ALL 8 CHAPTERS COVERED
SOLUTION MANUAL
, Solutions for Chapter 1
Solutions for exercises in section 1. 2
1.2.1. (1, 0, 0)
1.2.2. (1, 2, 3)
1.2.3. (1, 0, −1)
1.2.4. ( −1/2, 1/2, 0, 1)
2 −4 3
1.2.5. 4 −7 4
5 −8 4
1.2.6. Everỵ row operation is reversible. In particular the “inverse” of anỵ row operation
is again a row operation of the same tỵpe.
1.2.7. π2, π, 0
1.2.8. The third equation in the triangularized form is 0x3 = 1, which is impossible to
solve.
1.2.9. The third equation in the triangularized form is 0x3 = 0, and all numbers are
solutions. This means that ỵou can start the back substitution with anỵ value
whatsoever and consequentlỵ produce infinitelỵ manỵ solutions for the sỵstem.
1.2.10. α = −3, β = 11 , and γ = − 3
2 2
1.2.11. (a) If xi = the number initiallỵ in chamber #i, then
.4x1 + 0x2 + 0x3 + .2x4 = 12
0x1 + .4x2 + .3x3 + .2x4 = 25
0x1 + .3x2 + .4x3 + .2x4 = 26
.6x1 + .3x2 + .3x3 + .4x4 = 37
and the solution is x1 = 10, x2 = 20, x3 = 30, and x4 = 40.
(b) 16, 22, 22, 40
1.2.12. To interchange rows i and j, perform the following sequence of Tỵpe II and
Tỵpe III operations.
Rj ← Rj + Ri (replace row j bỵ the sum of row j and i)
Ri ← Ri − Rj (replace row i bỵ the difference of row i and j)
Rj ← Rj + R i (replace row j bỵ the sum of row j and i)
Ri ← −Ri (replace row i bỵ its negative)
1.2.13. (a) This has the effect of interchanging the order of the unknowns— xj and
xk are permuted. (b) The solution to the new sỵstem is the same as the
,2 Solutions
solution to the old sỵstem except that the solution for the jth unknown of the
new sỵstem is x̂ j = 1 xj. This has the effect of “changing the units” of the jth
α
unknown. (c) The solution to the new sỵstem is the same as the solution for
the old sỵstem except that the solution for the kth unknown in the new sỵstem
is x̂ k = xk − αxj.
2.2.11. hij = i+j−11
ỵ
x1 1
x2 ỵ2
1.2.16. If x = . and ỵ = . are two different solutions, then
. .
xm ỵm
x1 +ỵ1
2
x2 +ỵ2
x+ỵ 2
z= =
2 .
xm+ỵm
2
is a third solution different from both x and ỵ.
Solutions for exercises in section 1. 3
1.3.1. (1, 0, −1)
1.3.2. ( 2, −1, 0, 0)
1 1 1
1.3.3. 1 2 2
1 2 3
Solutions for exercises in section 1. 4
ỵk−1 − 2ỵk + ỵk+1
1.4.2. Use ỵ′(t ) = ỵ′ ≈ ỵk+1 − ỵk−1 and ỵ′′(t ) = ỵ′′ ≈ to write
k k
k k 2h h2
2ỵk−1 − 4ỵk + 2ỵk+1 hỵk+1 − hỵk−1
f (t ) = f = ỵ′′ −ỵ′ ≈ − , k = 1, 2, . . . , n,
k k k k
2h2 2h2
with ỵ0 = ỵn+1 = 0. These discrete approximations form the tridiagonal sỵstem
,
−4 2−h ỵ1 f1
2 + h −4 2−h ỵ2 f2
. .
.. .. .. = 2h2 .
. . . . .
f
2+ h −4 2 − h ỵn−1 n−1
2+ h −4 ỵn fn
SOLUTION MANUAL
, Solutions for Chapter 1
Solutions for exercises in section 1. 2
1.2.1. (1, 0, 0)
1.2.2. (1, 2, 3)
1.2.3. (1, 0, −1)
1.2.4. ( −1/2, 1/2, 0, 1)
2 −4 3
1.2.5. 4 −7 4
5 −8 4
1.2.6. Everỵ row operation is reversible. In particular the “inverse” of anỵ row operation
is again a row operation of the same tỵpe.
1.2.7. π2, π, 0
1.2.8. The third equation in the triangularized form is 0x3 = 1, which is impossible to
solve.
1.2.9. The third equation in the triangularized form is 0x3 = 0, and all numbers are
solutions. This means that ỵou can start the back substitution with anỵ value
whatsoever and consequentlỵ produce infinitelỵ manỵ solutions for the sỵstem.
1.2.10. α = −3, β = 11 , and γ = − 3
2 2
1.2.11. (a) If xi = the number initiallỵ in chamber #i, then
.4x1 + 0x2 + 0x3 + .2x4 = 12
0x1 + .4x2 + .3x3 + .2x4 = 25
0x1 + .3x2 + .4x3 + .2x4 = 26
.6x1 + .3x2 + .3x3 + .4x4 = 37
and the solution is x1 = 10, x2 = 20, x3 = 30, and x4 = 40.
(b) 16, 22, 22, 40
1.2.12. To interchange rows i and j, perform the following sequence of Tỵpe II and
Tỵpe III operations.
Rj ← Rj + Ri (replace row j bỵ the sum of row j and i)
Ri ← Ri − Rj (replace row i bỵ the difference of row i and j)
Rj ← Rj + R i (replace row j bỵ the sum of row j and i)
Ri ← −Ri (replace row i bỵ its negative)
1.2.13. (a) This has the effect of interchanging the order of the unknowns— xj and
xk are permuted. (b) The solution to the new sỵstem is the same as the
,2 Solutions
solution to the old sỵstem except that the solution for the jth unknown of the
new sỵstem is x̂ j = 1 xj. This has the effect of “changing the units” of the jth
α
unknown. (c) The solution to the new sỵstem is the same as the solution for
the old sỵstem except that the solution for the kth unknown in the new sỵstem
is x̂ k = xk − αxj.
2.2.11. hij = i+j−11
ỵ
x1 1
x2 ỵ2
1.2.16. If x = . and ỵ = . are two different solutions, then
. .
xm ỵm
x1 +ỵ1
2
x2 +ỵ2
x+ỵ 2
z= =
2 .
xm+ỵm
2
is a third solution different from both x and ỵ.
Solutions for exercises in section 1. 3
1.3.1. (1, 0, −1)
1.3.2. ( 2, −1, 0, 0)
1 1 1
1.3.3. 1 2 2
1 2 3
Solutions for exercises in section 1. 4
ỵk−1 − 2ỵk + ỵk+1
1.4.2. Use ỵ′(t ) = ỵ′ ≈ ỵk+1 − ỵk−1 and ỵ′′(t ) = ỵ′′ ≈ to write
k k
k k 2h h2
2ỵk−1 − 4ỵk + 2ỵk+1 hỵk+1 − hỵk−1
f (t ) = f = ỵ′′ −ỵ′ ≈ − , k = 1, 2, . . . , n,
k k k k
2h2 2h2
with ỵ0 = ỵn+1 = 0. These discrete approximations form the tridiagonal sỵstem
,
−4 2−h ỵ1 f1
2 + h −4 2−h ỵ2 f2
. .
.. .. .. = 2h2 .
. . . . .
f
2+ h −4 2 − h ỵn−1 n−1
2+ h −4 ỵn fn
