Let A be the matrix
Assuming that A (nxn) is diagonalizable, A = P-1DP, where P is an nxn matrix with n linearly independent eigenvectors as its columns and D is a diagonal matrix (with the same dimension as P) having the eigenvalues of A on the main diagonal (the entry in row i column i is the eigenvalue corresponding to the ith column of P).
Since A = P-1DP, A is similar to D, So trace of A = trace of D and detA = det(P-1DP) = (detP-1)(detD)(det(P) = (1/detP)(detD)(detP) = detD
Since D is a diagonal matrix, the trace of D is the sum of the diagonal entries = trace of A =33
The determinant of D is the product of the diagonal entries = detA = 6561
Now, (9)(9)(9)(3)(3) = 6561 and 9 + 9 + 9 + 3 + 3 = 33
So, 9 is an eigenvalue with multiplicity 3 and 3 is an eigenvalue with multiplicity 2 and A is a 5x5 matrix.