If $A = \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix}$ and $2A^{-1} = KI - A$, where K is a real number and I is the identity matrix of order 2, then the value of K is:

$A = \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix}, \quad 2A^{-1} = KI - A$

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$\det(A) = (2)(7) - (-3)(-4) = 14 - 12 = 2$

For a 2×2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the adjoint is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

$\text{adj}(A) = \begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}$

$A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A)$

$= \frac{1}{2}\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}$

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$2A^{-1} = 2 \times \frac{1}{2}\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}$

$= \begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}$

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$KI - A = K\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix}$

$= \begin{bmatrix} K-2 & 3 \\ 4 & K-7 \end{bmatrix}$

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Setting $2A^{-1} = KI - A$:

$\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} K-2 & 3 \\ 4 & K-7 \end{bmatrix}$

From position (1,1): $7 = K - 2 \implies K = 9$

From position (2,2): $2 = K - 7 \implies K = 9$

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Therefore, $K = 9$