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Gilbert_Strang-Linear_Algebra_and_Its_Applications_4ed_solutions.pdf
Solutions to Selected Exercises
Problem Set 1.2, page 9
s
-
1
1. The lines intersect at (x, y) = (3, 1). Then 3(column 1) + I(column 2) = (4, 4).
3. These "planes" intersect in a line in four-dimensional space. The fourth plane nor-
mally intersects that line in a point. An inconsistent equation like u + w = 5 leaves
no solution (no intersection).
+
-
r
5. The two points on the plane are (1, 0, 0, 0) and (0, 1, 0, 0).
7. Solvable for (3, 5, 8) and (1, 2, 3); not solvable for b = (3, 5, 7) orb = (1, 2, 2).
9. Column 3 = 2(column 2) - column 1.Ifb = (0, 0, 0), then (u, v, w) _ (c, -2c, c).
11. Both a = 2 and a = -2 give a line of solutions. All other a give x = 0, y = 0.
13. The row picture has two lines meeting at (4, 2). The column picture has 4(1, 1) +
.
t
i
r
-
i
1
-
1
2(-2, 1) = 4(column 1) + 2(column 2) = right-hand side (0, 6).
15. The row picture shows four lines. The column picture is infour-dimensional space.
No solution unless the right-hand side is a combination of the two columns.
17. If x, y, z satisfy the first two equations, they also satisfy the third equation. The line
L of solutions contains v = (1, 1, 0), w = (2, 1, 1), and u = 1v + 1w, and all
combinations cv + dw with c + d = 1.
2
2
2
r
-
1
19. Column 3 = column 1; solutions (x, y, z) _ (1, 1, 0) or (0, 1, 1) and you can add
C
3
'
any multiple of (-1, 0, 1); b = (4, 6, c) needs c = 10 for solvability.
21. The second plane and row 2 of the matrix and all columns of the matrix are changed.
The solution is not changed.
23. u = 0, v = 0; w = 1, because 1(column 3) = b.
Problem Set 1.3, page 15
1. Multiply by £ = z = 5, and subtract to find 2x + 3y = 1 and -6y = 6. Pivots
2, -6.
3. Subtract -1 times equation 1 (or add 1 times equation 1). The new second equation
is 3y = 3. Then y = 1 and x = 5. If the right-hand side changes sign, so does the
solution: (x, y) = (-5, -1).
l
f
.
2
-
-
S
I
2
N
5. 6x + 4y is 2 times 3x + 2y. There is no solution unless the right-hand side is
2. 10 = 20. Then all points on the line 3x + 2y = 10 are solutions, including (0, 5)
and (4, -1).
7. If a = 2, elimination must fail. The equations have no solution. If a = 0, elimination
stops for a row exchange. Then 3y = -3 gives y = -1 and 4x + 6y = 6 gives
x = 3.
9. 6x - 4y is 2 times (3x - 2y). Therefore, we need b2 = 2b1. Then there will be
infinitely many solutions. The columns (3, 6) and (-2, -4) are on the same line.
Solutions to Selected Exercises
429
2x - 3y = 3
11. 2x - 3y = 3
x = 3
y + z = 1 gives y + z = 1 and y= 1
2y - 3z = 2
z = 0
Subtract 2 x row 1 from row 2
Subtract 1 x row 1 from row 3
Subtract 2 x row 2 from row 3.
13. The second pivot position will contain -2 - b. If b = -2, we exchange with row
3. If b = -1 (singular case), the second equation is -y - z = 0. A solution is
(1, 1, -1).
- 5z =0
15. If row 1 = row 2, then row 2 is zero after the first step; exchange the zero row with
row 3 and there is no third pivot. If column 1 = column 2, there is no second pivot.
17. Row 2 becomes 3y - 4z = 5, then row 3 becomes (q + 4)z = t - 5. If q = -4,
the system is singular - no third pivot. Then, if t = 5, the third equation is 0 = 0.
Choosing z = 1, the equation 3y - 4z = 5 gives y = 3 and equation 1 gives
x=-9.
l
f
.
,
,
°
f
19. The system is singular if row 3 is a combination of rows 1 and 2. From the end view,
the three planes form a triangle. This happens if rows 1+ 2 = row 3 on the left-hand
side but not the right-hand side: for example, x + y + z = 0, x - 2y - z = 1,
2x - y = 9. No two planes are parallel, but still no solution.
