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INTRODUCTION
TO
LINEAR
ALGEBRA
Fifth Edition
MANUAL FOR INSTRUCTORS
Gilbert Strang
Massachusetts Institute of Technology
math.mit/linearalgebra
web.mit/18.
video lectures: ocw.mit
math.mit/∼gs
wellesleycambridge
email: linearalgebrabook@gmail
Wellesley - Cambridge Press
Box 812060
Wellesley, Massachusetts 02482
Problem Set 1, page 8
1 The combinations give (a) a line inR
3 (b) a plane inR
3 (c) all ofR
3 .
2 v+w= (2,3)andv−w= (6,−1)will be the diagonals of the parallelogram with
vandwas two sides going out from(0,0).
3 This problem gives the diagonalsv+wandv−wof the parallelogram and asks for
the sides: The opposite of Problem 2. In this examplev= (3,3)andw= (2,−2).
43 v+w= (7,5)andcv+dw= (2c+d, c+ 2d).
5 u+v= (− 2 , 3 ,1)andu+v+w= (0, 0 ,0)and 2 u+2v+w= (add first answers) =
(− 2 , 3 ,1). The vectorsu,v,ware in the same plane because a combination gives
(0, 0 ,0). Stated another way:u=−v−wis in the plane ofvandw.
6 The components of everycv+dwadd to zero because the components ofvand ofw
add to zero= 3andd= 9give(3, 3 ,−6). There is no solution tocv+dw= (3, 3 ,6)
because3 + 3 + 6is not zero.
7 The nine combinationsc(2,1) +d(0,1)withc= 0, 1 , 2 andd= (0, 1 ,2)will lie on a
lattice. If we took all whole numberscandd, the lattice would lie over the whole plane.
8 The other diagonal isv−w(or elsew−v). Adding diagonals gives 2 v(or 2 w).
9 The fourth corner can be(4,4)or(4,0)or(− 2 ,2). Three possible parallelograms!
10 i−j= (1, 1 ,0)is in the base (x-yplane).i+j+k= (1, 1 ,1)is the opposite corner
from(0, 0 ,0). Points in the cube have 0 ≤x≤ 1 , 0 ≤y≤ 1 , 0 ≤z≤ 1.
11 Four more corners(1, 1 ,0),(1, 0 ,1),(0, 1 ,1),(1, 1 ,1). The center point is(
1 2
,
1 2
,
1 2
).
Centers of faces are(
1 2
,
1 2
,0),(
1 2
,
1 2
,1)and(0,
1 2
,
1 2
),(1,
1 2
,
1 2
)and(
1 2
, 0 ,
1 2
),(
1 2
, 1 ,
1 2
).
12 The combinations ofi= (1, 0 ,0)andi+j= (1, 1 ,0)fill thexyplane inxyzspace.
13 Sum=zero vector. Sum=− 2 : 00 vector= 8: 00 vector. 2 : 00 is 30
◦ from horizontal
= (cos
π 6
,sin
π 6
) = (
√
3 / 2 , 1 /2).
14 Moving the origin to 6 : 00 addsj= (0,1)to every vector. So the sum of twelve vectors
changes from 0 to 12 j= (0,12).
26 Two equations come from the two components:c+ 3d= 14and 2 c+d= 8
solution isc= 2andd= 4. Then2(1,2) + 4(3,1) = (14,8).
27 A four-dimensional cube has 2 4 = 16corners and 2 ·4 = 8three-dimensional faces
and 24 two-dimensional faces and 32 edges in Worked Example2 A.
28 There are 6 unknown numbersv 1 , v 2 , v 3 , w 1 , w 2 , w 3. The six equations come from the
components ofv+w= (4, 5 ,6)andv−w= (2, 5 ,8). Add to find 2 v= (6, 10 ,14)
sov= (3, 5 ,7)andw= (1, 0 ,−1).
29 Fact : For any three vectorsu,v,win the plane, some combinationcu+dv+ewis
the zero vector (beyond the obviousc=d=e= 0). So if there is one combination
Cu+Dv+Ewthat producesb, there will be many more—just addc, d, eor 2 c, 2 d, 2 e
to the particular solutionC, D, E.
