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:
The equation that represents the total distance travelled by numbers of times he goes to work by Michael is y = 4*x
Step-by-step explanation:
Since Michael has to travel 4 km each time he goes to work if he goes to work 2 times he'll have to travel 8 km, if he goes 3 times he'll have to travel 12 km. If we keep doing this we'll realize that the distance travelled by Michael is given by the number of times he goes to work multiplied by 4. The equation that represents that is:
y = 4*x
I believe they all have y-intersects at the origin. From the looks of it, D and E are the same expression. I'm not sure if that is a typo or not.
Answer:
$54.25
Step-by-step explanation:
50.75 + 2.50 - 3.25 - 1.75 + 1.25 + 4.75 = ?
54.25
Answer:
Hi! The answer to your question is D. Equation B and Equation D
Step-by-step explanation:
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