Solution manual for linear algebra for engineers and scientists using matlab by hardy
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Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy
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Chapter 1
Linear Systems
1.1
Solving Linear Systems
1.
(x + 2)2 + (y − 1)2 = x2 + 4x + 4 + y 2 − 2y + 1 = 3 + x2 + y2 ⇒ 4x − 2y = −2 ⇒ 2x − y = −1.
Linear
2.
(x − y)2 = x2 − 2xy + y 2 = x2 + y2 ⇒ −2xy = 0 ⇒ xy = 0. Nonlinear.
3.
(x − y)(x + y) = x2 + xy − xy − y 2 = x2 + y2 ⇒ 2y 2 = 0 ⇒ y = 0. Linear.
4.
x + (x − 1)(y − 1) = x + xy − x − y + 1 = xy + y − 1 ⇒ −2y2 = −2 ⇒ y = 1. Linear.
5.
Pivot variables: x1 , x2 , x3 . Free variable: x4 . No variables uniquely determined. Let x4 = t,
where t is a real parameter. Back-substituting, third equation: x3 = 3 − 2t, second equation: x2 =
−1+x3 −2x4 = −1+(3−2t)−2t = 2−4t, first equation: x1 = 2x2 +10x4 = 2(2−4t)+10t = 4+2t.
Solution: (x1 , x2 , x3 , x4 ) = (4 + 2t, 2 − 4t, 3 − 2t, t).
6.
Pivot variables: x1 , x2 , x4 . Free variable: x3 . Uniquely determined: x4 . Let x3 = t, where t is a
real parameter. Back-substituting, third equation: x4 = 2.4, second equation: x2 = 1 + x3 − 7x4 =
1+t−7(2.4) = −15.8+t, first equation: x1 = 2−2x2 −x3 +x4 = 2−2(−15.8+t)−t+2.4 = 36−3t.
Solution: (x1 , x2 , x3 , x4 ) = (36 − 3t, −15.8 + t, t, 2.4).
7.
Pivot variables: x1 , x3 . Free variables: x2 , x4 . No variables uniquely determined. Let x2 = s and
x4 = t, where s, t are real parameters. Back-substituting, second equation: x3 = −1.5 + 0.5t, first
equation: x1 = 4 − 0.5s. Solution: (x1 , x2 , x3 , x4 ) = (4 − 0.5s, s, −1.5 + 0.5t, t).
8.
Pivot variables: x1 , x3 , x4 . Free variables: x2 , x5 . Uniquely determined: x3 . Let x2 = s and
x5 = t, where s, t are real parameters. Back-substituting, third equation: x4 = 2.0 − t, second
1
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ISM
Chapter 1 Linear Systems
equation: x3 = 2.1, first equation: x1 = 7.0+ x2 −2x3 −x4 = 7.0+s−2(2.1)− (2.0− t) = 0.8+s+t.
Solution: (x1 , x2 , x3 , x4 , x5 ) = (0.8 + s + t, s, 2.1, 2.0 − t, t).
9.
0
3x1
= 6.0 E1
L1
x1 − x2 = 5.5 E2
L2
−2
Solve E1 : x1 = 2.0. Solve E2 : −x2 = 5.5 − x1 =
3.5 ⇒ x2 = −3.5. Solution: (x1 , x2 ) = (2, −3.5).
−3.5
−4
Refer to Figure 1.1 [9]. Equations E1 , E2 are represented, respectively, by lines L1 , L2 in R2 .
−6
−1
0
2
4
6
Figure 1.1 [9]
10.
First equation: x1 = 32 , second equation: x2 = −x1 = − 23 , third equation: x3 = 1 − 3x1 + 3x2 =
1 − 2 − 2 = −3. Solution: (x1 , x2 , x3 ) = ( 32 , − 32 , −3).
Geometrically, the first equation is represented by a plane P1 in R3 parallel to the x2 x3 -plane, the
second equation is represented by a plane P2 passing through the x3 -axis, perpendicular to the
x1 x2 -plane, the third equation is represented by a plane P3 that intercepts the x1 -, x2 -, x3 -axes
at x1 = 13 , x2 = − 13 , x3 = 1, respectively. Planes P1 , P2 intersect in a line L and L intersects P3
in the point ( 23 , − 23 , −3).
11.
2
3
−1
1
Let (S) denote the linear system. M = [A| b] = 4
7
1
3 . In order to preserve
7 10 −4
4
integer coefficients, perform the following EROs on M: r2 − 2r1 → r2 , r3 − 3r1 → r3 , r1 ↔ r3 ,
r3 − 2r1 → r3 , r3 − r2 → r3 , to obtain an echelon form U for M and a linear system (S)′ equivalent
to (S).
x + x2 − x3 = 1
1
1 1 −1
1
x2 + 3x3 = 1
(S)′
3
1 = U
M s 0 1
0 0 −2
−2
− 2x3 = −2
Back-substituting, third equation: x3 = 1, second equation: x2 = 1 − 3x3 = −2, first equation:
x1 = 1 − x2 + x3 = 4. Solution: (x1 , x2 , x3 ) = (4, −2, 1). (S) is represented by three planes in R3
with a single point in common.
12.
3
3 1
−4.5
Let (S) denote the linear system. M = [A| b] = 1
1 1
0.5 . Perform the following
−2 −2 0
5.0
EROs on M: r1 ↔ r2 , r2 − 3r1 → r2 , r3 + 2r1 → r3 , r3 + r2 → r3 , to obtain an echelon form U
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1.1
for M and a linear system (S)′ equivalent to (S).
1 1
1
0.5
M s 0 0 −2
−6.0 = U
0 0
0
0
Solving Linear Systems 3
x + x2 + x3 = 0.5
1
(S)′
− 2x3 = −6.0 ,
0x1 + 0x2 + 0x3 =
0
the zero equation occuring in (S)′ because some equation in (S) is a linear combination of other
equations in (S). Pivot variables: x1 , x3 . Free: x2 . Uniquely determined: x3 . Let x2 = t be a real
parameter. Back-substituting, second equation: x3 = 3.0, first equation: x1 = 0.5 − x2 − x3 =
−2.5 − t. Solution: (x1 , x2 , x3 ) = (−2.5 − t, t, 3.0). (S) is represented by three planes in R3 that
intersect in a line.
13.
1
Let (S) denote the linear system. M = [A| b] = 3
1
13 . Perform the following
2 −3
6
1
9
4 8 −2
EROs on M: r2 − 3r1 → r2 , r3 − 4r1 → r3 , r3 − r2 → r3 , to obtain an echelon form U for M and
a linear system (S)′ equivalent to (S).
x + 2x2 − 3x3 =
1
1
1 2 −3
1
′
(S)
10 = U
10x3 = 10
M s 0 0 10
0 0
0
−5
0x1 + 0x2 + 0x3 = − 5 E3
Equation E3 in (S)′ indicates that (S) is inconsistent because (S)′ s (S). The solution set is
empty. (S) is represented by three planes in R3 that have no point in common.
14.
−3
0 1 −5
. Perform the following
1
1 3
0
EROs on M: r1 ↔ r2 , to obtain an echelon form U for M and a linear system (S)′ equivalent to
(S).
x1 + 3x2
= 1
1 3
0
1
=U
(S)′
Ms
0 1 −5
−3
x2 − 5x3 = −3
Let (S) denote the linear system. M = [A| b] =
Pivot variables: x1 , x2 . Free variables: x3 . Let x3 = t be a real parameter. Back-substituting,
second equation: x2 = −3 + 5t, first equation: x1 = 1 − 3x2 = 1 − 3(−3 + 5t) = 10 − 15t. Solution:
(x1 , x2 , x3 ) = (10 − 15t, −3 + 5t, t). (S) is represented by two (nonparallel) planes in R3 that
intersect in a line.