The example has 3 u− 2 v+w= 3(1,3)−2(2,7) + 1(1,5) = (0,0). It also has
− 2 u+ 1v+ 0w=b= (0,1). Adding givesu−v+w= (0,1). In this casec, d, e
equal 3 ,− 2 , 1 andC, D, E=− 2 , 1 , 0.
Could another example haveu,v,wthat could NOT combine to produceb? Yes. The
vectors(1,1),(2,2),(3,3)are on a line and no combination producesb. We can easily
solvecu+dv+ew= 0but notCu+Dv+Ew=b.
30 The combinations ofvandwfill the planeunlessvandwlie on the same line through
(0,0). Four vectors whose combinations fill 4 -dimensional space: one example is the
“standard basis”(1, 0 , 0 ,0),(0, 1 , 0 ,0),(0, 0 , 1 ,0), and(0, 0 , 0 ,1).
31 The equationscu+dv+ew=bare
2 c −d = 1
−c+2d −e= 0
−d+2e= 0
Sod= 2e
thenc= 3e
then 4 e= 1
c= 3/ 4
d= 2/ 4
e= 1/ 4
Problem Set 1, page 18
1 u·v=− 2 .4 + 2 = 0,u·w=−.6 + 1 = 1,u·(v+w) =u·v+u·w=
0 + 1,w·v= 4 + 6 = 10 =v·w.
2 ‖u‖= 1and‖v‖= 5and‖w‖=
√
- Then|u·v|= 0<(1)(5)and|v·w|= 10<
5
√
5 , confirming the Schwarz inequality.
3 Unit vectorsv/‖v‖= (
4 5
,
3 5
) = (0. 8 , 0 .6). The vectorsw,(2,−1), and−wmake
0
◦ , 90
◦ , 180
◦ angles withwandw/‖w‖= (1/
√
5 , 2 /
√
5). The cosine ofθis
v ‖v‖
·
w ‖w‖
= 10/ 5
√
5.
4 (a) v·(−v) =− 1 (b)(v+w)·(v−w) =v·v+w·v−v·w−w·w=
1+( )−( )−1 = 0 soθ= 90 ◦ (noticev·w=w·v) (c) (v− 2 w)·(v+2w) =
v·v− 4 w·w= 1−4 =− 3.
5 u 1 =v/‖v‖= (1,3)/
√
10 andu 2 =w/‖w‖= (2, 1 ,2)/ 3 .U 1 = (3,−1)/
√
10 is
perpendicular tou 1 (and so is(− 3 ,1)/
√
10 ).U 2 could be(1,− 2 ,0)/
√
5 : There is a
whole plane of vectors perpendicular tou 2 , and a whole circle of unit vectors in that
plane.
6 All vectorsw= (c, 2 c)are perpendicular tov. They lie on a line. All vectors(x, y, z)
withx+y+z= 0lie on aplane. All vectors perpendicular to(1, 1 ,1)and(1, 2 ,3)
lie on alinein 3 -dimensional space.
7 (a) cosθ=v·w/‖v‖‖w‖= 1/(2)(1)soθ= 60 ◦ orπ/ 3 radians (b)cosθ=
0 soθ = 90
◦ orπ/ 2 radians (c) cosθ = 2/(2)(2) = 1/ 2 soθ = 60
◦ orπ/ 3
(d)cosθ=− 1 /
√
2 soθ= 135
◦ or 3 π/ 4.
8 (a) False:vandware any vectors in the plane perpendicular tou (b) True:u·
(v+ 2w) =u·v+ 2u·w= 0 (c) True,‖u−v‖ 2 = (u−v)·(u−v)splits into
u·u+v·v= 2 whenu·v=v·u= 0.
9 Ifv 2 w 2 /v 1 w 1 =− 1 thenv 2 w 2 =−v 1 w 1 orv 1 w 1 +v 2 w 2 =v·w= 0: perpendicular!
The vectors(1,4)and(1,−
1 4 )are perpendicular.
