48
1 Mathematical Physics
on setting k = 1 and λ 1 = 5
−2C 11 + 2C 21 = 0
( 2 )
4C 11 − 4C 21 = 0
( 3 )
Thus C 21 = C 11 = a = 1
The substitution of k = 2 and λ 2 = −1 yields
(3 + 1)C 12 + 2C 22 = 0
4C 12 + 2C 22 = 0
or C 22 = −2C 12
We may set C 12 = 1 so that C 22 = −2
Thus C =
1 1
1 −2
(modal matrix)
The inverse of C is easily found to be
C
−1
=
2
3
1
3
1
3
−
1
3
Eigen vectors:
a 11 − λ a 12
a 21 a 22 − λ
x 1
x 2
= 0
Put λ = λ 1 = 5;
3 − 5 2
4 1 − 5
x 1
x 2
= 0 → −2x 1 + 2x 2 = 0 → x 1 = x 2
The normalized invariant vector is
1
√
2
1
1
Put λ = λ 2 = −1;
3 − (−1)
2
4
1− (−1)
x 1
x 2
= 0 → 4x 1 + 2x 2 = 0 →
x 2 = −2x 1
The second invariant eigen normalized eigen vector is
1
√
5
1
−2
(d) C
−1 AC =
2/3 1/3
1/3 −1/3
3 2
4 1
1 1
1 −2
=
5 0
0 −1
1.3.5 Maxima and Minima
1.33 Let f = y = x
3
− 3x + 3 = 0
Let the root be a
If x = a = −2, y = +1
If x = a = −3, y = −15
Thus x = a lies somewhere between −2 and −3.
For x = a − 2.1, y = 0.039, which is close to zero.
Assume as a first approximation, the root to be a = v + h
Put v = −2.1