A shear transformation with the x-axis invariant has the matrix form:

a) (i) The matrix representing the shear.

Step 1: A shear transformation with the x-axis invariant has the matrix form:

$$S = \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}$$

where $k$ is the shear factor.

Step 2: The original point A is $(2, 3)$ and its image A' is $(8, 3)$. Apply the shear matrix to A:

$$\begin{bmatrix} 8 \\ 3 \end{bmatrix} = \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 \\ 3 \end{bmatrix}$$

Step 3: Perform the matrix multiplication.
From the second row: $3 = 0 \cdot 2 + 1 \cdot 3 \implies 3 = 3$. This confirms the y-coordinate is invariant.
From the first row: $8 = 1 \cdot 2 + k \cdot 3$

$$8 = 2 + 3k$$ $$6 = 3k$$ $$k = 2$$

Step 4: Write the shear matrix.

$$S = \boxed{\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}}$$

a) (ii) The coordinates of B' and C'.

Step 1: Apply the shear matrix $S = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$ to point B(5, 2).

$$\begin{bmatrix} x'_{B} \\ y'_{B} \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 5 \\ 2 \end{bmatrix} = \begin{bmatrix} (1)(5) + (2)(2) \\ (0)(5) + (1)(2) \end{bmatrix} = \begin{bmatrix} 5 + 4 \\ 0 + 2 \end{bmatrix} = \begin{bmatrix} 9 \\ 2 \end{bmatrix}$$

So, B' is $\boxed{(9, 2)}$.

Step 2: Apply the shear matrix $S = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$ to point C(4, -1).

$$\begin{bmatrix} x'_{C} \\ y'_{C} \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 4 \\ -1 \end{bmatrix} = \begin{bmatrix} (1)(4) + (2)(-1) \\ (0)(4) + (1)(-1) \end{bmatrix} = \begin{bmatrix} 4 - 2 \\ 0 - 1 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \end{bmatrix}$$

So, C' is $\boxed{(2, -1)}$.

b) (i) Find the coordinates of triangle A''B''C''.

Step 1: Apply the transformation matrix $M = \begin{bmatrix} -2 & 0 \\ 0 & 1 \end{bmatrix}$ to A'(8, 3).

$$\begin{bmatrix} x''_{A} \\ y''_{A} \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 8 \\ 3 \end{bmatrix} = \begin{bmatrix} (-2)(8) + (0)(3) \\ (0)(8) + (1)(3) \end{bmatrix} = \begin{bmatrix} -16 \\ 3 \end{bmatrix}$$

So, A'' is $\boxed{(-16, 3)}$.

Step 2: Apply the transformation matrix $M = \begin{bmatrix} -2 & 0 \\ 0 & 1 \end{bmatrix}$ to B'(9, 2).

$$\begin{bmatrix} x''_{B} \\ y''_{B} \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 9 \\ 2 \end{bmatrix} = \begin{bmatrix} (-2)(9) + (0)(2) \\ (0)(9) + (1)(2) \end{bmatrix} = \begin{bmatrix} -18 \\ 2 \end{bmatrix}$$

So, B'' is $\boxed{(-18, 2)}$.

Step 3: Apply the transformation matrix $M = \begin{bmatrix} -2 & 0 \\ 0 & 1 \end{bmatrix}$ to C'(2, -1).

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