15.
5
3
Let (S) denote the linear system. M = [A| b] =
1
−2
2
1
7
2
3
. In order to preserve
2
1
3
9
1 11
7
integer coefficients, perform the following EROs on M: r1 ↔ r3 , r2 − 3r1 → r2 , r3 − 5r1 → r3 ,
r4 − 9r1 → r4 , r3 − 7r2 → r3 , r4 − 8r2 → r4 , r4 − r3 → r4 , to obtain an echelon form U for M
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ISM
Chapter 1 Linear Systems
and a linear system (S)′ equivalent to (S).
1
1
3
2
0 −1 −2
Ms
0
0
0
0
0
−3
=U
13
0
(S)′
0
x1 + x2 + 3x3 =
2
− x2 − 2x3 = − 3
0x1 + 0x2 + 0x3 =
0x1 + 0x2 + 0x3 =
13 E3
0 E4
Equation E4 in (S)′ shows that some equation in (S) is a linear combination of other equations in
(S) and E3 indicates that (S) is inconsistent because (S)′ s (S). (S) is represented by four planes
in R3 with no point in common.
16.
1 −1
1
1
3 −1
4 . Perform the following
Let (S) denote the linear system. M = [A| b] = 2
2
−1 −2
5
EROs on M: r2 − 2r1 → r2 , r3 + r1 → r3 , r3 − 35 r2 → r3 , to obtain an echelon form U for M and
a linear system (S)′ equivalent to (S).
x1 − x2 + x3 = 1
1 −1
1
1
′
5x2 − 3x3 = 2
(S)
M s 0
5 −3
2 = U
21
21
21
21
0
0
5
5
5 x3 = 5
Pivot variables: x1 , x2 , x3 . Free variables: none. All variables uniquely determined. Backsubstituting, third equation: x3 = 1, second equation: x2 = 51 (2 + 3x3 ) = 15 5 = 1, first equation:
x1 = 1 + x2 − x3 = 1 + 1 − 1 = 1. Solution: (x1 , x2 , x3 ) = (1, 1, 1). (S) is represented by three
planes in R3 with a single point in common.
17.
Let (S) denote the linear system. Then (S) is 3 × 4 and will be either inconsistent or consistent
with at least one free variable, indicating infinitely many solutions. Finding the reduced form M*
of M = [A| b], we have
1
4 −4
4
5
1
1 0
0 0
M = 2 −1
1 −1
1 s 0 1 −1 1
1 = M*
1
1
−1
1
2
0
0
0 0
0
Pivot variables: x1 , x2 . Free variables: x3 , x4 . Uniquely determined: x1 . Let x3 = s and x4 = t,
where s, t are real parameters. Then, reading the solution off from M*, we have x1 = 1, x2 =
1 + s − t. Solution: (x1 , x2 , x3 , x4 ) = (1, 1 + s − t, s, t).
18.
Let (S) denote the linear system. Then (S) is 4 × 3 and any one of the three possibilities for
solution may occur. Finding the reduced form M* of M = [A| b], we have
1 −2 1
2
1 0 0
0
2 −5 3
6
0
0 1 0
M=
s
= M*
1
2 2
4 0 0 1
2
2
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0 3
6
0 0
0
0
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1.1
Solving Linear Systems 5
Pivot variables: x1 , x2 , x3 . Free variables: none. All variables uniquely determined. Reading the
solution off from M*, we have x1 = x2 = 0, x3 = 2. Solution: (x1 , x2 , x3 ) = (0, 0, 2).
19.
Let (S) denote the linear system. Then (S) is 2 × 4 and will be either inconsistent or consistent
with at least one free variable, indicating infinitely many solutions. Finding the reduced form M*
of M = [A| b], we have
M=
−1
1
3
−1
−4
1
1
2
−1
3
s
0 − 21
1
1 − 32
0
11
2
5
2
−1
0
= M*
Pivot variables: x1 , x2 . Free variables: x3 , x4 . Uniquely determined: none. Let x3 = s and x4 = t,
where s, t are real parameters. Then, reading the solution off from M*, we have x1 = 5.5+0.5s+t,
x2 = 2.5 + 1.5t. Solution: (x1 , x2 , x3 ) = (5.5 + 0.5s + t, 2.5 + 1.5s, s, t).
20.
Let (S) denote the linear system. Then (S) is 4 × 5 and will be either inconsistent or consistent
with at least one free variable, indicating infinitely many solutions. Applying forward reduction
on M = [A| b] using the sequence of EROs shown below:
Clear below in column 1
r2 − 2r1 → r2 , r4 − r1 → r4
Clear below in column 3
r2 ↔ r3 , r3 − 2r2 → r3 , r4 + r2 → r4
Clear below in column 5
− 14 r3 → r3 , r4 + 2r3 → r4
and we have
1 2
2 4
M=
0 0
1 2
0
2
2
2
1
−1
−1
3
2
−2
−1
1
10
1
14 0
s
−1 0
3
0
2
0
0
0
0
−1
=U
1
0
2 0
1 −1 0
0
0
0 1
0 0
−6
and the last row in U indicates that (S) is inconsistent.
21.
Finding the reduced form M* of M = [A| b], we have
3
1 −2
0
3
1 0
M = 1
2
1 −2
4 s 0 1
0 −5
0
1
3
0 0
0 −0.6
0 −0.2
1
−1
2.8
−0.6 = M*
2.4
Pivot variables: x1 , x2 , x3 . Free variable: x4 . Let x4 = t, where t is a real parameter. Then,
reading the solution off from M*, we have x1 = 2.8 + 0.6t, x2 = −0.6 + 0.2t. General solution:
(x1 , x2 , x3 , x4 ) = (2.8 + 0.6t, −0.6 + 0.2t, 2.4 + t, t). We have x3 = 2.4 + t = 0 ⇒ t = −2.4 and
substituting gives the particular solution: (x1 , x2 , x3 ) = (1.36, −1.08, 0, −2.4).
22.
Free variables: x2 , x3 . Let x2 = s and x3 = t, where s, t are real parameters. Then x1 =
4 − 2s − 3t. General solution: (x1 , x2 , x3 ) = (4 − 2s − 3t, s, t). Setting s = 2 and t = −1
gives x1 = 4 − 4 + 3 = 3, showing that (3, 2, −1) is a particular solution. Setting x1 = p and
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Chapter 1 Linear Systems
x3 = q gives x2 = 2 − p − 32 q, giving the general solution (x1 , x2 , x3 ) = (p, 2 − 21 p − 32 q, q). For
any particular solution corresponding to the ordered pair (s, t), for fixed s, t, the unique ordered
pair (p, q) = (4 − 2s − 3t, t) corresponds to the same particular solution. Conversely, for any
particular solution corresponding to the ordered pair (p, q), for fixed p, q, the unique ordered pair
(s, t) = (2 − 21 p − 23 q, q) corresponds to the same particular solution.
23.
2 1
M = [A| b] = −1 1
28
−8 , where A is 3 × 3, M is 3 × 4 and b is 3 × 1. Solving the
13
−2
1 1
8
16
system by Gauss–Jordan elimination corresponds to reducing M to its reduced form M*.
1 0 5
12
4
M s 0 1 3
General solution: (x1 , x2 , x3 ) = (12 − 5t, 4 − 3t, t),
where t is a real parameter.
