In the diagram, A, B, C and D are points on a circle. AB||DC, AC and BD intersect at X and BDC = 40^. Find: (i) ACD. (ii) AXD.

To multiply matrices, we multiply the rows of the first matrix by the columns of the second matrix.

$$PQ = \begin{bmatrix} 2 & 4 \\ -9 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 3 & -2 \end{bmatrix}$$ $$PQ = \begin{bmatrix} (2)(1) + (4)(3) & (2)(-1) + (4)(-2) \\ (-9)(1) + (1)(3) & (-9)(-1) + (1)(-2) \end{bmatrix}$$ $$PQ = \begin{bmatrix} 2 + 12 & -2 - 8 \\ -9 + 3 & 9 - 2 \end{bmatrix}$$ $$PQ = \begin{bmatrix} 14 & -10 \\ -6 & 7 \end{bmatrix}$$

To multiply a matrix by a scalar, multiply each element in the matrix by the scalar.

$$2Q = 2 \begin{bmatrix} 1 & -1 \\ 3 & -2 \end{bmatrix}$$ $$2Q = \begin{bmatrix} 2 \times 1 & 2 \times (-1) \\ 2 \times 3 & 2 \times (-2) \end{bmatrix}$$ $$2Q = \begin{bmatrix} 2 & -2 \\ 6 & -4 \end{bmatrix}$$

To add matrices, add the corresponding elements.

$$PQ + 2Q = \begin{bmatrix} 14 & -10 \\ -6 & 7 \end{bmatrix} + \begin{bmatrix} 2 & -2 \\ 6 & -4 \end{bmatrix}$$ $$PQ + 2Q = \begin{bmatrix} 14+2 & -10+(-2) \\ -6+6 & 7+(-4) \end{bmatrix}$$ $$PQ + 2Q = \begin{bmatrix} 16 & -12 \\ 0 & 3 \end{bmatrix}$$

The final answer is $\boxed{\begin{bmatrix} 16 & -12 \\ 0 & 3 \end{bmatrix}}$.