A $2 \times 2$ real matrix $A$ has trace $\text{tr}(A) = 5$ and determinant $\det(A) = 6$. By the Cayley-Hamilton theorem, $A^2 = 5A - 6I$. If $A^3 = \alpha A + \beta I$ where $I$ is the $2 \times 2$ identity matrix, what is the value of the scalar coefficient $\alpha$?

Answer Key:

19.0

### 5-Stage Derivation & Analytical Solution: 1. **Stage 1: Characteristic Equation:** For any $2 \times 2$ matrix, $|A - \lambda I| = \lambda^2 - \text{tr}(A)\lambda + \det(A) = 0$. 2. **Stage 2: Cayley-Hamilton Theorem:** Every square matrix satisfies its own characteristic equation: $A^2 - 5A + 6I = 0 \implies A^2 = 5A - 6I$. 3. **Stage 3: Cubic Power Reduction:** * $A^3 = A \times A^2 = A(5A - 6I) = 5A^2 - 6A$. * Substituting $A^2 = 5A - 6I$: * $A^3 = 5(5A - 6I) - 6A = 25A - 30I - 6A = 19A - 30I$. 4. **Stage 4: Identification of Coefficients:** Comparing with $A^3 = \alpha A + \beta I$, we obtain $\alpha = 19$ and $\beta = -30$. 5. **Stage 5: Official Reference:** Erwin Kreyszig, *Advanced Engineering Mathematics*, Chapter 8.

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