Linear Algebra

For $n=3$, $p(\lambda) = -\lambda^3 + \text{tr}(A)\lambda^2 - M \lambda + \det(A)$ or in standard monic form $\lambda^3 - \text{tr}(A)\lambda^2 + \dots - \det(A) = 0$. The constant term is $(-1)^3 \det(A) = -\det(A) = -6 \implies \det(A) = 6$.