You find the eigenvalues of a matrix A by following these steps:
- Compute the matrix
, where I is the identity matrix (1s on the diagonal, 0s elsewhere) - Compute the determinant of A'
- Set the determinant of A' equal to zero and solve for lambda.
So, in this case, we have
![A = \left[\begin{array}{cc}1&-2\\-2&0\end{array}\right] \implies A'=\left[\begin{array}{cc}1&-2\\-2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]=\left[\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26-2%5C%5C-2%260%5Cend%7Barray%7D%5Cright%5D%20%5Cimplies%20A%27%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26-2%5C%5C-2%260%5Cend%7Barray%7D%5Cright%5D-%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Clambda%260%5C%5C0%26%5Clambda%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1-%5Clambda%26-2%5C%5C-2%26-%5Clambda%5Cend%7Barray%7D%5Cright%5D)
The determinant of this matrix is
Finally, we have
So, the two eigenvalues are
Answer: v = 2
Step-by-step explanation:
Your equation:
You can multiply both sides by 8 to eliminate the fraction, which gives you this:
6v - 5 = 7
Then, you can solve for it:
6v = 12
v = 2
can you please rotate the picture
The answer is 2.7 you do 18 times .15 and that is how i got the answer
