The Cayley-Hamilton theorem is a very efficient way to find the inverse of a matrix or higher powers of a matrix. For a $2 \times 2$ matrix $A$, the characteristic equation is always $\lambda^2 - (\text{tr}(A))\lambda + \det(A) = 0$.
We use the Cayley-Hamilton theorem, which states that every square matrix satisfies its own characteristic equation. The characteristic equation is given by $\det(A - \lambda I) = 0$. $(1-\lambda)(4-\lambda) - (2)(-1) = 0$ $4 - 5\lambda + \lambda^2 + 2 = 0$ $\lambda^2 - 5\lambda + 6 = 0$. According to the Cayley-Hamilton theorem, the matrix A satisfies this equation: $A^2 - 5A + 6I = 0$. To find an expression for $A^{-1}$, we multiply the entire equation by $A^{-1}$ (assuming A is non-singular, which it is since $\det(A) = 4 - (-2) = 6 \neq 0$). $A^{-1}(A^2 - 5A + 6I) = A^{-1}(0)$ $A^{-1}A^2 - 5A^{-1}A + 6A^{-1}I = 0$ $A - 5I + 6A^{-1} = 0$. Now, solve for $A^{-1}$: $6A^{-1} = 5I - A$ $A^{-1} = \frac{5}{6}I - \frac{1}{6}A$. We are given that $A^{-1} = \alpha I + \beta A$. Comparing the two expressions, we get: $\alpha = \frac{5}{6}$ $\beta = -\frac{1}{6}$ We need to calculate the value of $4(\alpha - \beta)$. $4(\alpha - \beta) = 4\left(\frac{5}{6} - \left(-\frac{1}{6}\right)\right)$ $= 4\left(\frac{5}{6} + \frac{1}{6}\right) = 4\left(\frac{6}{6}\right) = 4(1) = 4$.
