00:01
Hello friends, we are given with matrix which equal to 3 cross 3.
00:04
Therefore, that is entries are x 0 2 0 3 0 2 0 2.
00:09
We know that the characteristic equation is determinant of a minus lambda i equal to 0.
00:14
Therefore, we given with one of the eigenvalue lambda equal to 0.
00:19
Therefore, we substituting the values we get 6x plus 2 times of minus x which equal to 6x minus 12 which equal to 0.
00:29
Therefore, we get x equal to similarly we given with another matrix that is x 0 y 0 2 0 y 0 minus 2.
00:37
We are given with one of the eigenvalue lambda equal to minus 3 and product of 3 eigenvalues are minus 12.
00:44
Here we assume e1 e2 e3 are standard basis.
00:48
Therefore, aei is the column of a.
00:51
Therefore, we given with ae2 which is a 2 e2.
00:55
Therefore, 2 is the another eigenvalue.
00:57
We are given with minus 3 is a eigenvalue.
01:00
So, we get 2 eigenvalues which is 2 comma minus 3...