0:00
Hello.
00:01
So to find the eigenvalues of a, we solve the characteristic equation where we have the determinant of a minus lambda i is equal to zero, where i is the identity.
00:14
So in this case, we'd have the determinant we'd have of negative three minus lambda, 0, 2 as our first row, and then 1, and then negative 1 minus lambda, and then negative 1, minus lambda, and then negative 2, negative 1, negative lambda.
00:30
We'll take the determinant, set that equal to zero.
00:33
We calculate each minor and then simplify, and we get that our characteristic equation is going to be lambda cubed plus 4 lambda squared minus lambda minus 6 is equal to 0.
00:49
We can use the rational root theorem here to test roots plus or minus 1, 2, 3, and 6.
00:58
And we actually get that no rational root is going to work here.
01:09
So what we get is that the eigenvalues here of the matrix, they're going to be complex, they do not simplify nicely into rational numbers...