A $2 \times 2$ real matrix $A$ is given by $A = \begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix}$. Using the Cayley-Hamilton theorem, the matrix inverse $A^{-1}$ expressed as a linear combination $\alpha A + \beta I$ has coefficients $(\alpha, \beta)$ equal to:

(A) $\alpha = -\frac{1}{5}, \quad \beta = \frac{6}{5}$

(B) $\alpha = \frac{1}{5}, \quad \beta = -\frac{6}{5}$

(C) $\alpha = -\frac{1}{6}, \quad \beta = 1$

(D) $\alpha = \frac{1}{5}, \quad \beta = \frac{6}{5}$

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### 1. Strategy & Theoretical Principles The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation: $$p(\lambda) = \det(A - \lambda I) = 0$$ For a $2 \times 2$ matrix, the characteristic equation is: $$\lambda^2 - \text{tr}(A) \lambda + \det(A) = 0$$ By Cayley-Hamilton: $$A^2 - \text{tr}(A) A + \det(A) I = 0$$ ### 2. Analytical Formulation 1. Trace of $A$: $$\text{tr}(A) = 4 + 2 = 6$$ 2. Determinant of $A$: $$\det(A) = (4)(2) - (3)(1) = 8 - 3 = 5$$ Therefore, the characteristic matrix equation is: $$A^2 - 6A + 5I = 0$$ ### 3. Step-by-Step Calculation Multiply through by $A^{-1}$: $$A^{-1} (A^2 - 6A + 5I) = 0$$ $$A - 6I + 5 A^{-1} = 0$$ Rearranging for $A^{-1}$: $$5 A^{-1} = 6I - A \implies A^{-1} = -\frac{1}{5} A + \frac{6}{5} I$$ Comparing with $A^{-1} = \alpha A + \beta I$: $$\alpha = -\frac{1}{5}, \quad \beta = \frac{6}{5}$$ ### 4. Distractor Forensics - **Option A:** Correct coefficients derived directly from Cayley-Hamilton. - **Option B:** Sign error on both terms. - **Option C:** Divided by trace instead of determinant. ### 5. Exam Trap & Standard Reference - **Trap Alert:** Be careful with signs when rearranging $5A^{-1} = 6I - A$. - **Standard Reference:** Kreyszig, E., *Advanced Engineering Mathematics*, 10th Edition, Wiley, Chapter 8 (Matrices and Eigenvalue Problems).

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