Appendix A
A.1
(a) [A] + [B] =
4 0 2 0 6 0
1 8 2 4 3 12
Verify by multiplying [A] [A]–1 = [I]
A.3 [D]–1 =
[]
|[ ]|
T
C
D
[D] =
5 2 1
2 10 0
1 0 5
50 10 10
10 2 46
| [D] | = 5(50) + (2)(-10) + (1)(-10) = 220
5 1 1
22 22 22
1 6 1
22 55 110
1 1 23
22 110 110
50 10 10
110 24 2
220 10 2 46
A.4 Nonsense
A.5 [B] =
20
24
(1)
2 0 1 0
2 4 0 1
divide 1st row by 2
(2)
1
2
1 0 0
2 4 0 1
st row by -2 and add to row 2
1
2
1 0 0
631
2 1 1
5 5 5
122
555
1 0 0
1 0 5 0 0 1
c. Multiply R1 by -1 + Add to R3
2 1 1
5 5 5
1 2 2
5 5 5
2 4 1
5 5 5
1 0 0
0 9 1 0
0 4 0 1
d. Multiply R2 by 1/23 + Add to R3
2 1 1
5 5 5
1 2 2
5 5 5
110 5 1
23 23 23
1 0 0
0 0 1
e. Multiply R3 by 23/275 + Add to R2
2 1 1
5 5 5
23 276 23
1
5 55 275 275
110 5 1
23 23 23
1 0 0
0 0 1
f. Multiply R3 by -23/550 + Add to R1
23 23
21
5 110 550 550
23 276 23
1
5 55 275 275
110 5 1
23 23 23
10
0 0 1
g. Multiply R2 by -1/23 + Add to R1
511
22 22 22
46 23 276 23
1 0 0
22 55 110
23
11
22 110 110
A.7 Show that ([A] [B])T = [B]T [A]T by using
11 12
21 22
aa
aa
11 12 13
21 22 23
b b b
b b b
632
11 11 12 21 11 12 12 22 11 13 12 23
( ) ( ) ( ) ( ) ( ) ( )
a b a b a b a b a b a b
12 22
13 23
bb
12 22
aa
[B]T [A]T =
11 11 21 12 11 21 21 22
12 11 22 12 12 21 22 22
13 11 23 12 13 21 23 22
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
b a b a b a b a
b a b a b a b a
b a b a b a b a
=
11 11 12 21 21 11 22 21
11 12 12 22 21 12 22 22
11 13 12 23 21 13 22 23
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
a b a b a b a b
a b a b a b a b
a b a b a b a b
Answer: ([A] [B])T = [B]T [A]T
A.8 [T] =
CS
SC
CS
CS
CS
A.9 Show {X}T [A] {X} is symmetric. Given
{X} =
1
xy
x
, [A] =
ab
bc
1x
1
ay bx by cx x
22
ax bx bx c axy by bx cx
A.10 Evaluate [K] =
0[]
LT
B
E [B] dx, [B] =
11
LL
1
LL
LL
11
A.11 The following integral represents the strain energy in a bar
U =
0{}
2
LT
Ad
[B]T [D] [B] {d} dx
1
d
11
1
L
U =
0{}
2
LT
Ad
[B]T [D] [B] {d} dx
AL
AL
11
11
L
d
2
2
L L L
U =
1
21
2
11
2
d
dd
AEL
d
L L L
=
1
1 2 1 2
22
2
2
d
d d d d
AEL
d
LL
634
dU
22
1 1 2 1 2 2
d d d d d d
AEL
AE
2
2
dd
2
U
d
21
2
22
AE
L
dd
21
dd
L
12
dd
L
=
1
2
11
11
d
AE
d
L
dU
dd
=
1
2
11
11
d
AE
d
L
= [k]
1
2
d
d
= [k] {d} knowing that [k] =
11
11
AE
L
Thus
dU
dd
= [k] {d}