21. The fifth pivot is s . The nth pivot is ("+1)1) .
23. Triangular system
u+ v+ w= 2
2v + 2w = -2
2w= 2
u = 3
Solution v = -2.
w= 1
25. (u, v, w) = (3/2, 1/2,. -3). Change to +1 would make the system singular (2 equal
columns).
27. a = 0 requires a row exchange, but the system is nonsingular: a = 2 makes it singular
(one pivot, infinity of solutions); a = -2 makes it singular (one pivot, no solution).
29. The second term be + ad is (a + b)(c + d) - ac - bd (only 1 additional
n
-
'
A
.
'
multiplication).
.
,
31. Elimination fails for a = 2 (equal columns), a = 4 (equal rows), a = 0 (zero
i
v
.
column).
Problem Set 1.4, page 26
1.
17
4 ,
17
5
-2 ,
3.
4
[f].
With sides to (2, 1) and (0, 3), the parallelogram goes to (2, 4).
3. Inner products 54 and 0, column times row gives -6 -10 -2
3
21
5
35
1
7
5. Ax = (0, 0, 0), so x = (2, 1, 1) is a solution; other solutions are cx = (2c, c, c).
'
-
`
7. Examples: Diagonal
1
0
0
0
2
0
0
0
7
, symmetric
1
3
4
skew-symmetric
4
0
3
-3 0
0 .
-4 0 0
4
3
2 0 , triangular
0
7
1
0
0
3
2
0
4
0
7
430
Solutions to Selected Exercises
9. (a) all
(b) fil = ail/all
(d) second pivot a22 -
a21-a12.
all
(c) new aid is aid - ail`a11
all
11. The coefficients of rows of B are 2, 1, 4 from A. The first row of AB is [6 3].
1
],B=[g jc=[? ]D=AE=F=[
0
13. A= [1
15. AB1=B1Agivesb=c=0.AB2=B2Agivesa=d.SoA=aI.
17. A(A + B) + B(A + B), (A + B)(B + A), A2 + AB + BA + B2 always equal
-1].
+
-
I
1
(A + B)2.
19b] p q] _ [a+ [b[r s] -
p + br
21. AA; B_ 1 ()C = 0 o] = zero matrix.
c +dr c +ds
aq + bsc
d
d
r
s
23. E32E21b = (1, -5, -35) but E21E32b = (1, -5, 0). Then row 3 feels no effect
from row 1.
25. Changing a33 from 7 to 11 will change the third pivot from 5 to 9. Changing a33
from 7 to 2 will change the pivot from 5 to no pivot.
D
C
B
27. To reverse E31i add 7 times row 1 to row 3. The matrix is R31 = 0
7
1
0
1
0
0
0 .
1
29. E13=
1
0
0
31. E21 has f21
a+ b+ c= 4
33. a + 2b + 4c = 8
a+3b+9c=14
1
1
0
1 0 ; 0
0
1
1
0
1
0
1
2
0 ; E31 E13 = 0
1
1
0
1
0
1
0 . Test on the identity matrix!
1
E32 has 32 = - , E43 has f43 = - .Otherwise the E's match I .
I
M
A
gives
a=2
b = 1.
c=1
35. (a) Each column is E times a column of B.
[1 2 4- 2 4 8Rows
(b) EB = [11
2
Ol]
2
1
4
of EB are combinations of rows of B, so they are multiples of [1 2 4].
37. (row 3) x is
39. BA=3Iis5by5,AB=5Iis3by3,ABD=5Dis3byl,ABD:No,A(B+C):
a3 j xj, and (A2)11 = (row 1)
(column 1) =
No.
41. (a) B = 41.
(b) B = 0.
(d) Every row of B is 1, 0, 0.
0
0
(c) B = 0 1
0
1
1
0
0
43. (a) mn (every entry).
(b) mnp.
(c) n3 (th is is n2 dot product s).
1
45. 2 3 3 0] + 4 [1 2 1] = 6 6 0 + 4 8 4 = 10
2
1
6
6
0
1
2
1
7
l 0
[
0
01
]
3
3
0
3
3
14
8
0
4
1
0
47. A times,B is A
1
Solutions to Selected Exercises
431
I B, [-] [ I
I ],
I
I I [_].
