Respuesta :
Answer:
λ = 8,-5,0
Step-by-step explanation:
given matrix,
A = [tex]\left[\begin{array}{ccc}-5&-4&-9\\0&0&2\\0&0&8\end{array}\right][/tex]
det (A − λI) = 0
A − λI = 0
[tex]\left[\begin{array}{ccc}-5&-4&-9\\0&0&2\\0&0&8\end{array}\right][/tex]-λ [tex]\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right][/tex]=0
[tex]\left[\begin{array}{ccc}-5-\lambda&-4&-9\\0&-\lambda&2\\0&0&8-\lambda\end{array}\right][/tex]=0
now finding determinant
( 8-λ){ (-5-λ)(-λ)- 0} = 0
λ = 8,-5,0
The possible values of λ are -5, 0, and 8.
Given
A matrix A, the eigenvalues of A, called are scalars who comply with the relation:
Eigenvalues of a Matrix
There is a vector such that for some scalar, then is called the eigenvalue of with corresponding (right) eigenvector,
det (A − λI) = 0
Where I refer to the identity matrix.
The identity matrix is;
[tex]\rm I=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right][/tex]
Then,
The value of λ is;
[tex]\rm A-\lambda I= 0\\\\\left[\begin{array}{ccc}-5&-4&-9\\0&0&2\\0&0&8\end{array}\right] - \lambda\rm \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]=0\\\\\left[\begin{array}{ccc}-5&-4&-9\\0&0&2\\0&0&8\end{array}\right] -\rm \left[\begin{array}{ccc}\lambda &0&0\\0&\lambda &0\\0&0&\lambda \end{array}\right]=0\\\\ -5-\lambda =0 , \ \lambda =-5\\\\8-\lambda =0, \ \lambda =8\\\\0-\lambda =0, \ \lambda =0[/tex]
Hence, the possible values of λ are -5, 0, and 8.
To know more about the matrix click the link given below.
https://brainly.com/question/4470545