212 v·w≤ 2 ‖v‖‖w‖leads to‖v+w‖
2 =v·v+2v·w+w·w≤ ‖v‖
2 +2‖v‖‖w‖+
‖w‖
2 . This is(‖v‖+‖w‖)
2 . Taking square roots gives‖v+w‖ ≤ ‖v‖+‖w‖.
22 v 2 1 w 2 1
- 2v 1 w 1 v 2 w 2 +v 2 2 w 2 2 ≤v 2 1 w 2 1 +v 2 1 w 2 2 +v 2 2 w 2 1 +v 2 2 w 2 2 is true (cancel 4 terms)
because the difference isv
2 1 w
2 2 +v
2 2 w
2 1 − 2 v 1 w 1 v 2 w 2 which is(v 1 w 2 −v 2 w 1 )
2 ≥ 0.
23 cosβ=w 1 /‖w‖andsinβ=w 2 /‖w‖. Thencos(β−a) = cosβcosα+sinβsinα=
v 1 w 1 /‖v‖‖w‖+v 2 w 2 /‖v‖‖w‖=v·w/‖v‖‖w‖. This iscosθbecauseβ−α=θ.
24 Example 6 gives|u 1 ||U 1 | ≤
1 2
(u 2 1
+U
2 1 )and|u 2 ||U 2 | ≤
1 2
(u 2 2
+U
2 2 ). The whole line
becomes. 96 ≤(.6)(.8) + (.8)(.6)≤
1 2
(. 6
2 +. 8 2 ) +
1 2
(. 8
2 +. 6 2 ) = 1. True:. 96 < 1.
25 The cosine ofθisx/
√
x 2 +y 2 , near side over hypotenuse. Then|cosθ|
2 is not greater
than 1:x
2 /(x
2 +y
2 )≤ 1.
26–27(with apologies for that typo !) These two lines add to 2 ||v||
2 + 2||w||
2 :
||v+w||
2 = (v+w)·(v+w) =v·v+v·w+w·v+w·w
||v−w||
2 = (v−w)·(v−w) =v·v−v·w−w·v+w·w
28 The vectorsw= (x, y)with(1,2)·w=x+ 2y= 5lie on a line in thexyplane. The
shortestwon that line is(1,2). (The Schwarz inequality‖w‖ ≥v·w/‖v‖=
√
5 is
an equality whencosθ= 0andw= (1,2)and‖w‖=
√
5 .)
29 The length‖v−w‖is between 2 and 8 (triangle inequality when‖v‖= 5and‖w‖=
3 ). The dot productv·wis between− 15 and 15 by the Schwarz inequality.
30 Three vectors in the plane could make angles greater than 90
◦ with each other: for
example(1,0),(− 1 ,4),(− 1 ,−4). Four vectors couldnotdo this ( 360 ◦ total angle).
How many can do this inR 3 orR n ? Ben Harris and Greg Marks showed me that the
answer isn+ 1 vectors from the center of a regular simplex inR
n to itsn+ 1
vertices all have negative dot products. Ifn+2vectors inR
n had negative dot products,
project them onto the plane orthogonal to the last one. Now you haven+ 1vectors in
R
n− 1 with negative dot products. Keep going to 4 vectors inR 2 : no way!
31 For a specific example, pickv= (1, 2 ,−3)and thenw= (− 3 , 1 ,2). In this example
cosθ= v·w/‖v‖‖w‖ =− 7 /
√
14
√
14 =− 1 / 2 andθ = 120
◦ . This always
happens whenx+y+z= 0:
v·w=xz+xy+yz=
1
2
(x+y+z)
2 −
1
2
(x
2 +y
2 +z
2 )
This is the same asv·w= 0−
1
2
‖v‖‖w‖.Thencosθ=
1
2
.
32 Wikipedia gives this proof of geometric mean G = 3
√
xyz ≤arithmetic mean
A= (x+y+z)/ 3. First there is equality in casex=y =z. OtherwiseAis
somewhere between the three positive numbers, say for examplez < A < y.
