00:02
Given matrix a equals 1, negative 3, 3, 3, negative 5, 3, and 6, negative 6, 4, in part one we find the characteristic polynomial of a, in two we find all eigenvalues and eigenvectors corresponding to each eigenvalue, and in part three we discuss about a being diagonalizable, and in the case that is true we find matrices b and d such that a equals p times d times inverse of p.
00:33
So in part one we find the characteristic polynomial of a and for that we know that polynomial which we call p of lambda is defined as the determinant of the matrix a minus lambda times the identity 3 by 3, and that is the determinant of the matrix 1 minus lambda, negative 3, 3, then 3 minus 5 minus lambda, the second entry of that row, and 3, then 6, negative 6, 4 minus lambda, and this determinant here can be developed by the second, sorry the first, by the first row, and then the coefficient 1 minus lambda, it keeps its sign because it's in position 1 ,1 and the indices 1 and 1 add up to 2 which is an even number, so that coefficient times the determinant of the sub matrix we get when we remove the first column and the first row, and we remove those column and row because element 1 minus lambda which we are using now is in position 1 ,1...