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Unformatted text preview: roof of .AB/c D A.B c / used the column rule for matrix multiplication—this rule is clearly linear, column by column. Even for nonlinear transformations, A.B.c // would be the “composition” of A with B
(applying B then A). This composition A ı B is just AB for matrices. One of many uses for the associative law: The leftinverse B = rightinverse C from
B D B.AC / D .BA/C D C . Problem Set 2.5, page 89
1A 1 D 0
1
3 1
4 0 and B 1 D 1
2 1 0
1
2 and C 2 A simple row exchange has P 2 D I so P P 1 1 1 D 7
5 D P . Here P = “transpose” of P , coming in Section 2:7. 1
4
.
3
" D #
001
1 0 0 . Always
010 Solutions to Exercises 18
1
x
:5
t
:2
5
2
1
3
D
and
D
so A D
. This question solved
y
:2
z
:1
2
1
10
1
AA D I column by column, the main idea of GaussJordan elimination. 4 The equations are x C 2y D 1 and 3x C 6y D 0. No solution because 3 times equation 1 gives 3x C 6y D 3.
a
for any a. And also U .
1 1
5 An upper triangular U with U D I is U D
0
2 6 (a) Multiply AB D AC by A
as B x
x C has the form 1 to ﬁnd B D C (since A is invertible) (b) As long
y
11
, we have AB D AC for A D
.
y
11 7 (a) In Ax D .1; 0; 0/, equation 1 C equation 2 equation 3 is 0 D 1
(b) Right
(c) Row 3 becomes a row of zeros—no third pivot. sides must satisfy b1 C b2 D b3 8 (a) The vector x D .1; 1; 1/ solves Ax D 0 (b) After elimination, columns 1
and 2 end in zeros. Then so does column 3 D column 1 C 2: no third pivot.
1 9 If you exchange rows 1 and 2 of A to reach B , you exchange columns 1 and 2 of A
1 to reach B
2 1 . In matrix notation, B D PA has B
3
2
0
0
0 1=5
0 1=4 0 7
60
6
and B 1 D 4
10 A 1 D 4
0
1=3 0
05
1=2 0
0
0
block of B ). DA
3
4
0
0 1 1 P
2
3
0
0 1 D A P for this P .
3
0
0
0
07
(invert each
6
55
7
6
10
A then certainly A C B = zero matrix is not invertible. (b) A D
00
00
are both singular but A C B D I is invertible.
and B D
01 11 (a) If B D 12 Multiply C D AB on the left by A 13 M 1 CM
14 B 1 DC
A. 1 1 DA 1 B 1 1
1 0
1 A
1 1 and on the right by C 1 : Then A 1 D BC so multiply on the left by C and the right by A W B 1 DA 1
10
: subtract column 2 of A
11 1 1 1 .
D from column 1. 1
.
d
b
ad b c
0
The inverse of each matrix is
ab
D
.
16
the other divided by ad b c
cd
c
a
0
ad b c
#
#
"
"
#"
#"
1
1
1
1
1
1
1
1
11
D E.
D
17 E32 E31 E21 D
0
11
1
11
1
1
"
#
1
Reverse the order and change 1 to C1 to get inverses E211 E311 E321 D 1 1
D
111
L D E 1 . Notice the 1’s unchanged by multiplying in this order. 15 If A has a column of zeros, so does BA. Then BA D I is impossible. There is no A 18 A2 B D I can also be written as A.AB/ D I . Therefore A 1 is AB . Solutions to Exercises 19 19 The .1; 1/ entry requires 4a bD
aD 1
5
2
.
6 and a D 2
.
5 3b D 1; the .1; 2/ entry requires 2b a D 0. Then
For the 5 by 5 case 5a 4 b D 1 and 2b D a give b D 1 and
6 20 A ones.4; 1/ is the zero vector so A cannot be invertible.
21 Six of the sixteen 0 1 matrices are invertible, including all four with thre...
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This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.
 Spring '12
 Minki
 Mass

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