00:01
Hi from the question given that from the given matrix a is equal to 0 1 minus 2 minus 3.
00:11
So here we need to write the characteristic polynomial and the eigenvalues.
00:17
So characteristic polynomial is a minus lambda i is equal to 0.
00:21
So determinant of minus lambda 1 minus 2 minus 3 minus lambda is equal to 0.
00:31
So minus lambda times of minus 3 minus lambda plus 2 which is equal to 0.
00:38
For further simplification 3 lambda plus lambda square plus 2 is equal to 0.
00:45
Therefore the characteristic polynomial is lambda square plus 3 lambda plus 2 is equal to 0.
00:52
Now we need to find the eigenvalues.
00:55
So for further simplification first we need to factor out we have taken lambda square plus lambda plus 2 lambda plus 2 is equal to 0.
01:04
Then we factor out lambda times of lambda plus 1 plus 2 times of lambda plus 1 which is equal to 0.
01:11
So that the eigenvalues are lambda is equal to negative 1 and minus 2.
01:17
Now let us move on to the second question.
01:19
In the second question the given matrix a is 5 0 0 1 2 1 1 1 2.
01:39
Now we need to find the characteristic polynomial.
01:42
So a minus lambda i is equal to determinant of 5 minus lambda 0 0 1 2 minus lambda sorry 1 not 0.
01:55
So 1 1 1 2 minus lambda that could be equal to 0...