0 0 0
0
Particular solutions in nonnegative integers must satisfy x1 = 12 − 5t ≥ 0, x2 = 4 − 3t ≥ 0 and
4
t ≥ 0. Hence 12
5 ≥ t ≥ 0 and 3 ≥ t ≥ 0 so that 1 ≥ t ≥ 0, where t is an integer. Particular
solutions: (x1 , x2 , x3 ) = (12, 4, 0), (7, 1, 1).
24.
7 14 −2 5
3 6 −1 2
15
, where A is 2 × 4, M is 2 × 5 and b is 2 × 1. Solving the
6
system by Gauss–Jordan elimination corresponds to reducing M to its reduced form M*.
M = [A| b] =
Ms
1
0
2 0
0 1
1
1
3
3
General solution:
(x1 , x2 , x3 , x4 ) = (3 − 2s − t, s, 3 − t, t), where s, t
are real parameters.
Particular solutions in nonnegative integers must satisfy x1 = 3 − 2s − t ≥ 0, x3 = 3 − t ≥ 0, s ≥ 0,
and t ≥ 0. Hence 3 ≥ t ≥ 0 ⇒ t = 0, 1, 2, 3. Particular solutions (x1 , x2 , x3 , x4 ) are:
t = 0,
t = 1,
t = 2,
t = 3,
25.
x1
x1
x1
x1
= 3 − 2s ≥ 0 ⇒ 32 ≥ s ≥ 0
= 2 − 2s ≥ 0 ⇒ 1 ≥ s ≥ 0
= 1 − 2s ≥ 0 ⇒ 12 ≥ s ≥ 0
= − 2s
⇒
⇒
⇒
⇒
s = 0, 1,
s = 0, 1,
s = 0,
s=0
⇒
⇒
⇒
(3, 0, 3, 0), (1, 1, 3, 0)
(2, 0, 2, 1), (0, 1, 2, 1)
(1, 0, 1, 2)
(0, 0, 0, 3).
Let (S) denote the system. First, assume α = 0 and apply forward reduction on M = [A| b].
α 1
1 α
1
1
1
1
α
1
1
1
s
r1 ↔ r3
r2 − r1 → r2
r3 − αr1 → r3
s
r3 + r2 → r3
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1
0
1
α−1
0
1−α
1
1
0
0
1
0
α
1−α
1−α
2
1−α
α
α−1
1−α
0
(1 − α)(2 + α)
(1.1)
1
0
1−α
(1.2)
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1.1
Solving Linear Systems 7
If α = 1, then the right side of (1.1) shows that (S) has infinitely many solutions. Now assume
α = 1 and reduce further to obtain (1.2), noting that 1 − α2 + 1 − α = 2 − α − α2 = (1 − α)(2 + α).
If α = −2, then (S) has a unique solution because A s I3 and if α = −2, the (S) is inconsistent.
If α = 0, then the reduced form of M is
1 0 0
0.5
M* = 0 1 0
0.5
0
0 1
0.5
Summary. (a) α = −2, no solutions. (b) α = 1 and α = −2, unique solution. (c) α = 1, infinitely
many solutions.
26.
Let (S) denote the system. Apply forward reduction
1 2 2
1
s
1
0 1 α
r3 + r1 → r3
−1 1 α
α
s
r3 − 3r2 → r3
on M = [A| b].
1
1 2
2
α
1
0 1
0 3 α+2
α+1
1
1 2
2
α
0
0 1
0 0 2(1 − α)
α−2
(a) α = 1, no solution, (b) α = 1, unique solution, (c) No value α gives infinitely many solutions.
27.
Apply forward and backward reduction on the augmented matrix M = [A| b] of (S).
1 1 −1
1
−1
1
1 1 0 0
M = 1 1
0 −1
0 s 0 0 1 0
3 = M*
2 2 −2
5
1
0 0 0 1
1
Pivot variables: x1 , x3 , x4 . Free variable: x2 . Let x2 = t, where t is a real parameter. Using M*,
the general solution is: (x1 , x2 , x3 , x4 ) = (1 − t, t, 3, 1). Any particular solution in positive integers
must satisfy x1 = 1 − t > 0, where t is an integer. But then 1 > t and so there are none.
28.
Substituting x1 = −1, x2 = 1, x3 = 0 into E1 gives −3(−1) + 5(1) − 7(0) = 8 = −2. The particular
solution fails to satisfy E1 and so is not a particular solution to (S).
Apply forward and backward reduction on the augmented matrix M = [A| b] of (S).
M=
−3
5 −7
4 −6
8
−2
2
s
1
0
0 −1
1 −2
−1
−1
= M*
Pivot variables: x1 , x2 . Free variable: x3 . Let x3 = t, where t is a real parameter. Using M*, the
general solution is: (x1 , x2 , x3 ) = (−1 + t, −1 + 2t, t). For a particular solution in positive integers
we must have: x1 = −1 + t > 0 ⇒ t > 1 and x2 = −1 + 2t > 0 ⇒ t > 0.5. Hence t = 2, 3, . . .
determines all particular solutions in positive integers. The condition x3 = t < 3 ⇒ t = 2 giving
the single solution (1, 3, 2).
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29.
ISM
Chapter 1 Linear Systems
M is 4 × 4, A is 4 × 3. (S)
x1 + 2x2 + 4x3 =
2x1 + 4x2 + 8x3 =
3
6
−x1 + 6x2 + 8x3 = −15
4x1 + 2x2 + 7x3 = 21
Perform forward reduction on M, maintaining integer coefficients.
1
2
M=
−1
4
3
6
−15
2 4
4 8
6 8
2 7
21
s
r2 − 2r1 → r2
r3 + r1 → r3
r4 − 4r1 → r4
s
r2 ↔ r4
− 31 r3 → r3
r3 − 4r2 → r3
1 2
0 2
0 0
1
0
0
2
0
4
0
8
0 −6
12
−9
0 0
4
3
0
3
0
−12
9
3
−3
=U
0
0
0
Pivot variables: x1 , x2 . Free variables: x3 . Let x3 = t, where t is a real parameter. Reading
off solutions from U, we have 2x2 = −3 − 3t ⇒ x2 = −1.5 − 1.5t and x1 = 3 − 2x2 − 4t =
3 − 2(−1.5 − 1.5t) − 4t = 6 − t. General solution: (x1 , x2 , x3 ) = (6 − t, −1.5 − 1.5t, t).
30.
M is 3 × 6 and A is 3 × 5. (S)
x1 − x2
+ x4 + x5 = 2
−x1 + 2x2 − x3 + x4 + x5 = 2
5x1 − 8x2 + 3x3 − x4 − x5 = 0
Perform forward reduction on M: Eliminate below
eliminate below in column 2: r3 + 3r2 → r3 .
2
1 −1
0
1
1
2 −1
1
1
2 s
M = −1
5 −8
3 −1 −1
0
s
Row 3 of U show that (S) is inconsistent.
31.
in column 1: r2 + r1 → r2 , r3 − 5r1 → r3 ,
1
0
0
1
0
0
−1
(S)
Perform forward reduction on M.
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2x2 + 2x3
1
1
−3
−1
3
2
−6
2
−6
−1
0
1 1
2
−1
0
2 2
0 0
4 = U
2
1
0
= 0
3x3 + 3x4 = 3
x1
−5x1 − 5x2
1
M is 4 × 5, A is 4 × 4. We have
2
0
− x4 = 2
= 15
.
