Determinant of Two By Two Matrix Given Traces

Let $A$ be a $2 \times 2$ matrix, and let $\lambda_1, \lambda_2$ be the eigenvalues of $A$.

We are given:

1. The trace of $A$, which is the sum of its eigenvalues: $\text{trace}(A) = \lambda_1 + \lambda_2 = 3$

2. The trace of $A^3$, which is the sum of the cubes of its eigenvalues: $\text{trace}(A^3) = \lambda_1^3 + \lambda_2^3 = -18$

We need to find the determinant of $A$, which is given by the product of its eigenvalues: $\det(A) = \lambda_1 \lambda_2$

Using the algebraic identity for the sum of cubes: $\lambda_1^3 + \lambda_2^3 = (\lambda_1 + \lambda_2)^3 - 3\lambda_1 \lambda_2 (\lambda_1 + \lambda_2)$

Substituting the known values into the equation: $-18 = (3)^3 - 3\det(A)(3)$

Simplifying the right-hand side: $-18 = 27 - 9\det(A)$

Rearranging the terms: $9\det(A) = 27 + 18$ $9\det(A) = 45$ $\det(A) = 5$

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Alternative Method (Using the Cayley-Hamilton Theorem):

By the Cayley-Hamilton theorem, every square matrix satisfies its own characteristic equation. For a $2 \times 2$ matrix $A$: $A^2 - \text{trace}(A)A + \det(A)I = 0$

Let $d = \det(A)$. Substituting $\text{trace}(A) = 3$: $A^2 - 3A + dI = 0 \implies A^2 = 3A - dI$

Multiplying both sides by $A$: $A^3 = 3A^2 - dA$

Substituting $A^2 = 3A - dI$: $A^3 = 3(3A - dI) - dA = (9 - d)A - 3dI$

Taking the trace on both sides: $\text{trace}(A^3) = (9 - d)\text{trace}(A) - 3d \cdot \text{trace}(I)$

Since $I$ is a $2 \times 2$ identity matrix, $\text{trace}(I) = 2$. Substituting the given values: $-18 = (9 - d)(3) - 3d(2)$ $-18 = 27 - 3d - 6d$ $-18 = 27 - 9d$ $9d = 45$ $d = 5$

Thus, the value of the determinant of $A$ is $5$.