49. The (2, 2) block S = D - CA-1 B is the Schur complement: blocks in d - (cb/a).
51. A times X = [x1 x2 x3] will be the identity matrix I = [Ax, Axe Ax3].
53.
[a+b a+b
(c + d c + dj agrees with [a + c b + d] when b = c and a = d.
ra+c b+d
55. 2x + 3y + z + 5t = 8 is Ax = b with the 1 by 4 matrix A = [2 3 1 5]. The
solutions x fill a 3D "plane" in four dimensions.
57. The dot product [1 4 5] y = (1 by 3)(3 by 1) is zero for points (x, y, z) on a
x
z
plane x + 4y + 5z = 0 in three dimensions. The columns of A are one-dimensional
vectors.
59. A * v = [3 4 5f and v' * v = 50; v * A gives an error message.
8
3
61. M= 1 5
7
6
4
9
2
5+u
5-u-v
5+v
5-u+v
5+u-v
5
5-v
5+u+v ;
5-u
M3(1, 1, 1) = (15, 15, 15); M4(1, 1, 1, 1) = (34, 34, 34, 34) because the numbers
1 to 16 add to 136, which is 4(34).
Problem Set 1.5, page 39
1. U is nonsingular 'when no entry on the main diagonal is zero.
3.
0
1
1
0
0
0
0
1
1
0
1
0-
0
1
-2,
-1 1
0
= 0 1 0 ; -2 1
0
1
2
0
-1 1-1 1
(E-1F j 1G_1)(GFE) -_- E-1F-1FE = E-lE = I; also (GFE)(E-1F-1G-1) =I.
2
2
-1
2
3
7; after elimination, 0
-1
0
0
1 0= I also.
-1 -1 1
0
1
-
0
2
-
r.
1
-
-
/
0
1
-
+
O
0
3
7
-1
U
v
w
2
N
O
0
O
0
K
3
5
0
3
5
0
0
0
1
1
-1
1
3
0
1
5. LU=
7. FGH=
1
0
0
0
2
0
0
0
1
2
1
0
00 ;HGF= 4
1
8
0
2
4
0
0
1
2
0
0
1
9. (a) Nonsingular when dld2d3 0 0.
[01
downward: Lc = b gives c = 1. Then
(b) Suppose d3 ; 0: Solve Lc = b going
d1
0
0
-d1
d2
0
0
-d2
d3
01
u
v = 10 gives
w
1
l/d3
x= 1/d3
1/d3
1
432
Solutions to Selected Exercises
5
11. Lc = b going downward gives c = -2 ; Ux = c upward gives x = -2
0
0
s
'
'
2
13. Permutation
rows 2 and 3
permutation
rows 1 and 2
15. PA =LDUis
PA = LDUis
17. L becomes
1
1
2
1
0
0
r
l
o
0
1
0
1
0
0
0
1
0
0
0
1
1
0
0
'
-
1
,
0
0
0
0
1
'
1,
0
1
0
0
0
0
00
0
0
1
0
1
0
0
1
2
1
2
1
2
u
v = -3
4
w
u
v
w
1
0
3
-
1
1
1
.
1
1
0
1 = 0 1
4
2
3
1
1
2
4 2 = 1
2
1
1
0
1
0
0
0
1
0
0
1
0
0
o 0
0
0
n
1
'
-
-1
o
0
+
1
0
0
1
0
0
0
0
-1 0
0
0
1
0
0
1
0
0
0
1
0
2
1
0
1
1
1
1
0
0
0
0 . MATLAB and other codes use PA = LU.
n
i
t
1
19. a = 4 leads to a row exchange; 3b + lOa = 40 leads to a singular matrix; c = 0
r
-
1
leads to a row exchange; c = 3 leads to a singular matrix.
21. 231 = 1 and 232 = 2 (233 = 1): reverse steps to recover x + 3y + 6z = 11 from
Ux = c: 1 times (x + y + z = 5) + 2 times (y + 2z = 2) + 1 times (z = 2) gives
x+3y+6z=11.
`
/
'
23.
