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:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 9
$A = \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix}$
Given: $2A^{-1} = K I - A$
Multiply both sides by $A$: $2I = K A - A^2$
$\Rightarrow A^2 - K A + 2I = 0$
Compute $A^2$:
$A^2 = \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix} \cdot \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix} = \begin{bmatrix} 2\cdot2 + (-3)\cdot(-4) & 2\cdot(-3) + (-3)\cdot7 \\ -4\cdot2 + 7\cdot(-4) & -4\cdot(-3) + 7\cdot7 \end{bmatrix}$
$= \begin{bmatrix} 4 + 12 & -6 - 21 \\ -8 - 28 & 12 + 49 \end{bmatrix} = \begin{bmatrix} 16 & -27 \\ -36 & 61 \end{bmatrix}$
Now compute $A^2 - K A + 2I$:
$\begin{bmatrix} 16 & -27 \\ -36 & 61 \end{bmatrix} - K \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix} + \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = 0$
$\Rightarrow \begin{bmatrix} 18 & -27 \\ -36 & 63 \end{bmatrix} - K \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix} = 0$
Equating entries:
$18 - 2K = 0 \Rightarrow K = 9$