00:01
To determine the general solution to dx equals a, given the matrix a.
00:07
First, we have to find the determinant of a minus lambda i to get the eigenveillance of a.
00:14
So i skipped this process because i assume everyone watching this video can figure out the characteristic polynomial and solve for the roots to get these three eigenvectors, eigenvalues.
00:29
So for lambda 1 equals minus 5, we have a minus 5 i times v equals 0.
00:45
So you get 2 plus 5 to 7, minus 4 plus 5 is 1, and minus 3 plus 5 is 2.
01:06
1, 1, 2.
01:24
So we're going to reduce this we should get.
01:31
And again, i'm not going to take the time to show up.
01:35
All the steps so i assume anyone watching this video can reduce this matrix by themselves so when you reduce this matrix you should get one zero one over three zero one two over three and zero zero zero zero so we have a resolution x is equal to if first we have x1 is equal to minus one over three x3 and we have x2 is equal to minus 2 over 3 x3 we have x3 3 the solution is x equal to minus 1 over 3 x3 minus 2 over 3 x3 x 3 so you can write a basis minus 1 over 3 minus 2 over 3 1 and we can choose x3 to be equal to minus 3 to get our first eigen vector v1 to be equal to 1, 2, minus 3.
03:28
So let's do the same for lambda 2 equals 2.
03:34
Well first, the corresponding solution to this eigenvector v1 is equal to the homogenous solution h1 of t is equal to e to the lambda 1t times v1, e to the minus 5t times minus times 1, 2 ,000, 2, minus 3.
04:01
Now let's work to solve for lambda 2 which is equal to 2.
04:11
So here of a minus 2 i times v equals 0.
04:19
So we get 2 minus 2 0.
04:37
Then we have minus 2 1.
04:45
Then we have 1 minus 6 1.
04:52
And the last row we have 2 2 minus 5.
05:01
And again i'm going to skip the work of reducing this matrix...