00:01
It is given in the problem that the matrix, 43, 18, minus 90, minus 38 has eigenvalues 7 and minus 2 and we have to find the eigenvectors.
00:11
So we firstly find the eigenvector at the eigenvalue lambda is equal to 7.
00:20
So for that we need to solve 43 minus lambda 18 minus 90 and minus 38 minus lambda multiplied by the 10.
00:36
By matrix x1 x2 and it is equal to matrix 0.
00:43
In place of lambda we put the eigenvalues and then we solve for x1 x2.
00:49
Firstly we take lambda is equal to 7 so we have matrix 43 minus 7, 18 minus 19 minus 38 minus 7 multiplied by matrix x1 x2 is equal to matrix 0.
01:05
This can also be written as matrix 13.
01:09
36, 18, minus 90, minus 45 multiplied by x1 x2 is equal to matrix 0 and then we use reduced row echelon form of the above matrix.
01:30
Therefore, by reduced roichelon form of matrix, we will get matrix 1, when divided by 2, 0 ,000 ,000.
01:45
0 multiplied by matrix x1 x2 this is equal to matrix 0 0 now we take the values we take x2 is equal to t so we have x1 is equal to minus t divided by 2 therefore the vector x is equal to matrix minus t divided by 2 t is equal to matrix minus t divided by 2 t that is equal to minus 1 divided by 2 1 multiplied by t so we have found the eigenvector at lambda is equal to 7 and we find the eigenvector at lambda is equal to 2 minus 2 so for that we have matrix 43 minus minus 2 that is 43 plus 2 18 minus 19 minus 38 minus minus 2 that is minus 38 plus 2, multiplied by matrix x1 x2 is equal to matrix 0...