Please help me with the following!?

Since A has a characteristic equation of degree 5 (from adding the exponents of the factors in the characteristic polynomial), the matrix A has 5 eigenvalues.

Three of these eigenvalues is the triple eigenvalue 1 (from the factor
(λ - 1)^3); let x and y be the other two.

Since the trace of A is the sum of the eigenvalues of A, we have
1 + 1 + 1 + x + y = 18 ==> x + y = 15.

Since the determinant of A is the product of the eigenvalues of A, we have
1 * 1 * 1 * x * y = 54 ==> xy = 54.

Noting that y = 15 - x, we obtain x(15 - x) = 54.
==> x^2 - 15x + 54 = 0
==> (x - 6)(x - 9) = 0
==> x = 6 or 9 [and so y = 9 or 6, respectively].

Thus, the characteristic polynomial of A is
(λ - 1)^3 (λ - 6)(λ - 9) = (λ - 1)^3 (λ^2 - 15λ + 54).

That is, a = -15 and b = 54.

I hope this helps!