00:01
Given the matrix as (-1, 0, 6), (-3, 4), (-3, 0, 0, 5), to find the eigenvalues associated with the given eigenvectors.
00:13
So, for eigenvalues we must have determinant a -lambda equal to 0.
00:18
So, we must have (-1, 3), (-0, 4), (-0, 6), (-3, 5), so, on taking the determinant we have y -1 -lambda, (-4,-) lambda and 5 -lambda is equal to 0.
00:35
So, we have lambda is equal to 1, lambda is equal to minus 4 and lambda is equal to 5.
00:41
So, for eigenvectors we have a v1 is equal to lambda v1.
00:45
So, for lambda is equal to 1 we can find (-1, 0, 6), (-3, 0, 0, 5), v1, v2, v3 is equal to 1 times v1, v2, v3.
00:57
Now, we can compare very easily and we can find the eigenvectors with this.
01:04
So, also eigenvectors are given to us.
01:08
So, first eigenvector is (-1, 0, 1).
01:11
So, this corresponds to lambda is equal to 1.
01:15
Now, next is 0, 0, minus 1.
01:18
So, this corresponds to lambda is equal to 5 and last is (-1, 1, 1).
01:24
So, this corresponds to lambda is equal to minus 4...