Answer:
Step-by-step explanation:
<u>Eigenvalues of a Matrix</u>
Given a matrix A, the eigenvalues of A, called 
are scalars who comply with the relation:
Where I is the identity matrix
![I=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=I%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
The matrix is given as
![A=\left[\begin{array}{cc}3&5\\8&0\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%265%5C%5C8%260%5Cend%7Barray%7D%5Cright%5D)
Set up the equation to solve
![det\left(\left[\begin{array}{cc}3&5\\8&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda \end{array}\right]\right)=0](https://tex.z-dn.net/?f=det%5Cleft%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%265%5C%5C8%260%5Cend%7Barray%7D%5Cright%5D-%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Clambda%260%5C%5C0%26%5Clambda%20%5Cend%7Barray%7D%5Cright%5D%5Cright%29%3D0)
Expanding the determinant
![det\left(\left[\begin{array}{cc}3-\lambda&5\\8&-\lambda\end{array}\right]\right)=0](https://tex.z-dn.net/?f=det%5Cleft%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3-%5Clambda%265%5C%5C8%26-%5Clambda%5Cend%7Barray%7D%5Cright%5D%5Cright%29%3D0)
Operating Rearranging
Factoring
Solving, we have the eigenvalues
Step-by-step explanation:
the answer to your question with roster method is
(-5,-4,-3,-2 and -1)
Answer:
6 , 7 , 8 , 9...
Step-by-step explanation:
2x > 10
x > 5
then
x = 6 , 7 , 8, 9...
I think that I am right and also think this will helpful to you.....
Answer:
i belive it is the 3rd on
Step-by-step explanation:
sorry if wrong
