Given matrix A = 2x & 2 3 & x .

Here are the solutions to the questions:

1. Matrix A
Given matrix $A = \begin{pmatrix} 2x & 2 \\ 3 & x \end{pmatrix}$.

i) To find the positive value of $x$ for which the determinant of A is 12:
The determinant of a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $ad - bc$.
For matrix A, the determinant is:

$$\det(A) = (2x)(x) - (2)(3) = 2x^2 - 6$$

We are given that $\det(A) = 12$:

$$2x^2 - 6 = 12$$ $$2x^2 = 12 + 6$$ $$2x^2 = 18$$ $$x^2 = \frac{18}{2}$$ $$x^2 = 9$$ $$x = \pm\sqrt{9}$$ $$x = \pm 3$$

Since we need the positive value of $x$:

$$x = \boxed{3}$$

ii) To write $A^{-1}$:
First, substitute $x=3$ into matrix A:

$$A = \begin{pmatrix} 2(3) & 2 \\ 3 & 3 \end{pmatrix} = \begin{pmatrix} 6 & 2 \\ 3 & 3 \end{pmatrix}$$

The determinant of A is 12 (as found in part (i)).
The inverse of a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $\frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$.
Using this formula for matrix A:

$$A^{-1} = \frac{1}{12} \begin{pmatrix} 3 & -2 \\ -3 & 6 \end{pmatrix}$$ $$A^{-1} = \boxed{\begin{pmatrix} \frac{3}{12} & -\frac{2}{12} \\ -\frac{3}{12} & \frac{6}{12} \end{pmatrix} = \begin{pmatrix} \frac{1}{4} & -\frac{1}{6} \\ -\frac{1}{4} & \frac{1}{2} \end{pmatrix}}$$

2. Solve the equation $x^2 - 4x - 2 = 0$, giving your answers correct to 2 decimal places.
This is a quadratic equation in the form $ax^2 + bx +