Throughout, we use x instead of lambda, for typographical reasons.

Note: The characteristic polynomial does not come with "= 0" attached to it. One sets it equal to zero to solve for its root, which are the eigenvalues. When simply asked what the characteristic polynomial is, do not give the answer with "= 0" in it.

1. [5] (a) What is the characteristic polynomial of the matrix A =

```
5  2  7
0  4  6    ?
0  0  3

```
If it is convenient, you may leave your answer as product. What are the eigenvalues?

Solution: The characteristic polynomial is (x-5)(x-4)(x-3). The eigenvalues are 5, 4, and 3.

2. [5] What is the characteristic polynomial of the matrix B =

```
2  -1
2   5      ?

```
Solution. This is the determinant of xI - B =
```
x-2  1
-2 x-5

```
or (x-2)(x-5) - 1(-2) = x^2 - 7x + 10 + 2 = x^2 - 7x + 12. (This factors (x-3)(x-4), but that was not required.)

Alternate solution. Trace B = 2+5 = 7, and det B = 2(5) - (-1)2 = 12. Thus, it is x^2 - (trace B)x + det B = x^2 - 7x + 12.