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

1

The correct answer is Option (1) → 1

Given

$A=\begin{bmatrix}3 & -2 \\ 4 & -2\end{bmatrix}$, and $A^2 = K A - 2I$.

Compute $A^2$:

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

$A^2 = \begin{bmatrix} 3\cdot3 + (-2)\cdot4 & 3\cdot(-2) + (-2)\cdot(-2)\\ 4\cdot3 + (-2)\cdot4 & 4\cdot(-2) + (-2)\cdot(-2) \end{bmatrix}$

$A^2 = \begin{bmatrix} 9 - 8 & -6 + 4\\ 12 - 8 & -8 + 4 \end{bmatrix} = \begin{bmatrix} 1 & -2\\ 4 & -4 \end{bmatrix}$

Now compare with $KA - 2I$:

$KA - 2I = K\begin{bmatrix}3 & -2 \\ 4 & -2\end{bmatrix} - 2\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$

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

Equate with $A^2$:

$3K - 2 = 1$

$3K = 3$

$K = 1$

Final Answer: $K = 1$