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:
i would make a number line up to 10. the with a hollow dot on -4 have a line facing left
Answer:
false
Step-by-step explanation:
it fails the vertical line test, (x cannot be equal to 2 or more y values or its not a function)
1 yard = 3 feet = 36 in
so 1 in = 1/36 yard, 1 ft = 1/3 yard.
The volume of the layer of the topsoil, is:
5 in * 23 ft * 17 ft = 5 *(1/36)*23*(1/3)*17*(1/3) yard cubed =
(5*23*17)/(36*3*3)= 1955/324=6.034 yard cubed.
The store only sells increments of 1/4 = 0.25 cubic yards.
we need 6.034/0.25=24.136
So we need to buy 25 (24 will not be enough) increments of 1/4 cubic yards.
Answer: 25 increments of 1/4 cubic yard, that is 25*0.25=6.25 cubic yards.
