If $A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$, $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $A² = KA - 2I$, then the value of K is

Given $A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$, $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $A^2 = KA - 2I$.

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$A^2 = A \times A$

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

Position (1,1): $(3 \times 3) + (-2 \times 4)$

$= 9 - 8$

$= 1$

Position (1,2): $(3 \times -2) + (-2 \times -2)$

$= -6 + 4$

$= -2$

Position (2,1): $(4 \times 3) + (-2 \times 4)$

$= 12 - 8$

$= 4$

Position (2,2): $(4 \times -2) + (-2 \times -2)$

$= -8 + 4$

$= -4$

Therefore $A^2 = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}$

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

$2I = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$

$KA - 2I = \begin{bmatrix} 3K & -2K \\ 4K & -2K \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$

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

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Since $A^2 = KA - 2I$:

$\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3K-2 & -2K \\ 4K & -2K-2 \end{bmatrix}$

From position (1,1):

$1 = 3K - 2$

$3 = 3K$

$K = 1$

Therefore, $K = 1$