Use the known inequalityg≤afor thetwopositive numbersxandy+z−A. Their
meana =
1 2
(x+y+z−A)is
1 2
(3A−A) =same asA! Soa ≥gsays that
A
3 ≥g 2 A=x(y+z−A)A. But(y+z−A)A= (y−A)(A−z) +yz > yz.
Substitute to findA 3 > xyz=G 3 as we wanted to prove. Not easy!
There are many proofs ofG= (x 1 x 2 · · ·xn)
1 /n ≤A= (x 1 +x 2 +· · ·+xn)/n. In
calculus you are maximizingGon the planex 1 +x 2 +· · ·+xn=n. The maximum
occurs when allx’s are equal.
33 The columns of the 4 by 4 “Hadamard matrix” (times
1 2
) are perpendicular unit
vectors:
1
2
H=
1
2
1 1 1 1
1 −1 1 − 1
1 1 − 1 − 1
1 − 1 −1 1
.
34 The commandsV=randn(3,30);D=sqrt(diag(V
′ ∗V));U=V\D;will give
30 random unit vectors in the columns ofU. Thenu ′ ∗Uis a row matrix of 30 dot
products whose average absolute value should be close to 2 /π.
Problem Set 1, page 29
13 s 1 + 4s 2 + 5s 3 = (3, 7 ,12). The same vectorbcomes fromStimesx= (3, 4 ,5):
7 All three rows are perpendicular to the solutionx(the three equationsr 1 ·x= 0and
r 2 ·x= 0andr 3 ·x= 0tell us this). Then the whole plane of the rows is perpendicular
tox(the plane is also perpendicular to all multiplescx).
8
x 1 −0 =b 1
x 2 −x 1 =b 2
x 3 −x 2 =b 3
x 4 −x 3 =b 4
x 1 =b 1
x 2 =b 1 +b 2
x 3 =b 1 +b 2 +b 3
x 4 =b 1 +b 2 +b 3 +b 4
=
1 0 0 0
1 1 0 0
1 1 1 0
1 1 1 1
b 1
b 2
b 3
b 4
=A
− 1 b
9 The cyclic difference matrixChas a line of solutions (in 4 dimensions) toCx= 0 :
1 0 0 − 1
−1 1 0 0
0 −1 1 0
0 0 −1 1
x 1
x 2
x 3
x 4
=
0
0
0
0
whenx=
c
c
c
c
= any constant vector.
10
z 2 −z 1 =b 1
z 3 −z 2 =b 2
0 −z 3 =b 3
z 1 =−b 1 −b 2 −b 3
z 2 = −b 2 −b 3
z 3 = −b 3
=
− 1 − 1 − 1
0 − 1 − 1
0 0 − 1
b 1
b 2
b 3
= ∆
− 1 b
11 The forward differences of the squares are(t+ 1)
2 −t
2 =t
2 + 2t+ 1−t
2 = 2t+ 1.
Differences of thenth power are(t+ 1)
n −t
n =t
n −t
n +nt
n− 1 +· · ·. The leading
term is the derivativent
n− 1 . The binomial theorem gives all the terms of(t+ 1)
n .
12 Centered difference matrices ofeven sizeseem to be invertible. Look at eqns. 1 and 4 :
0 1 0 0
−1 0 1 0
0 −1 0 1
0 0 −1 0
x 1
x 2
x 3
x 4
=
b 1
b 2
b 3
b 4
First
solve
x 2 =b 1
−x 3 =b 4
x 1
x 2
x 3
x 4
=
−b 2 −b 4
b 1
−b 4
b 1 +b 3
13 Odd size: The five centered difference equations lead tob 1 +b 3 +b 5 = 0.
x 2 =b 1
x 3 −x 1 =b 2
x 4 −x 2 =b 3
x 5 −x 3 =b 4
−x 4 =b 5
Add equations 1 , 3 , 5
The left side of the sum is zero
The right side isb 1 +b 3 +b 5
There cannot be a solution unlessb 1 +b 3 +b 5 = 0.
14 An example is(a, b) = (3,6)and(c, d) = (1,2). We are given that the ratiosa/cand
b/dare equal. Thenad=bc. Then (when you divide bybd) the ratiosa/bandc/d
must also be equal!
