# Matrix characteristic Equation:Eigen Values

653 views

The characteristic equation can be rewritten as :

λ3  +  2 λ2  + 2 λ + 1  =  0

==>   λ ( λ2  +  2 λ + 1)  + 1(λ + 1)  =  0

==>    λ (λ + 1)2  + 1(λ + 1)  =  0

==>    (λ + 1) (λ+ λ + 1)   =  0

Solving which we get λ = -1 , ω  , ω2  where ω  , ωare cube roots of unity ..

As we know :

Modulus of each of cube roots of unity  =  | ω |  =  | ω|  =  1

Also we know ,

Eigen values of matrix satisfies the corresponding characteristic equation and if all eigen values have modulus value = 1 , then the matrix is said to be orthogonal.

which is the case here..

Hence C) is the correct answer..

selected
0
0
didnt get it

## Related questions

1
197 views
I have learned a shortcut for finding eigen values and characteristic equation. It is as follows. $\lambda ^{_{3}}- \alpha \lambda ^{2}+ \beta \lambda - \gamma =0$ where $\alpha =$ trace of 3*3 matrix $\beta =$ ... in the below problem I'm stuck, Could any one help me find why I'm stuck . I was getting answer using this trick for almost all problems
Given that a matrix $A_{3\times3},$which is not idempotent matrix.And $A^{3}=A.$ Then find them, $1)$ Eigen Values $2)$ Trace of the matrix$=$Sum of Leading Diagonal Elements$=\sum a_{ij},$ where $i=j$ $3)Det(A)$