4
−10
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Solving Linear Systems 9
1.1
0
2 2
0
0
M=
1
0 3
0 0
3
−1
−5 −5 0
0
0
3
2
15
s
r1 ↔ r3
r4 + 5r1 → r4
r2 ↔ r3
1
0
0
0
2
0
0
1
0
0
0
s
r4 + 52 r2 → r4
r4 − 53 r3 → r4
−5
0 0
2 2
0 3
0 0
0 −1
2
0 −5
25
2
0
3
2
3
0
3
0
3
−1
0
3
−10
20
Back-substitution: −10x4 = 20 ⇒ x4 = −2, 3x3 = 3 − 3x4 = 3 − 3(−2) = 9 ⇒ x3 = 3,
2x2 = −x3 = −3, x1 = 2 + x4 = 0. General solution: (x1 , x2 , x3 , x4 ) = (0, −3, 3, −2).
32.
x1 − x2
M is 2 × 5, A is 2 × 4. (S)
+ x4 + x5 = 2
−x1 + 2x2 − x3 + x4 + x5 = 2
Performing forward reduction using r2 + r1 → r2 , we have
M=
1
−1
−1
0 1
2 −1 1
1
1
2
2
s
1
0
−1
1
0 1
−1 2
1
2
2
4
Pivot variables: x1 , x2 . Free variables: x3 , x4 , x5 . Let x3 = r, x4 = s, x5 = t, where r, s, t
are real parameters. Back-substituting, we have x2 = 4 + x3 − 2x4 − 2x5 = 4 + r − 2s − 2t,
x1 = 2 + x2 − x4 − x5 = 2 + (4 + r − 2s − 2t) − s − t = 6 + r − 3s − 3t. General solution:
(x1 , x2 , x3 , x4 , x5 ) = (6 + r − 3s − 3t, 4 + r − 2s − 2t, r, s, t). Note that (S) is a subsystem of the
linear system in Exercise 30. This problem illustrates the fact that deleting a single equation from
an inconsistent linear system may (or may not) result in a consistent subsystem. See Exercise 60.
33.
Let (S) denote the linear system. Performing forward and backward elimination on (S) using
augmented matrices, we have
3
1
M = −3
2
5
−9
16
2
6
16
s
r2 + 3r1 → r2
r3 − 2r1 → r3
s
r3 + r2 → r3
1
6 r2 → r2
r1 − 5r2 → r1
1
0
0
1
0
0
5
6
6
0
1
0
2
12
12
−8
2
Unique solution: (x, y) = (−8, 2). Refer to Figure 1.1 [33].
Equations E1 , E2 , E3 in (S) are represented, respectively,
by lines L1 , L2 , L3 in R2 that intersect in a single point.
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0
2
L3
1
L2
L1
0
−1
−10
−8
−6
−4
−2
Figure 1.1 [33]
0
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Chapter 1
34.
Rearranging the system into standard form, we have
2x + 4y = − 6 E1
. Perform forward and back(S)
7x + 5y = 6 E2
ward elimination using augmented matrices. The EROs
are chosen to preserve integer coefficients.
M=
ISM
Linear Systems
2 4
7 5
−6
6
0
L2
−1
s
r2 − 3r1 → r2
r1 ↔ r2
s
1
r
18 2 → r2
r1 + 7r2 → r1
1 −7
0
18
24
−54
−2
L1
1 0
0 1
3
−3
−3
−4
0
Unique solution: (x, y) = (3, −3). Refer to Figure 1.1 [34].
Equations E1 , E2 in (S) are represented, respectively, by
lines L1 , L2 in R2 that intersect in a single point.
35.
2y =
The system is (S)
0
2
2 −4
s
r1 ↔ r2
r1 + 2r2 → r1
s
1
2 r1 → r1
1
2 r2 → r2
4
−6
The system is (S)
6x −
9y =
4
2
0
0
2
0
L1
2
3
4
L2
1
1
0
1.5
0
1
2
E1
. Perform for−x + 1.5y = 2.5 E2
ward and backward elimination using augmented matrices.
−1
2 −4
s
2.5
−1 1.5
3
2
M = 1
r1 ↔ r2
0
0
15
4
5
c r2 + 6r1 → r2
Row 2 in the last matrix shows that (S) is inconsistent.
Refer to Figure 1.1 [36]. Equations E1 , E2 in (S) are represented, respectively, by parallel lines L1 , L2 in R2 with
no point in common.
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3
3
0
0
Unique solution: (x, y) = (1.5, 2). Refer to Figure 1.1 [35].
Equations E1 , E2 in (S) are represented, respectively, by
lines L1 , L2 in R2 that intersect in a single point.
36.
2
4 E1
. Perform for2x − 4y = − 5 E2
ward and backward elimination using augmented matrices.
M=
1
Figure 1.1 [34]
1
2
1.5
Figure 1.1 [35]
3
L2
2
L1
1
0
0
1
2
Figure 1.1 [36]
3
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37.
1.1 Solving Linear Systems 11
1
Perform forward elimination using augmented matrices.
2 −4
M = 1
3
4
5
s
1 1.5
r1 ↔ r2
2
0
−10
r2 − 2r1 → r2
c
0 −7
r3 − 4r1 → r3
1
3
s
0 −10
7
r2 → r3
r3 − 10
0
0
−1
2.5
−5
c−8
2
L2
0.5
L1
−5 = U
c − 4.5
0
L3
0
0.5
1
Figure 1.1 [37]
(S) is consistent if and only if c = 4.5 and for this value of c the unique solution (x1 , x2 ) = (0.5, 0.5)
is obtained by performing backward elimination on the linear system with augmented matrix U.
Refer to Figure 1.1 [37]. Equations E1 , E2 , E3 in (S) are represented, respectively, by lines L1 ,
L2 , L3 in R2 intersecting in a single point.
38.
Perform forward elimination using augmented matrices.
1
M = −1
2
1
2
3
s
1 1
r1 ↔ r2
0 3
r2 + r1 → r2
c2
0 1
r3 − 2r1 → r3
1 1
s
0 3
r3 − 13 r2 → r3
0 0
1
2
1
3
c2 − 2
2
1
3 = U
c2 − 3
1.6667
L3
L2
1
0.5
L1
0
−1
0
1
Figure 1.1 [38]
√
(S) is consistent if and only if c = ± 3. For these values of c the unique solution (x1 , x2 ) = (1, 1)
is obtained by performing backward elimination on the linear system with augmented matrix U.
Refer to Figure 1.1 [38]. Equations E1 , E2 , E3 in (S) are represented, respectively, by lines L1 ,
L2 , L3 in R2 that intersect in a single point.
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Chapter 1
39.
Substitute x = 1, y = 2 and x = −1, y = 0.5 into mx+ b =
y to obtain a 2 × 2 linear system
m+b= 2
(S)
−m + b = 0.5
Perform forward and backward elimination on (S) using
augmented matrices.
M=
1
−1
ISM
Linear Systems
1
1
2
0.5
s
r2 + r1 → r2
1
2 r2 → r2
r1 − r2 → r1
1
0
0
1
0.75
1.25
2
L
1.25
=U
0
−2 −1.6
−1
The equation of the line L is y = 0.75x + 1.25.
40.
