A shear transformation S maps A(2,3) onto A'(8,3). Find the matrix S. Find the images of B(5,2) and C(4,-1) under S.

Step 1: Find the matrix representing the shear (a)(i).
A shear transformation with the x-axis invariant has the matrix form $S = \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}$.
Given that A(2,3) is mapped onto A'(8,3):

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

This gives the equations:
$1(2) + k(3) = 8 \implies 2 + 3k = 8$
$0(2) + 1(3) = 3 \implies 3 = 3$
From the first equation:
$3k = 8 - 2$
$3k = 6$
$k = 2$
The matrix representing the shear is:

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

Step 2: Find the coordinates of B' and C' (a)(ii).
Using the shear matrix $S = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$:
For B(5,2):

$$\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 5 \\ 2 \end{bmatrix} = \begin{bmatrix} 1(5) + 2(2) \\ 0(5) + 1(2) \end{bmatrix} = \begin{bmatrix} 5 + 4 \\ 2 \end{bmatrix} = \begin{bmatrix} 9 \\ 2 \end{bmatrix}$$

So, B' is $\boxed{(9,2)}$.
For C(4,-1):

$$\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 4 \\ -1 \end{bmatrix} = \begin{bmatrix} 1(4) + 2(-1) \\ 0(4) + 1(-1) \end{bmatrix} = \begin{bmatrix} 4 - 2 \\ -1 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \end{bmatrix}$$

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

Step 3: Find the coordinates of triangle A''B''C'' (b)(i).
Triangle A''B''C'' is the image of A'B'C' under the transformation matrix $M = \begin{bmatrix} -2 & 0 \\ 0 & 1 \end{bmatrix}$.
Coordinates of A' are (8,3), B' are (9,2), and C' are (2,-1).
For A'(8,3):

$$\begin{bmatrix} -2 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 8 \\ 3 \end{bmatrix} = \begin{bmatrix} -2(8) + 0(3) \\ 0(8) + 1(3) \end{bmatrix} = \begin{bmatrix} -16 \\ 3 \end{bmatrix}$$

So, A'' is $\boxed{(-16,3)}$.
For B'(9,2):

$$\begin{bmatrix} -2 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 9 \\ 2 \end{bmatrix} = \begin{bmatrix} -2(9) + 0(2) \\ 0(9) + 1(2) \end{bmatrix} = \begin{bmatrix} -18 \\ 2 \end{bmatrix}$$

So, B'' is $\boxed{(-18,2)}$.
For C'(2,-1):

$$\begin{bmatrix} -2 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 \\ -1 \end{bmatrix} = \begin{bmatrix} -2(2) + 0(-1) \\ 0(2) + 1(-1) \end{bmatrix} = \begin{bmatrix} -4 \\ -1 \end{bmatrix}$$

So, C'' is $\boxed{(-4,-1)}$.

Step 4: Describe the transformation M fully (b)(ii).
The transformation matrix is $M = \begin{bmatrix} -2 & 0 \\ 0 & 1 \end{bmatrix}$.
This matrix represents a stretch with a scale factor of -2 parallel to the x-axis (with the y-axis as the invariant line). This can also be described as a reflection in the y-axis combined with a stretch of factor 2 parallel to the x-axis.

Step 5: Determine a single matrix that maps triangle A''B''C'' onto triangle ABC (c).
The transformation from ABC to A''B''C'' is given by the product of the matrices $M$ and $S$, in that order: $MS$.

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