Let $$A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$$. If $$A^{-1} = \alpha I + \beta A$$, $$\alpha, \beta \in R$$, $$I$$ is a $$2 \times 2$$ identity matrix, then $$4(\alpha - \beta)$$ is equal to:

We are given the matrix $$A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$$ and it is stated that its inverse can be expressed in the form $$A^{-1} = \alpha I + \beta A$$, where $$\alpha, \beta \in \mathbb{R}$$ and $$I$$ is the $$2 \times 2$$ identity matrix. Our task is to find the value of $$4(\alpha - \beta)$$.

First, we explicitly find $$A^{-1}$$. For a general $$2 \times 2$$ matrix $$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$, the formula for the inverse (when the determinant is non-zero) is

$$\left(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\right)^{-1} \;=\; \dfrac{1}{ad-bc}\; \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.$$

For our specific matrix, we have $$a = 1, \; b = 2, \; c = -1, \; d = 4$$. We first compute the determinant:

$$\det(A) \;=\; ad - bc \;=\; (1)(4) - (2)(-1) \;=\; 4 + 2 \;=\; 6.$$

Since the determinant is non-zero, the inverse exists. Substituting these values into the formula gives

$$A^{-1} \;=\; \dfrac{1}{6}\; \begin{bmatrix} 4 & -2 \\ 1 & 1 \end{bmatrix}.$$ That is,

$$A^{-1} \;=\; \begin{bmatrix} \dfrac{4}{6} & -\dfrac{2}{6} \\[4pt] \dfrac{1}{6} & \dfrac{1}{6} \end{bmatrix} \;=\; \begin{bmatrix} \dfrac{2}{3} & -\dfrac{1}{3} \\[4pt] \dfrac{1}{6} & \dfrac{1}{6} \end{bmatrix}.$$

Now we impose the condition $$A^{-1} = \alpha I + \beta A$$. We first write expressions for $$I$$ and $$A$$:

$$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}.$$

Multiplying $$A$$ by the scalar $$\beta$$ and then adding $$\alpha I$$, we obtain

$$\alpha I + \beta A \;=\; \alpha \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \beta \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix} \;=\; \begin{bmatrix} \alpha + \beta & 2\beta \\ -\beta & \alpha + 4\beta \end{bmatrix}.$$

This matrix must equal $$A^{-1}$$, namely

$$\begin{bmatrix} \alpha + \beta & 2\beta \\ -\beta & \alpha + 4\beta \end{bmatrix} \;=\; \begin{bmatrix} \dfrac{2}{3} & -\dfrac{1}{3} \\ \dfrac{1}{6} & \dfrac{1}{6} \end{bmatrix}.$$

Because two matrices are equal iff all corresponding entries are equal, we equate the entries one by one:

1. From the (1,1) positions: $$\alpha + \beta \;=\; \dfrac{2}{3}.$$

2. From the (1,2) positions: $$2\beta \;=\; -\dfrac{1}{3}.$$ Solving this gives $$\beta = \dfrac{-\frac{1}{3}}{2} = -\dfrac{1}{6}.$$

3. Using $$\beta = -\dfrac{1}{6}$$ in the equation $$\alpha + \beta = \dfrac{2}{3},$$ we find

$$\alpha = \dfrac{2}{3} - \beta = \dfrac{2}{3} - \left(-\dfrac{1}{6}\right) = \dfrac{2}{3} + \dfrac{1}{6} = \dfrac{4}{6} + \dfrac{1}{6} = \dfrac{5}{6}.$$

We have therefore obtained $$\alpha = \dfrac{5}{6}$$ and $$\beta = -\dfrac{1}{6}$$. The quantity required is $$4(\alpha - \beta)$$, so we compute

$$\alpha - \beta \;=\; \dfrac{5}{6} - \left(-\dfrac{1}{6}\right) \;=\; \dfrac{5}{6} + \dfrac{1}{6} \;=\; \dfrac{6}{6} \;=\; 1.$$

Now multiply by 4:

$$4(\alpha - \beta) = 4 \times 1 = 4.$$

Hence, the correct answer is Option D.