution set can be described by the paramet
ric equations x = -10 + 13t, x = - 5 + 13t, x = 2- t, x = t.
1 2 3 4
Alternate solution via Gauss-Jordan (starting from the original matrix and reducing further):
0 8 -5
[~ ~]
1 4 - 9
0 1 l
Add - 4 times row 3 to row 2. Add -8 times row 3 to row 1.
-10]
0 0 -13
[~
1 0 -13 -5
0 1 1 2
From this we conclude (as before) that x1 = -10 + 13t, x2 ~ -5 + 13t, x3 = 2- t, x4 = t.
Exercise Set 2.2 31
17. The corresponding system of equations is
x1 + 7xz - 2x3 - 8xs = - 3
X3 + x4 + 6xs = 5
+
X4 3xs ,... 9
St<Uting with the last equation aml working up, it follows that x" = 9- 3xs, X3 = 5- :t4 - 6xs =
=
5-(9- 3xs}-6xs = -4- 3xs, and Xt = - 3-1x2 + 2x3 + 8xs - 3-7x'l + 2( -4-3xs) + 8xs =
-11 - ?x2 + 2xs. Fina.lly, assigning arbitrary 'Values to xz and :ts, the solution set can be described
by
Xt = -11- 7s + 2t, Xz = S, X3 = - 4-3t, X4 = 9 - 3t, Xs = t
18. The corresponding system
X1 - 3x2 + 7XJ = 1
+ =
:tz 4x3 0
0 = 1
is inconsistent since there are no values of x1, x2, and x3 which satisfy the third equation.
10. The corrcspouding system is
x1 + x2 - Sx3 + 2x4 = 1
X2 + 4x3 = 3
X4 = 2
Starting with the last equation, we have x4 = 2, x2 = 3 - 4x3. X1 = 1 - x2 + 3x3 - 2x4 =
l - (3- 4x3) + 3x3 - 2{2) = - 6 + 7x3. Thus, making x3 inLo a parameter, the solution set can be
described by the p<Uame~t:ic equations
Xl = - 6 + 7t, X2 = 3 - 4t, X:~ = t, X4 = 2
20. The correspondi.ng system is
x1 + 5x3 + 3x4 = 2
x2 - 2x3 + 4x4 = - 7
X3 + X4 = 3
Thus x4 is a free. vc:~.riable and, setting x4 = t, we have :r:, = 3- t, :r.2 = - 7 + 2(3 - t) --4t =
- l - 6l, a.nd x = 2- 5(3- t)- 3t = - 13 + 2t.
1
= = Hs - = =
21. Starting with the first equation and working down, we have x1 2, xz x1) S(S- 2) l,
= !-
and X3 = l(12 - J:r1 - 2:r2) i(12-6- 2} =
4 - 2x 4 + 2
22. :r1=- l,x2 = 1 = - -=2,x3=5-x - 4x2=5+1-8 = - 2
1
3 3
23. The augmented matrix of the system is
Add row 1 to row 2. Add -3 times row 1 to row 3.
[ 1
~
-1
-10