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 _____

Let the eigenvalues of the $$2 \times 2$$ matrix $$A$$ be $$\alpha$$ and $$\beta$$. For any square matrix, the trace equals the sum of its eigenvalues and the determinant equals the product of its eigenvalues:

$$\alpha + \beta = \text{tr}(A) = 3$$ $$-(1)$$
$$\alpha \beta = \det(A) = d \; (\text{say})$$ $$-(2)$$

The trace of $$A^3$$ is the sum of the cubes of the eigenvalues, because cubing a matrix cubes each eigenvalue:

$$\text{tr}(A^3) = \alpha^3 + \beta^3 = -18$$ $$-(3)$$

Use the algebraic identity $$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta)$$. Substitute the known sum from $$(1)$$ and the product from $$(2)$$:

$$\alpha^3 + \beta^3 = (3)^3 - 3d(3) = 27 - 9d$$ $$-(4)$$

Equate $$(3)$$ and $$(4)$$ to solve for $$d$$:

$$27 - 9d = -18$$
$$-9d = -45$$
$$d = 5$$

Therefore, the determinant of $$A$$ is $$5$$.

Answer: $$5$$