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:
a)c
Step-by-step explanation:
i dunno about b
I just agree with the other comment
Answer:
<em>P= 5.5 cm</em>
Step-by-step explanation:
<u>The Perimeter of Plane Shapes</u>
If we are given a plane shape, the perimeter is the measurement of the boundary of it. In case the shape is made of lines, the perimeter is the sum of the lengths of all of them.
We are given an irregular shape, we only need to add all the sides. But we don't have all the individual sides. Let's write the expression for the perimeter.
P=AE+ED+DC+CB+BA
We have: AE + DC = 1 1/5 cm, AB = 1 3/4 cm, DE = 1 1/4 cm, and BC = 1 3/10 cm. Note we don't have AE or DC, but we have their sum, so
P=AE+DC+ED+CB+BA
Recall that AB=BA, the order of the letters is not important. The perimeter is

We add the integer parts separate from the fractions

Now we use the common denominator (Least Common Multiple or LCM ) which is 20:

Operating

Simplifying