0
1
Figure 1.1 [39]
Substitute x = −1, y = 1, z = 0 and x = 0, y = 2, z = 1 and x = 4, y = 5, z = 2 into ax+by +c = z
to obtain a 3 × 3 linear system
−a + b + c = 0
(S)
b+c=1
4a + 5b + c = 2
We will solve by Gauss-elimination beginning with
nation:
s
0 r3 + 4r1 → r3
−1 1 1
1
− r1 → r1
0 2 1
1
4 5 1
2
2 r2 → r2
r3 − 9r2 → r3
the augmented matrix for (S). Forward elimi-
1
0
0
−1
1
0
−1
0
0.5 = U
−2.5
0.5
0.5
4
− 2.5
0.5
0
z-values
=
Back-substitution: Using U, we have c =
−5, b = 0.5 − 0.5c = 0.5 + 0.5(−5) = 3, a =
b + c = 3 − 5 = −2. The equation of the plane P is
z = −2x+3y −5, alternatively, 2x−3y +z −5 = 0.
Refer to Figure 1.1 [40]. The plane intercepts the
z-axis in the point (0, 0, −5). The point (3, 0, −11)
lies in P and in the xz-plane. The point (0, 3, 4)
lies in P and in the yz-plane.
−5
P
−11
0
0
1
y-values
1
2
2
3
3
Figure 1.1 [40]
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x-values
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42.
The values x1 = 0.7, x2 = 0 do not satisfy equation E1
and so (0.7, 0) is not a particular solution to (S). Substituting the values x1 = 1.2, x2 = −1 into each equation
of (S) in turn, we have E1 : 1.2 + 1.5(−1) = −0.3, E2 :
2(1.2) + (−1) = 1.4, E3 : 3(1.2) + 2.5(−1) = 1.1. All equations in (S) are satisfied and so (1.2, −1) is a particular
solution. Refer to Figure 1.1 [41]. Equations E1 , E2 , E3
are represented, respectively, by lines L1 , L2 , L3 that intersect in the single point (1.2, −1) in R2 . It follows that
(1.2, −1) is the unique solution to (S) and there is only one
particular solution in this case.
0
L3
−1
L1
L2
−2
0
(b) Then (S) is consistent if and only if
c − 3a − b = 0, or alternatively, −3a − b + c = 0.
When this algebraic condition is met, (S) has
infinitely many solutions.
(c) Refer to Figure 1.1 [42]. The equation −3x−
y +z = 0 is represented by a plane P in R3 passing
through the origin (0, 0, 0). The coordinates of any
point (a, b, c) lying in P make (S) consistent, and
conversely, any values of (a, b, c) which make (S)
consistent correspond to a point on P .
Exercises 43-44. Denote the points by D, E, F . The perpendicular bisectors of the chords DE and EF meet at
a unique point C forming isosceles triangles △DEC and
△EF C with base angles α and β, respectively, and common side EC. Hence r = |DC| = |EF | = |F C| is the
radius of the circle passing through the three points.
Full file at />
1 1.2
2
Figure 1.1 [41]
(a) Performing forward elimination on (S) using augmented
a
1
2 3
1
s
M = 2
4 7
b
r2 − 2r1 → r2
0
r3 − 5r1 → r3
5 10 16
0
c
1
s
0
r3 − r2 → r3
0
matrices, we have
2
2 3
0 1
b − 2a
0 1
c − 5a
2
3
0
0
1
0
2
b − 2a
c − 3a − b
2
z-values
41.
1.1 Solving Linear Systems 13
0
P
−2
−1
−1
0
0
y-values
1
x-values
1
Figure 1.1 [42]
E
D
α
α
β
β
C
F
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Chapter 1
ISM
43.
Substitute (x, y) = (0, −1), (1, 0), (2, 2) into the equation ax + by + c = −(x2 + y 2 ) to obtain a 3 × 3
linear system
− b + c = −1
(S)
a
+ c = −1
2a + 2b + c = −8
Linear Systems
Performing forward and backward elimination on (S) using augmented matrices, we have
0
1
2
−1
0
1
1
2
1
−1
−1
−8
s
r1 ↔ r2
r3 − 2r1 → r3
− r2 → r2
0
0
−1
1
0
0
1
0
1
0
s
r3 − 2r1 → r3
s
r2 + r3 → r2
r1 − r3 → r1
0
2
−1
0
−1
1
−6
−1
1
−8
−1
1
7
−7 ,
0 0
1 0
0
−1
1
1
−1
0 1
−8
showing that (S) has a unique solution (a, b, c) = (7, −7, −8) giving x2 + y 2 + 7x − 7y − 8 = 0 as
the equation of the circle. Completing the square, we have
x2 + y 2 + 7x − 7y − 8
Hence p = −3.5, q = −3.5 and r2 =
44.
x+
7
2
2
=
x+
7
2
2
=
−
7
2
2
+ y−
65
so that r =
2
+ y−
7
2
2
−
7
2
2
−
7
2
2
−8
65
=0
2
65/2 ≃ 5.7.
Substitute (x, y) = (1, 1), (−1, 1), (−1, 2) into the equation ax + by + c = −(x2 + y2 ) to obtain a
3 × 3 linear system
a + b + c = −2
(S)
−a + b + c = −2
−a + 2b + c = −5
Performing forward and backward elimination
1 1 1
−2
1
s
−2
r2 + r1 → r2
−1 1 1
0
r3 + r1 → r3
−1 2 1
−5
0
s
1
1
2 r2 → r2
0
r3 − 3r1 → r3
0
− r3 → r3
Full file at />
on (S) using augmented matrices, we have
−2
1 1
2 2
−4
3 2
−7
s
1 1
−2
1 0 0
r2 − r3 → r2
1 1
−2
0
1 0
r1 − r3 → r1
0 1
1
0 0 1
r1 − r2 → r1
0
−3 ,
1
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1.1 Solving Linear Systems 15
showing that (S) has a unique solution (a, b, c) = (0, −3, 1) giving x2 + y2 − 3y + 1 = 0 as the
equation of the circle. Completing the square, we have
(x − 0)2 + y −
3
2
2
x2 + y2 − 3y + 1 =
(x − 0)2 + y −
3
2
2
=
Hence p = 0, q = 1.5 and r 2 =
45.
5
so that r =
4
2
−
3
2
−
5
=0
4
+1
5/4 ≃ 1.1180.
Denote the system by (S). Setting X = xy, Y = y2 , Z = yz in (S) gives a linear system (S)′ ,
where
X + Y + 2Z = 9
′
(S)
2X + 4Y − 3Z = 1
3X + 6Y − 5Z = 0
′
Forward and backward elimination on (S) using augmented matrices gives
1 1
2
9
1
1 0 0
1 s 0 1 0
2
⇒
X = xy = 1, Y = y 2 = 2, Z = zy = 3.
2 4 −3
3 6
−5
0
0 0
1
3
√
√
1
3
3
1
1 √ 3
Solving: y = ± 2 and then x = ± √ , z = ± √ . Solutions: √ , 2, √ , − √ , − 2, − √ .
2
2
2
2
2
2
46.
Denote the system by (S). Setting X = sin α, Y = cos β, Z = tan γ in (S) gives a linear system
(S)′ , where
X + 4Y + Z = 2
′
(S)
X + 6Y − Z = 5
4X + 18Y + Z = 12
′
Forward and backward elimination on (S) using augmented matrices gives
2
1
1 4
1
1 0 0
X = sin α = 1
5 s 0 1 0
0.5
⇒
Y = cos β = 0.5 .
1 6 −1
Z = tan γ = −1
4 18
1
12
0 0 1
−1
1
tan γ
1
0.5
sin α
cos β
0
0
−1
tan γ
0
π/2
2.3562
π
−1
0
1.0472
Figure 1.1 [46]
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π
5.2360 2π
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Chapter 1
ISM
Linear Systems
≃ 1.0472, 5π
3 ≃
π π 3π
≃ 2.3562. There are two solutions: (α, β, γ) =
, ,
and
2 3 4
For the specified ranges, sin α = 1 ⇒ α = π/2 ≃ 1.5708, cos β = 0.5 ⇒ β =
5.2360, tan γ = −1 ⇒ γ =
π 5π 3π
,
,
.
