082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

8.11 Eigenvalues and eigenvectors 301
Example8.44 Find the eigenvalues λinthe system
( ( ) ( 4 1 x x
= λ
3 2)
y y)
Solution We form the characteristic equation, |A − λI| = 0.Now
( ) 4 − λ 1
A−λI=
3 2−λ
Then
|A−λI|=(4−λ)(2−λ)−3=λ 2 −6λ+5
Solving the characteristic equation, λ 2 −6λ +5 = 0,gives
λ=1 or 5
There aretwo eigenvalues, λ = 1, λ = 5.
The process of finding the characteristic equation and eigenvalues of a matrix has
been illustrated using 2 × 2 matrices. This same process can be applied to a square
matrix of any size.
Example8.45 Find (a)the characteristic equation (b)the eigenvalues ofAwhere
⎛ ⎞
1 2 0
A = ⎝−1−1 1⎠
3 2−2
Solution (a) We need tocalculate |A − λI|. Now
⎛
⎞
1−λ 2 0
A−λI= ⎝ −1 −1−λ 1 ⎠
3 2 −2−λ
and ∣∣∣∣∣∣ 1 −λ 2 0
∣ ∣ ∣∣∣
−1 −1−λ 1
3 2 −2−λ∣ = (1−λ) −1 − λ 1
∣∣∣ 2 −2−λ∣ −2 −1 1
3 −2−λ∣
= (1 − λ)[(−1 − λ)(−2 − λ) −2]
−2[−1(−2 − λ) −3]
Upon simplification thisreduces to −λ 3 −2λ 2 + λ +2.Hence
yields
|A−λI|=0
−λ 3 −2λ 2 +λ+2=0
which maybe written as
λ 3 +2λ 2 −λ−2=0
The characteristic equation is λ 3 +2λ 2 − λ −2 = 0.