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

300 Chapter 8 Matrix algebra
which yields
λ=−1 or 5
The given system has non-trivial solutions when λ = −1 and λ = 5. These are the
eigenvalues.
If A is a 2 × 2 matrix, the characteristic equation will be a polynomial of degree 2,
that is a quadratic equation in λ, leading to two eigenvalues. IfAis a 3 ×3 matrix, the
characteristicequationwillbeapolynomialofdegree3,thatisacubic,leadingtothree
eigenvalues. In general ann×n matrix gives rise to a characteristic equation of degree
n and hence toneigenvalues.
The characteristic equation of a square matrixAisgiven by
|A−λI|=0
Solutions of this equation are the eigenvalues of A. These are the values of λ for
whichAX = λX has non-trivial solutions.
Example8.43 Determine the characteristic equation and eigenvalues, λ, inthe system
( ( ) ( 3 1 x x
= λ
−1 5)
y y)
Solution InthisexampletheequationshavebeenwritteninmatrixformwithA =
characteristic equation is given by
|A−λI| =0
( ) ( )∣ 3 1 1 0 ∣∣∣
∣ − λ = 0
−1 5 0 1
∣ 3 − λ 1
−1 5−λ∣ = 0
(3−λ)(5−λ)+1 =0
λ 2 −8λ+16 =0
( ) 3 1
.The
−1 5
The characteristic equation is λ 2 − 8λ + 16 = 0. Solving the characteristic equation
gives
λ 2 −8λ+16 =0
(λ−4)(λ−4) =0
λ = 4 (twice)
There isone repeated eigenvalue, λ = 4.