2 3 4
47.
3π
4
π
3
Denote the system by (S). Setting X = x cos α and Y = x in (S) gives a linear system (S)′ , where
(S)′
2X + 3Y = 8
3X + 4Y = 11
Forward and backward elimination on (S)′ using augmented matrices gives
2
3
3
4
8
11
s
1 0
0 1
1
2
⇒
Figure 1.1, [46] shows that cos α = 0.5 implies α =
integer. There are infinitely many solutions.
48.
π
3
X = x cos α = 1
.
Y =x=2
± 2πk, 5π
3 ± 2πk, where k is a nonnegative
(a) Assume m = 0, for otherwise no action is taken and (S2 ) is identical to (S1 ). Perform
E1 − mE2 → E1 on (S1 ) to obtain
(S2 )
(p1 − mq1 )x1 + (p2 − mq2 )x2 + (p3 − mq3 )x3 = b1 − mb2
q1 x1 +
q2 x3 +
q3 x3 =
b2
The second equation in (S2 ) is satisfied by the solution (x1 , x2 , x3 ) = (a1 , a2 , a3 ) and so is the first
because
(p1 − mq1 )a1 + (p2 − mq2 )a2 + (p3 − mq3 )a3
=
p1 a1 + p2 a2 + p3 a3 − m(q1 a1 + q2 a2 + q3 a3 )
=
b1 − mb2
(b) The solution set to the linear system (S2 ) obtained by performing E1 ↔ E2 is clearly the
same as (S1 ). More generally, rearranging (permuting) the equations in a linear system does not
change the solution set.
(c) Performing cE1 → E1 on (S1 ) we obtain a linear system (S2 ) whose first equation is satisfied
by the solution (x1 , x2 , x3 ) = (a1 , a2 , a3 ) because cp1 a1 +cp2 a2 +cp3 a3 = c(p1 a1 +p2 a2 +p3 a3 ) = cb1 .
Note that scaling by c = 0 turns E1 into a zero which is satisfied for any x1 , x2 , x3 .
49.
The parametric form of the equation of L, given in (1.30) on page 16, is (x, y, z) = (1 − 0.6t, 1 −
0.4t, t), where t is a real parameter. (a) Setting x = 0 gives t = 53 and so L passes through the
plane x = 0 (yz-plane) in the point 0, 13 , 53 . (b) Setting y = 0 gives t = 52 and so and so L passes
through the plane y = 0 (xz-plane) in the point − 12 , 0, 25 .
50.
The problem stated in the Nine Chapters (200–100 B .C .) is roughly this: There are three bundles
of corn, labeled B1 , B2 , B3 , respectively. Each bundle contains a certain (unknown) number of
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1.1 Solving Linear Systems 17
measures of corn. Three bundles of B1 , two of B2 and one of B3 contain 39 measures, two of B1 ,
three of B2 and one of B3 contain 34 measures, and one of B1 , two of B2 and three of B3 contain
26 measures. How many measures of corn are contained in one bundle of each type?
Let x1 , x2 , x3 denote the measures of corn contained in one bundle of each type B1 , B2 , B3 ,
respectively. The problem as stated translates into the linear system (S). Performing forward
elimination on (S) using augmented matrices, and maintaining integer coefficients, we have
3 2
1
2 3
1 2
1
3
s
1
2
3
r1 ↔ r3
34
0
−1
−5
r2 − 2r1 → r2
26
0 −4 −8
r3 − 3r1 → r3
39
s
1
− r2 → r2
−18
0
− r3 → r3
−39
0
r3 − 4r2 → r3
26
2
3
1
5
0 12
26
18
33
Back-substitution gives x3 = 33
12 = 2.75, x2 = 18−5x3 = 18−5(2.75) = 4.25, x1 = 26−2x2 −3x3 =
26 − 2(4.25) − 3(2.75) = 9.25. Unique solution: (x1 , x2 , x3 ) = (9.25, 4.25, 2.75) (measures).
51.
Apply Gauss elimination using augmented matrices.
2 3
1
1
4 7
5
1
s
r2 − 4r1 → r2
2
3 1
1
0
1 3
5
Pivot variable: x1 . Free variable: x2 . Let x2 = t, where t is a real parameter. Back-substituting,
we have x2 = 5 − 3t, x1 = 12 − 32 x2 − 12 t = 12 − 32 (5 − 3t) − 21 t = −7 + 4t. The equation of the line
in parametric form is x1 = −7 + 4t, x2 = 5 − 3t, x3 = t.
52.
Apply Gauss elimination using augmented matrices.
2
4
−3
−5
0
1
2
6
s
r2 − 2r1 → r2
2
0
−3
1
0
1
2
2
Pivot variables: x1 , x2 . Free variable: x3 . Let x3 = t, where t is a real parameter. Backsubstituting, we have x2 = 2 − t, x1 = 1 + 23 x2 = 1 + 32 (2 − t) = 4 − 32 t. The equation of the line
in parametric form is x1 = 4 − 32 t, x2 = 2 − t, x3 = t.
53.
True. Each variable is a pivot variable.
54.
False. For example, the equation 0x1 + 0x2 = 0 has no pivot variables because there are no nonzero
coefficients.
55.
False. For example, the linear system
x1 + x2 = 1
2x1 + 2x2 = 2
is equivalent to
x1 + x2 = 1
0x1 + 0x2 = 0
,
which has only one pivot variable. However,
x1 + x2 = 1
x1 + 2x2 = 2
is equivalent to
x1 + x2 = 1
0x1 + x2 = 1
,
which has two pivot variables. A consistent 2 × 2 linear system has two pivot variables.
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56.
Chapter 1
ISM
Linear Systems
False. For example, the linear system
57.
x1 + x2 + x3
2x1 + 2x2 + 2x3
=1
= 2 has two free variables.
−x1 − x2 − x3 = −1
x + x2 = 1
1
False. For example, the linear system
2x1 + 3x2 = 3 is consistent, having solution (x1 , x2 ) =
3x1 + 3x2 = 3
(0, 1).
58.
False. See Illustration 1.2.
59.
True. Although scaling an equation can provide useful simplification during forward elimination,
scaling only becomes necessary during back-substitution or backward elimination.
60.
False. See Exercise 32.
61.
False. According to the Computational Note on page 14, Gauss–Jordan elimination is about 30%
less efficient than Gauss–elimination for solving large-scale linear systems.
62.
True. The matrix contains m rows.
63.
True. Just as a linear equation in three variables has two degrees of freedom and is represented by
a plane in R3 , so a linear equation in eight variables has seven degrees of freedom and is represented
by a hyperplane in R8 .
64.
True. Each linear equation in a linear system can be regarded as a constraint on the solution to
the system. The more equations there are, the more constraints are imposed on the solution set.
65.
True. Such a linear system is called homogeneous (see Section 1.2) and is always consistent, having
the zero solution x1 = x2 = x3 = x4 = 0. There will be at least one free variable because there
can be at most two pivot variables.
USING MATLAB
27 6
.
,
13 13
66.
(x1 , x2 ) =
67.
Apply forward and backward elimination on (S). Using gauss.m with the format rat option, the
4 × 6 augmented matrix M of (S) changes into the augmented matrix M* of an equivalent system
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1.1 Solving Linear Systems 19
(S)′ .
