00:01
This problem gives us a matrix and asks us to find its eigenvalues and eigenvectors.
00:05
We do this by first finding characteristic polynomial, which is found by taking the determinant of a minus lambda times i, which will give us the determinant of matrix 2 minus lambda, negative 3, 3, 2 minus lambda.
00:23
Taking the determinant, we get 2 minus lambda square minus 3 times negative 3, which will, simplify out to lambda squared minus 4 lambda plus 13.
00:37
Then using the quadratic equation we solve for lambda, so negative negative 4 is 4 plus or minus the square root and negative 4 squared, which is 16, minus 4 times a, which is 1, times c, which is 13.
00:53
Then we divide by 2a, so 2 times 1.
00:57
And this becomes 4 plus or minus the square root of negative 36 divided by 2, which is equivalent to 2 plus or minus 3i.
01:09
So our two eigenvalues are not real, and there are 2 plus 3i and 2 minus 3i.
01:14
To find the corresponding eigenvectors, we have to find a vector x such that a minus lambda i times x is equal to 0.
01:25
So first for a lambda equal to 2 plus 3i, we'll plug that in, and that gives us the matrix 2 minus 2 minus 3i, negative 3, 3, and 2 minus 2 minus 3i, times the eigenvector x1, which is equivalent to negative 3i, negative 3, negative 3i, times a1, b1, equal to negative 3i, negative 3i, 0 ,000, 0 ,000, 0 ,000, 0 ,000 ,000, 5 ,000 ,000 ,000 ,000.
01:57
To 0 -0...