A linear transformation T maps A(2, 1) to A'(4, 1). If the transformation matrix T has the form [1 k; 0 1], find T and the images of B(3, 2), C(3, 4), D(2, 3) under T.

The transformation T is a shear with the x-axis invariant. A shear matrix with the x-axis invariant has the form $\begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}$.
Given A(2, 1) is mapped to A'(4, 1).

$$\begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 4 \\ 1 \end{bmatrix}$$

This gives two equations:
$1(2) + k(1) = 4 \Rightarrow 2 + k = 4 \Rightarrow k = 2$
$0(2) + 1(1) = 1 \Rightarrow 1 = 1$ (consistent)
So, the transformation matrix T is:

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

Let the transformation matrix H be $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$.
Given A'(4, 1) is mapped to A''(0, 1) and B'(7, 2) is mapped to B''(-12, -15).
For A':

$$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 4 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$

This gives:

1. $4a + b = 0$
2. $4c + d = 1$

For B':

$$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 7 \\ 2 \end{bmatrix} = \begin{bmatrix} -12 \\ -15 \end{bmatrix}$$

This gives: 3) $7a + 2b = -12$ 4) $7c + 2d = -15$

From (1), $b = -4a$. Substitute into (3):
$7a + 2(-4a) = -12$
$7a - 8a = -12$
$-a = -12 \Rightarrow a = 12$
Substitute $a = 12$ into $b = -4a$:
$b = -4(12) = -48$

From (2), $d = 1 - 4c$. Substitute into (4):
$7c + 2(1 - 4c) = -15$
$7c + 2 - 8c = -15$
$-c + 2 = -15$
$-c = -17 \Rightarrow c = 17$
Substitute $c = 17$ into $d = 1 - 4c$:
$d = 1 - 4(17) = 1 - 68 = -67$

So, the transformation matrix H is:

$$\mathbf{H} = \boxed{\begin{bmatrix} 12 & -48 \\ 17 & -67 \end{bmatrix}}$$