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 +