9 (a) Ax= (18, 5 ,0)and (b)Ax= (3, 4 , 5 ,5).
10 Multiplying as linear combinations of the columns gives the sameAx= (18, 5 ,0)and
(3, 4 , 5 ,5). By rows or by columns: 9 separate multiplications whenAis 3 by 3.
11 Axequals(14,22)and(0,0)and ( 9 ,7).
12 Axequals(z, y, x)and(0, 0 ,0)and ( 3 , 3 ,6).
13 (a) xhasncomponents andAxhasmcomponents (b) Planes from each equation
inAx=bare inn-dimensional space. The columns ofAare inm-dimensional space.
142 x+3y+z+5t= 8isAx=bwith the 1 by 4 matrixA= [ 2 3 1 5 ]:one row. The
solutions(x, y, z, t)fill a 3D “plane” in 4 dimensions. It could be called ahyperplane.
15 (a) I=
1 0
0 1
=“identity” (b) P=
0 1
1 0
=“permutation”
1690
◦ rotation fromR=
0 1
−1 0
, 180 ◦rotation fromR 2 =
−1 0
0 − 1
=−I.
17 P =
0 1 0
0 0 1
1 0 0
produces
y
z
x
andQ=
0 0 1
1 0 0
0 1 0
recovers
x
y
z
. Qis the
inverseofP. Later we writeQP=IandQ=P
− 1 .
18 E=
1 0
−1 1
andE=
1 0 0
−1 1 0
0 0 1
subtract the first component from the second.
19 E=
1 0 0
0 1 0
1 0 1
andE
− 1 =
1 0 0
0 1 0
−1 0 1
,Ev=
3
4
8
andE
− 1 Evrecovers
3
4
5
.
20 P 1 =
1 0
0 0
projects onto thex-axis andP 2 =
0 0
0 1
projects onto they-axis.
The vectorv=
5
7
projects toP 1 v=
5
0
andP 2 P 1 v=
0
0
.
21 R=
1
2
√
2 −
√
2
√
2
√
2
rotates all vectors by 45◦. The columns ofRare the results
from rotating(1,0)and(0,1)!
22 The dot productAx= [ 1 4 5 ]
x
y
z
= (1 by 3)(3 by 1)is zero for points(x, y, z)
on a plane in three dimensions. The 3 columns ofAare one-dimensional vectors.
23 A= [ 1 2 ; 3 4 ]andx= [ 5 −2 ]
′ or[ 5 ; −2 ]andb= [ 1 7 ]
′ or[ 1 ; 7 ].
r=b−A∗xprints as two zeros.
24 A∗v= [ 3 4 5 ]
′ andv ′ ∗v= 50. Butv∗Agives an error message from 3 by 1
times 3 by 3.
25 ones(4,4)∗ones(4,1) =column vector[ 4 4 4 4 ]
′ ;B∗w= [ 10 10 10 10 ]
′ .
26 The row picture has two lines meeting at the solution ( 4 , 2 ). The column picture will
have4(1,1) + 2(− 2 ,1) = 4 (column 1)+ 2 (column 2)=right side(0,6).
27 The row picture shows2 planesin3-dimensional space. The column picture is in
2-dimensional space. The solutions normally fill aline in 3 -dimensional space.
28 The row picture shows fourlinesin the 2D plane. The column picture is infour-
dimensional space. No solution unless the right side is a combinationofthe two columns.
29 u 2 =
. 7
. 3
andu 3 =
. 65
. 35
.
The components add to 1. They are always positive.
Their components still add to 1.
30 u 7 andv 7 have components adding to 1 ; they are close tos= (. 6 , .4).
. 8. 3
. 2. 7
. 6
. 4
=
. 6
. 4
=steady states. No change when multiplied by
. 8. 3
. 2. 7
.
