The trace of a square matrix is defined to be the sum of its diagonal entries. If $A$ is a $2 \times 2$ matrix such that the trace of $A$ is $3$ and the trace of $A^{3}$ is $-18$, then the value of the determinant of $A$ is ________

$\operatorname{tr}(A) = 3,\qquad \operatorname{tr}(A^3) = -18$ $\text{To find: } \det(A)$ $\text{Plan: } \operatorname{tr}(A)=\lambda_1+\lambda_2,\ \ \det(A)=\lambda_1\lambda_2,\ \ \operatorname{tr}(A^3)=\lambda_1^3+\lambda_2^3$ $\lambda_1 + \lambda_2 = 3,\qquad \lambda_1\lambda_2 = \det(A) = p$ $\operatorname{tr}(A^3) = \lambda_1^3 + \lambda_2^3 = -18$ $\lambda_1^3 + \lambda_2^3 = (\lambda_1 + \lambda_2)^3 - 3\lambda_1\lambda_2(\lambda_1 + \lambda_2)$ $-18 = 3^3 - 3p(3) = 27 - 9p$ $9p = 45 \Rightarrow p = 5$ $\det(A) = p = 5$ $\boxed{5}$