0
1
2
3
M=
−4 −13
0
−1
−2 2
6 3
−1
6
2 1
2 1
2
−3
0
11
−15
1
s
1
0
0
0
0
0
0
1 −2
0
0 −2/3
= M*
0
1/3
1
0
0
0
1
0
0
0
6
6
and we solve (S)′ which has the same solution set as (S). M* shows that (S)′ has pivot variables x1 , x2 , x4 , x5 and one free variable x3 . Let x3 = t, where t is a real parameter. Then
(x1 , x2 , x3 , x4 , x5 ) = (6 − 6t, − 32 + 2t, t, 13 , 0). The variables x4 and x5 are uniquely determined.
The system (1.24) on page 14 is identical to (S) except that the variable x5 (and its coefficients in
all equations) is missing. As x5 = 0, the solution to (1.24) is (x1 , x2 , x3 , x4 ) = (6− 6t, − 23 +2t, t, 13 ).
68.
Apply forward and backward elimination on (S). Using gauss.m with the format rat option, the
5 × 6 augmented matrix M of (S) changes into the augmented matrix M* of an equivalent system
(S)′ .
1 1
0
0 30 −497
1
1 2
12
48
100
2 6
24
96
50
1
0 0 −75
0 0
2
249 = M*
0 3
4
6
24
M=
0
1 23
s 0 0
0
4
−1
−1
0
−12
−48
−250
0
0 0
0
0 0
2
2 11
28 102
−76
0 0
0
0 0
0
and we solve (S)′ which has the same solution set as (S). M* shows that (S)′ has pivot variables
x1 , x3 , x4 and free variables x2 , x5 . Let x3 = s and x5 = t, where s, t is are real parameters. Then
3
(x1 , x2 , x3 , x4 , x5 ) = (−497 − s, s, −75, 249
4 − 2 t, t). The variable x3 is uniquely determined.
69.
Solving the system
x + 2y = 2 E1
x + y = 3 E2
by Gauss elimination, we have (x1 , x2 ) = (4, −1). In order to plot the lines and their intersection
point we need to define a (row) vector x of x-values that includes the value x = 4 (contrast with
text, page 20). Then y1 and y2 are vectors whose components are y-values corresponding to each
value x in the vector x.
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Chapter 1
ISM
Linear Systems
The command
title(′ Two lines in 2-space′ , ′ FontSize′ ,10)
adds a title to the plot and the commands
Two lines in 2-space
1
L2
gtext(′ L 1′ )
gtext(′ L 2′ )
manually position labels on the lines.
0
L1
−1
−2
−2
71.
Figure 1.1 [69]
4
5
70.
Download fig01101.m
72.
We are given that a + b + c = 14 and a − b + c = 0. (a) We have n = 100a + 10b + c and
m = 100a + 10c + b and so n − m = 27 ⇒ b − c = 3 and we obtain a 3 × 3 linear system
a + b + c = 14 E1
(S)
(1.3)
a − b + c = 0 E2 ,
b − c = 3 E3
where a, b, c are integers in the range
14
1
1
1
s
1
0 r2 − r1 → r2
1 −1
− 12 r2 → r2
0
1 −1
3
Download fig01102.m
0, 1, . . . , 9, where a = 0. Forward elimination on (S) gives
1 1
1
14
14
1 1 1
s
0
7
7 = U
0 1
0 1 0
r3 − r2 → r3
0 1 −1
3
0 0 1
4
Back-substitution on the linear system with augmented matrix U gives the unique solution (a, b, c) =
(3, 7, 4) and so n = 374.
(b) If n − m = k, then 9b − 9c = k implies that b − c = k9 and this equation replaces equation E3
in (1.3) to form a new system (S)′ . The same sequence of EROs are used to reduce the augmented
matrix of (S)′ . We have
1
1
1
1 1 1
14
14
1
0 s 0 1 0
7
1 −1
k
0
1 −1
0 0 1
7 − k9
9
We must have 0 ≤ 7 − k9 and so 0 ≤ k ≤ 63 and k must be a multiple of 9. The solutions in terms
of k and n are given below.
(k, n) = (9, 176),
(18, 275),
(27, 374),
(36, 473),
(45, 572),
(54, 671),
(63, 770)
For each value k = 1, 2, . . . , test if the triples of nonnegative integers (a, b, c) satisfy the constraints
a + b + c = 14, b = a + c and that 9b − 9c = k is an integer. The range for b and c is 0, 1, . . . , 9 and
so k = 9b − 9c ≤ 9(9) − 9(0) = 81 gives an upper bound for k.
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Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy
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1.2
1.2 Echelon Forms, Rank 21
E chelon For ms, Rank
Exercises 1–8. Pivot positions in the matrix A are located using any echelon form for A. The pivots in
these positions are nonzero numbers that are necessarily equal to 1 (circled) in the reduced form A*.
We find r = rank A by counting the number of nonzero rows in an echelon form for A and this integer
coincides with the number of pivots and number of pivot columns in A.
1.
2.
The matrix A =
1
0
0 0
column: 1, rank r = 1.
is already in reduced form and so A* = A. Pivot position: (1, 1), pivot
The matrix is in echelon form, but not in reduced form. Performing backward reduction, we have
A=
1
2
0
2
s
→ r2
r1 − 2r2 → r1
1
2 r2
1
0
0
1
= A*
Pivot positions: (1, 1) and (2, 2), pivot columns: 1 and 2, rank r = 2.
3.
A is not in echelon form— forward reduction is required. Performing forward reduction, then
backward reduction, we have
A=
1 2
2 0
s
r2 − 2r1 → r2
1
0
0
−4
s
− 14 r2 → r2
r1 − 2r2 → r1
1
0
0
1
= A*
Pivot positions: (1, 1) and (2, 2), pivot columns: 1 and 2, rank r = 2.
4.
A is not in echelon form— forward reduction is required. Performing forward reduction (no
backward reduction required), we have
2 −3
−6
9
s
r2 + 3r1 → r2
1
2 r1 → r1
1 − 32
0
0
= A*
Pivot positions: (1, 1), pivot column: 1 and rank r = 1.
5.
A is in echelon form but not in reduced form— the pivot should have value 1. Computing the
reduced form, we have
2 2
1 1
s
= A*
A=
1
0 0
r
→
r
0 0
1
2 1
Pivot positions: (1, 1), pivot column: 1, rank r = 1.
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6.
Chapter 1
ISM
Linear Systems
A is not in echelon form— forward reduction is required. Performing forward reduction, then
backward reduction, we have
A=
0
2
−4
0
3
2
s
r1 ↔ r2
4
0 4
0 −4 3
s
− 41 r2 → r2
1
2 r1 → r1
1
0
0
2
1
− 34
= A*
Pivot positions: (1, 1) and (2, 2), pivot columns: 1 and 2, rank r = 2.
7.
A is not in echelon form— forward reduction
backward reduction, we have
1 1 1 5
1 1
s
A = 2 2 2 14 r2 − 2r1 → r2 0 0
r3 − 6r1 → r3
6 6 6 30
0 0
is required. Performing forward reduction, then
1
1 5
s
1
0 4
0
4 r2 → r2
r1 − 5r2 → r1
0 0
0
0
0 0
1 = A*
0 0
Pivot positions: (1, 1) and (2, 4), pivot columns: 1 and 4, rank r = 2.
8.
A is not in echelon form— forward
reduced form, we have
0 1 0 1
s
A = 1 0 1 1
r1 ↔ r2
r3 − r2 → r3
0 1 1 1
10.
11.
1 1
If A =
1 1
rank r = 1.