31 M=
8 3 4
1 5 9
6 7 2
=
5 +u 5 −u+v 5 −v
5 −u−v 5 5 +u+v
5 +v 5 +u−v 5 −u
;M 3 (1, 1 ,1) = (15, 15 ,15);
M 4 (1, 1 , 1 ,1) = (34, 34 , 34 ,34)because1 + 2 +· · ·+ 16 = 136which is4(34).
56 x+ 4yis 2 times 3 x+ 2y. There is no solution unless the right side is 2 ·10 = 20.
Then all the points on the line 3 x+2y= 10are solutions, including(0,5)and(4,−1).
The two lines in the row picture are the same line, containing all solutions.
6 Singular system ifb= 4, because 4 x+ 8yis 2 times 2 x+ 4y. Theng= 32makes the
lines 2 x+ 4y= 16and 4 x+ 8y= 32become thesame: infinitely many solutions like
(8,0)and(0,4).
7 Ifa= 2elimination must fail (two parallel lines in the row picture). The equations
have no solution. Witha= 0, elimination will stop for a row exchange. Then 3 y=− 3
givesy=− 1 and 4 x+ 6y= 6givesx= 3.
8 Ifk= 3elimination must fail: no solution. Ifk=− 3 , elimination gives0 = 0in
equation 2: infinitely many solutions. Ifk= 0a row exchange is needed: one solution.
9 On the left side, 6 x− 4 yis 2 times(3x− 2 y). Therefore we needb 2 = 2b 1 on the right
side. Then there will be infinitely many solutions (two parallel lines become one single
line in the row picture). The column picture has both columns along the same line.
10 The equationy= 1comes from elimination (subtractx+y= 5fromx+ 2y= 6).
Thenx= 4and 5 x− 4 y= 20−4 =c= 16.
11 (a) Another solution is
1 2 (x+X, y+Y, z+Z). (b) If 25 planes meet at two points,
they meet along the whole line through those two points.
12 Elimination leads to this upper triangular system; then comes back substitution.
2 x+ 3y+ z= 8
y+ 3z= 4
8 z= 8
gives
x= 2
y= 1 If a zero is at the start of row 2 or row 3,
z= 1 that avoids a row operation.
13 2 x− 3 y = 3
4 x− 5 y+ z= 7
2 x− y− 3 z= 5
gives
2 x− 3 y= 3
y+ z= 1
2 y+ 3z= 2
and
2 x− 3 y= 3
y+ z= 1
− 5 z= 0
and
x= 3
y= 1
z= 0
Here are steps 1 , 2 , 3 : Subtract 2×row 1 from row 2, subtract 1×row 1 from row 3,
subtract 2×row 2 from row 3
14 Subtract 2 times row 1 from row 2 to reach(d−10)y−z= 2. Equation (3) isy−z= 3.
Ifd= 10exchange rows 2 and 3. Ifd= 11the system becomes singular.
15 The second pivot position will contain− 2 −b. Ifb=− 2 we exchange with row 3.
Ifb=− 1 (singular case) the second equation is−y−z= 0. But equation(3)is the
same so there is aline of solutions(x, y, z) = (1, 1 ,−1).
16 (a)
Example of
2 exchanges
0 x+ 0y+ 2z= 4
x+ 2y+ 2z= 5
0 x+ 3y+ 4z= 6
(exchange 1 and 2, then 2 and 3)
(b)
Exchange
but then
breakdown
0 x+ 3y+ 4z= 4
x+ 2y+ 2z= 5
0 x+ 3y+ 4z= 6
(rows 1 and 3 are not consistent)
17 If row1 =row 2 , then row 2 is zero after the first step; exchange the zero row with row
3 and row 3 has no pivot. If column2 =column 1 , then column 2 has no pivot.
18 Examplex+ 2y+ 3z= 0, 4 x+ 8y+ 12z= 0, 5 x+ 10y+ 15z= 0has 9 different
coefficients but rows 2 and 3 become0 = 0: infinitely many solutions toAx= 0 but
almost surely no solution toAx=bfor a randomb.
19 Row 2 becomes 3 y− 4 z= 5, then row 3 becomes(q+ 4)z=t− 5. Ifq=− 4 the
system is singular—no third pivot. Then ift= 5the third equation is0 = 0which
allows infinitely many solutions. Choosingz= 1the equation 3 y− 4 z= 5givesy= 3
and equation 1 givesx=− 9.
