Given that matrix K = [[8, 9x], [1, x]], find the (i) value of x for which the determinant of K is 2, (ii) inverse of K.

For a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is $ad - bc$.

$$\det(K) = (8)(x) - (9x)(1)$$ $$\det(K) = 8x - 9x$$ $$\det(K) = -x$$

$$-x = 2$$ $$x = -2$$

The value of $x$ is $\boxed{-2}$.

a) (ii) Find the inverse of $K$.

$$K = \begin{pmatrix} 8 & 9(-2) \\ 1 & -2 \end{pmatrix}$$ $$K = \begin{pmatrix} 8 & -18 \\ 1 & -2 \end{pmatrix}$$

From part (i), we know $\det(K) = -x$.

$$\det(K) = -(-2) = 2$$

Alternatively, using the matrix with $x=-2$:

$$\det(K) = (8)(-2) - (-18)(1)$$ $$\det(K) = -16 - (-18)$$ $$\det(K) = -16 + 18 = 2$$

For a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the inverse is $\frac{1}{\det(K)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$.

$$K^{-1} = \frac{1}{2} \begin{pmatrix} -2 & -(-18) \\ -1 & 8 \end{pmatrix}$$ $$K^{-1} = \frac{1}{2} \begin{pmatrix} -2 & 18 \\ -1 & 8 \end{pmatrix}$$

$$K^{-1} = \begin{pmatrix} \frac{-2}{2} & \frac{18}{2} \\ \frac{-1}{2} & \frac{8}{2} \end{pmatrix}$$ $$K^{-1} = \boxed{\begin{pmatrix} -1 & 9 \\ -\frac{1}{2} & 4 \end{pmatrix}}$$