Where did you get 18

The original matrix was $K = \begin{pmatrix} 8 & 9x \\ 1 & x \end{pmatrix}$.
In part a) (i), we found $x = -2$.
When we substitute $x = -2$ into the element $9x$, we get:

$$9x = 9(-2) = -18$$

So, the matrix $K$ became:

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

For a matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the inverse is $\frac{1}{\det(K)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$.
From our matrix $K = \begin{pmatrix} 8 & -18 \\ 1 & -2 \end{pmatrix}$, we have $b = -18$.
Therefore, the $-b$ term in the inverse formula becomes:

$$-b = -(-18) = 18$$

This is why the adjoint matrix was $\begin{pmatrix} -2 & 18 \\ -1 & 8 \end{pmatrix}$.
Finally, this 18 was divided by the determinant (which was 2) to get 9 in the final inverse matrix.