, then A* =
1 0
0 1
0 0
1
1
0
0
1
1 1
s
0 1
0
r1 − r3 → r1
1 0
0
1 1
1 1
1 2 , then A* = 0 0
1 1 1
0 0
columns and rank r = 2.
1
If A = 1
0 1
1
0 1 = A*
0
1 0
0
1 , which shows that columns 1 and 3 in A are pivot
0
Note that
A=
3
s
4r1 → r1
2
3 r2 → r2
2
4
4
8
2
8
3
3
8 r1
3
8 r2
s
→ r1
→ r2
4 3
=B
2 1
and so A s B. It can be easily shown that A s I2 s B and so rank A = rank B = 2.
12.
One approach is to show that the reduced forms of A and B are identical.
1 1
0
A = 1 1
2 2
1
1
Full file at />
s
1
r2 − r1 → r2
1
0
r3 − 2r1 → r3
1
0
r3 − r2 → r3
0
0
, which shows that column 1 in A is a pivot column and
1
0
reduction is required. Computing an echelon form and the
Pivot positions: (1, 1), (2, 2), (3, 3), pivot columns: 1, 2, 3, rank r = 3.
9.
1 1
1
0 0
0
1 1 = A*
0
0 0
Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy
Full file at />ISM
1.2 Echelon Forms, Rank 23
2
2
B= 8
−2
8
−2
−9
4
6
−9
4
6
s
r2 − 4r1 → r2
r3 + r1 → r3
1
40 r2 → r2
s
−3 r3 → r3
r3 + 9r2 → r3
1
2 r1 → r1
2
0
0
2 −9 −9
0
1
1
0 −3 −3
1 1
0
0
0 0
0 0
1
1 = B*
0
0
To show that A s B directly, perform the sequence of EROs that changes A into A* followed by the
sequence of inverse EROs that changes B* into B, applied in reverse order: 2 r1 → r1 , r3 −9r2 → r3 ,
− 31 r3 → r3 , 40 r2 → r2 , r3 − r1 → r3 , r2 + 4r1 → r2 . We have r = rank A = rank B = 2.
13.
(a) We have
1
A = 0
0
3
1
4 −8
2
4
1
s
r2 ↔ r3 0
1
0
2 r3 → r3
3
2
1
4
s
r3 + r1 → r3
2 −4
1 3
0 2
1 5
1
4 = B
−3
(b) Performing the sequence of EROs that changes A into B in reverse order: r3 − r1 → r3 ,
2r3 → r3 , r2 ↔ r3 shows that B s A.
14.
We have A s A* and B s B*, where A* and B* are the reduced forms of A and B, respectively.
If A* = B*, then A s A* = B* s B, which shows that A s B. We have
1 3
A = 0 4
0 2
s
1 3
→ r2
−8
0 1
r3 − 2r2 → r3
0 0
4
1
8 r3 → r3
1
1
4 r2
s
1 0
r2 + 2r3 → r2
−2
0 1
r1 − r3 → r1
1
0 0
r1 − 3r2 → r1
1
0
0 = A*
1
Also
2
B = 1
1
s
6
2
1 3
1
r1 → r1
7 −7 2
0 4
r2 − r1 → r2
5
4
0 2
r3 − r1 → r3
1
−8
3
s
1 3
→ r2
0 1
r3 − 2r2 → r3
0 0
1
7 r3 → r3
1
4 r2
1
−2 s U
1
and U s B* = A* as seen in the calculation A s A* above.
15.
We have
6 −6
6
[A| b] = 2
4 −6
10 −5
5
Full file at />
s
1
6
→ r1
12 r2 − 2r1 → r2 0
r3 − 10r1 → r3
30
0
r2 ↔ r3
1
6 r1
−1
5
6
1
−5
−8
1
20
10
Solution Manual for Linear Algebra for Engineers and Scientists Using Matlab by Hardy
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Chapter 1
ISM
Linear Systems
s
1 −1
1
→ r2
1 −1
r3 − 6r2 → r3 0
− 21 r3 → r3
0
0
1
r2 ↔ r3
1
5 r2
s
1
1 0
r2 + r3 → r2
4
0 1
r1 − r3 → r1
7
0 0
r1 + r2 → r1
5
11
0
0
1
7
Unique solution (x1 , x2 , x3 ) = (5, 11, 7).
16.
We have
2 −2
3
1
[A| b] =
4 −1
1
3
3
−5
1
−13
3
0
3
−6
s
r1 ↔ r4
r2 − 3r1 → r2
r3 − 4r1 → r3
r4 − 2r1 → r4
s
13
r3 − 8 r2 → r3
r4 − 78 r2 → r4
s
− 49 r3 → r3
r4 − 43 r3 → r4
s
r2 − 34r3 → r2
r1 + 13r3 → r1
− 81 r2 → r2
r1 − 3r2 → r1
1
3
0
−8
0 −13
0 −7
1
53
29
−6
34
− 49
0
1
3 −13
− 34
−6
34
1
0
0
0
1
0 1
0 0
0
1
2
1
0
18
1
1 0
0 0
18
− 94
− 34
0
0 −8
0
0
0
0
15
3 −13
0 −8
0
0
−6
18
27
−13
34
0
Unique solution: (x1 , x2 , x3 ) = (1, 2, 1).
17.
We have
2
3
1
M = [A| b] = 1
5
1
−1
1
α
11
6
β
s
r1 ↔ r2
r2 − 2r1 → r2
r3 − 5r1 → r3
s
r3 + 6r2 → r3
1
1 1
1
0 1
0 0
−1
1
6
1
−1
β − 30
1
−1
0
0 −6 α − 5
α − 11
6
−1
β − 36
(a) If α = 11 and β = 36, then 2 = rank A < rank M = 3 and (S) is inconsistent.
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1.2 Echelon Forms, Rank 25
(b) For α = 11 and any β, we have 3 = rank A = rank M = n the number of unknowns and so
there is a unique solution for every β.
(c) If α = 11 and β = 36, then 2 = rank A = rank M and so there are two pivots and one free
variable, giving infinitely many solutions.
18.
We have
2
1
0
1
M = [A| b] = 1
4 −1
1
α
β
2
1
s
r2 − r1 → r2
r3 − 4r1 → r3
s
r3 − 9r2 → r3
1
2
0 −1
0 −9
1
2
0
−1
0
0
1
0
β
2
0
α−4
0
β
1
0
α−4
2 − 9β
2
, then 2 = rank A < rank M = 3 and so (S) is inconsistent.
9
(b) If α = 4, then 3 = rank A = rank M = n the number of unknowns and so there is a unique
solution for any β.
2
(c) If α = 4 and β = , then 2 = rank A = rank M and so there are two pivots and one free
9
variable, giving infinitely many solutions.
(a) If α = 4 and β =
19.
We have
1 2
2
1
M = [A| b] = 0 1
−1 1
p
p
1
5
s
r3 + r1 → r3
s
r3 − 3r2 → r3
1 2
0
0
1
0
0
1
2
1
3
p
p+2
2
2
1
0
p
2 − 2p
1
6
1
1
3
(a) No solutions when p = 1, (b) unique solution when p = 1, (c) infinitely many solutions for no
p.
20.
We have
1
M = [A| b] = 2
−1
−2
p
3
6
3 p−3
1
6
0
s
r2 − 2r1 → r2
r3 + r1 → r3
s
r2 ↔ r3
Full file at />
1
0
−2
p+4
0
3
0
0
1
p
1
0
−2
1
3
p
p+4
0
1
4
1
1
1
4
