48 1 Mathematical Physics
on settingk=1 andλ 1 = 5− 2 C 11 + 2 C 21 =0(2)
4 C 11 − 4 C 21 =0(3)ThusC 21 =C 11 =a= 1
The substitution ofk=2 andλ 2 =−1 yields
(3+1)C 12 + 2 C 22 = 0
4 C 12 + 2 C 22 = 0
orC 22 =− 2 C 12
We may setC 12 =1 so thatC 22 =− 2
ThusC=(
11
1 − 2
)
(modal matrix)
The inverse ofCis easily found to beC−^1 =
( 2
31
3
1
3 −1
3)
Eigen vectors:(
a 11 −λ a 12
a 21 a 22 −λ)(
x 1
x 2)
= 0
Putλ=λ 1 =5;(
3 − 52
41 − 5
)(
x 1
x 2)
= 0 →− 2 x 1 + 2 x 2 = 0 →x 1 =x 2The normalized invariant vector is√^12(
1
1
)
Putλ=λ 2 =−1;(
3 −(−1) 2
41 −(−1)
)(
x 1
x 2)
= 0 → 4 x 1 + 2 x 2 = 0 →
x 2 =− 2 x 1
The second invariant eigen normalized eigen vector is√^15(
1
− 2
)
(d)C−^1 AC=(
2 / 31 / 3
1 / 3 − 1 / 3
)(
32
41
)(
11
1 − 2
)
=
(
50
0 − 1
)
1.3.5 MaximaandMinima
1.33 Letf=y=x^3 − 3 x+ 3 = 0
Let the root bea
Ifx=a=− 2 ,y=+ 1
Ifx=a=− 3 ,y=− 15
Thusx=alies somewhere between−2 and−3.
Forx=a− 2 .1,y= 0 .039, which is close to zero.
Assume as a first approximation, the root to bea=v+h
Putv=− 2. 1