The characteristic polynomial

Characteristic polynomial. $A$ is upper triangular with diagonal entries $1,3,4$, so
$$\chi_A(\lambda) = (1-\lambda)(3-\lambda)(4-\lambda).$$
Expanding:
$$(1-\lambda)(3-\lambda)=3-4\lambda+\lambda^2,$$
$$(3-4\lambda+\lambda^2)(4-\lambda)=12-3\lambda-16\lambda+4\lambda^2+4\lambda^2-\lambda^3=12-19\lambda+8\lambda^2-\lambda^3.$$
So $\chi_A(\lambda)=-\lambda^3+8\lambda^2-19\lambda+12$.

Check. Trace: coefficient of $\lambda^2$ is $8=\textrm{tr}(A)$. Constant term: $-(-1)^3\cdot 12 = 12=\det A$. Correct.

Conclusion. $\chi_A(\lambda)=-(\lambda-1)(\lambda-3)(\lambda-4)$.