20 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 rows1+2=row 3 on the left side but not the right side:
x+y+z= 0,x− 2 y−z= 1, 2 x−y= 4. No parallel planes but still no solution. The
three planes in the row picture form a triangular tunnel.
21 (a) Pivots 2 ,
3 2
,
4 3
,
5 4
in the equations 2 x+y= 0,
3 2
y+z= 0,
4 3
z+t= 0,
5 4
t= 5
after elimination. Back substitution givest= 4, z =− 3 , y= 2, x=− 1. (b) If
the off-diagonal entries change from+1to− 1 , the pivots are the same. The solution is
(1, 2 , 3 ,4)instead of(− 1 , 2 ,− 3 ,4).
22 The fifth pivot is
6 5
for both matrices (1’s or− 1 ’s off the diagonal). Thenth pivot is
n+ n
.
(a) Some linear combination of the 100 rows isthe row of 100 zeros.
(b) Some linear combination of the 100columnsisthe column of zeros.
(c) A very singular matrix has all ones:A=ones(100). A better example has 99
random rows (or the numbers 1 i ,... , 100 i in those rows). The 100th row could
be the sum of the first 99 rows (or any other combination of those rows with no
zeros).
(d) The row picture has 100 planesmeeting along a common line through 0. The
column picture has 100 vectors all in the same 99-dimensional hyperplane.
Problem Set 2, page 66
1 E 21 =
1 0 0
−5 1 0
0 0 1
, E 32 =
1 0 0
0 1 0
0 7 1
, P=
1 0 0
0 0 1
0 1 0
0 1 0
1 0 0
0 0 1
=
0 1 0
0 0 1
1 0 0
.
2 E 32 E 21 b= (1,− 5 ,−35)butE 21 E 32 b= (1,− 5 ,0). WhenE 32 comes first, row 3
feels no effect from row 1.
3
1 0 0
−4 1 0
0 0 1
,
1 0 0
0 1 0
2 0 1
,
1 0 0
0 1 0
0 −2 1
M=E 32 E 31 E 21 =
1 0 0
−4 1 0
10 −2 1
.
ThoseE’s are in the right order to giveM A=U.
4 Elimination on column 4: b =
1
0
0
E 21
→
1
− 4
0
E 31
→
1
− 4
2
E 32
→
1
− 4
10
. The
originalAx=bhas becomeUx=c= (1,− 4 ,10). Then back substitution gives
z=− 5 , y=
1 2
, x=
1 2
.This solvesAx= (1, 0 ,0).
5 Changinga 33 from 7 to 11 will change the third pivot from 5 to 9. Changinga 33 from
7 to 2 will change the pivot from 5 tono pivot.
6 Example:
2 3 7
2 3 7
2 3 7
1
3
− 1
=
4
4
4
. If all columns are multiples of column 1 , there
is no second pivot.
7 To reverseE 31 ,add 7 times row 1 to row 3. The inverse of the elimination matrix
E=
1 0 0
0 1 0
− 7 0 1
isE − 1 =
1 0 0
0 1 0
7 0 1
. Multiplication confirmsEE − 1 =I.
8 M =
a b
c d
andM*=
a b
c−ℓa d−ℓb
. detM*=a(d−ℓb)−b(c−ℓa)
reduces toad−bc! Subtracting row 1 from row 2 doesn’t changedetM.
9 M=
1 0 0
0 0 1
−1 1 0
. After the exchange, we needE 31 (notE 21 ) to act on the new row 3.
10 E 13 =
1 0 1
0 1 0
0 0 1
;
1 0 1
0 1 0
1 0 1
;E 31 E 13 =
2 0 1
0 1 0
1 0 1
.Test on the identity matrix!
11 An example with two negative pivots isA=
1 2 2
1 1 2
1 2 1
. The diagonal entries can
change sign during elimination.
12 The first product is
9 8 7
6 5 4
3 2 1
rows and
also columns
reversed.
The second product is