1
0
0 -2 1
1
~
1
-0
2 1
N
0
°
1
1
1
1
A= 0 2 3 =U.
. 0 0 -6
A= 2 1 0 U=E211E321U=LU.
1
0
0
o
0
o
m
2
1
2 by 2: d = 0 not allowed;
1 1 2= 2 1
0
1
1
1
m n l
1
2
1
25.
27.
A= 0 3 9 hasL=landD=
2
4
0 0
8
7
o
a
S
g
e
f h
i
12
d=1, e=1, then 2=1
f = 0 is not allowed
no pivot in row 2.
1
7
A= L U has U = A (pivots on
2
the diagonal); A= LDU has U= D-lA = 0 1
0
0
1
4
3
1
with is on the diagonal.
a
b
b
b
a
b
a
b
c
c
c d
a
1
'
-
F
1
1
1
1
1
1
1
1
1-
a
a
a
b-a b-a b-a
d-c
c-b c-b Need b
a
1 11
1
1
+
1
-
F
1
-
1
0
ra
a
1 = LIU; a a+b
b
0
1
0
b
b+c
( same U).
0
1
0
1 0 c = 5 gives c = 1
4
4
1
1
6
rl
0
1
1
1 1 x= 1
1 0 0 1
= (same L)
b
3
gives x = 0
1
1
0
a
a
c.
cj
d
a
a
a
a
a
1
1
0
1
1
29.
31.
33.
35.
37.
A = LU.
The 2 by 2 upper submatrix B has the first two pivots 2, 7. Reason: Elimination on
A starts in the upper left corner with elimination on B.
R
D
C
1
1
1
1
1
,
-
,
1
2
3
4
5
1
3
6
10
15
1
1
1
1
.
-
1
a
'
.
-
-
+
1
4
10
20
35
1
5
15
35
70
Fl
1
1
1
2
1
1
2
3
4
1
3
1
6 4
1
1
l
4
3
3, 61
4
1
1
Pascal's triangle in L and U.
MATLAB's lu code will wreck
the pattern. chol does no row
exchanges for symmetric
matrices with positive pivots.
39. Each new right-hand side costs only n2 steps compared to n3/3 for full elimination
v
.
,
A\b.
41. 2 exchanges; 3 exchanges; 50 exchanges and then 51.
0
1
43. P= 0 0
0
1
0
1
0
0
o
1
; P1 = 0 0
o
m
0
1
0
0
0
1 and P2 = 0 1
0
0
1
1
0
0
(P2 gives a column exchange).
45. There are n! permutation matrices of order n. Eventually two powers of P must be
G
1
.
the same: If Pr = Ps then Pr-s = I. Certainly r - s < n!
P= [P2
0
N
.
LLL
l
P3J
is 5 by 5 with P2 =
0
.
N
r
L1
L
]'3=[? 0 1 andP6=l.
0
0
010
47. The solution is x = (1, 1, ... , 1). Then x = Px.
Problem Set 1.6, page 52
0
1. A11 = 2
3. A-1 = BL
A21
1
0 '
C-1; A-1 = U-1L-1 P.
0
1 A31' [-sin6 cos8'
sin 01
cos 8
_
434
Solutions to Selected Exercises
5.
7.
A(AB) = (move parentheses) = (A2)(B) = I.
1/2
1/2
-!/2 1/2
1/2
/3/21' [1 0all have A2 = I.
0
1
9. If row 3 of A-1 were (a, b, c, d), then A-1 A = I would give 2a = 0, a + 3b = 0,
4a +f 8b = 1. This has no solution.
11. (a)
+
-
M
10 01l 1+ 0
10
01
01
(b)
[0
0
0
0
] + [0
01] - [0
1]
-Oll
1 _
0
-1 0] - -1 1
1
1
(B-' + A-')-' = B(A + B)-'A.
6
13. ATB = 8; BTA = 8; ABT = [2 2]; BAT = [6
15. (a) n(n + 1)/2 entries on and above diagonal.
p
(
'
2
2]°
(b) (n - 1)n/2 entries above
diagonal.
17. (a) The inverse of a lower (upper) triangular matrix is still lower (upper) triangu-
lar. Multiplying lower (upper) triangular matrices gives a lower (upper) triangular
(b) The main diagonals of Li 1 L2 D2 and D1 U, U2 1 are the same as those
matrix.
of D2 and D1 respectively. L- 1L2D2 = D1 U,U2so we have D1 = D2. By com-
paring the off-diagonals of L i1 L2 D2 = D1 U1 U2both matrices must be diagonal.
Li1L2D2 = D2, D1U1U2 1 = Dl, D, is invertible so L11L2 = I, U,UU l = I.
Then L, = L2, U1 = U2-
19.
1
3
5
0
1
1
0
0
1
1
0
0
0
3
0
0
0
2
1
0
0
5
3
1 1;
0
1 b/a
1
1
[b/a
a
0
11 [0 d - (b2/a)] [0
0
] -
1
LDLT.
21. From B(I - AB) = (I - BA)B we get (I - BA)-1 = B(I - AB)-1B-1, an
explicit inverse provided B and I - AB are invertible. Second approach: If I - BA
is not invertible, then BAx = x for some nonzero x. Therefore ABAx = Ax, or
ABy = y, and I - AB could not be invertible. (Note that y = Ax is nonzero, from
BAx = x.)
23.
25.
[z] °
.1]
so A-1 = 0 2
.2]'
[y]
(a) In Ax = (1, 0, 0), equation 1 + equation 2 - equation 3 is 0 = 1.
right-hand sides must satisfy b, + b2 = b3.
third pivot.
(b) The
(c) Row 3 becomes a row of zeros-no
1].
27. If B exchanges rows 1 and 2 of A, then B-1 exchanges columns 1 and 2 of A-'.
29. If A has a column of zeros, so does BA. So BA = I is impossible. There is no A-1.
F1
[ I
31.
+
-
W
F
-1
1
1r1
1JI
L 1
1
1J
1
lrl
rl
rl
i
1] =[ 0 -1
1
= E;
1]
then 1
1
1
1
1
is L = E-',after reversing the order of these 3 elementary matrices
and changing -1 to +1.
Solutions to Selected Exercises
435
33. A * ones(4,1) gives the zero vector, so A cannot be invertible.
35.
37.
1
2
[
+
-
r
[3
I
0
0
3
7
1
0
8
0
a
1
0
b
c
1
0
1
01]
1
0
0
39.
[
2
0
2
2
0
1
1
0
[
2
0
1
0
3
1
1
0
-2 1
O
l -
2
[0
-
1]
01
[0
8
o
-3 -1]
o
w
1 0 - 0 1
0 0
a
1
0
1
0
1
0
0
0
0
1
0
0 -
1 O]
0
0
1
0
0
1
1 0 -b
-c
0
0
"
1
-
r
0
1
1 -a ac-b
-c
0
0
1
0
0 -
[o
1
1
1/2
0
= [I A-']
.
1
43. A-1 =
1
0
0
LO
diagonal.
I0],
45.
[-C
41. Not invertible for c = 7 (equal columns), c = 2 (equal rows), c = 0 (zero column).
1 OH 01
0
1
1
1
0
1
0
0
. The 5 by 5 A-1 also has is on the diagonal and super-
[-DACA-1 D 1]>
and [-D
47. For Ax = b with A = ones(4, 4) = singular matrix and b = ones(4, 1), A\b
will pick x = (1, 0, 0, 0) and pinv(A) * b will pick the shortest solution x =
(1, 1, 1, 1)/4.
01
49. AT = [0 3], A-1 = [_1
1/3], (A'-')T = (AT)-1 = [1
-3]' AT = A and
then A-1 = c [O
_1] = (A-')T = (AT)-1.
51.
((AB)-1)'r = (B`'A-I)T = (A-1)T(B-I)T; (U`1)T is lower triangular.
53. (a) xTAy = a22 = 5.
(b) xT A = [4
5
6].
(c) Ay = [].
55. (Px)T(Py) = xT PT Py = xTy because pTp = I ; usually Px y = X. PT y
0
0
1
1
0
0
0
1
0
1
2
3
1
2
1,
21
3
0
0
1
1
0
0
0
1
0
1
1
2
57. PAPT recovers the symmetry.
59. (a) The transpose of RTAR is RTATRTT = RTAR = n by n.
(b) (RTR)jJ = (column j of R) (column j of R) = length squared of column j.
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