A triangle ABC with vertices A(1, -1), B(3, -1), C(1, 3) is mapped to A'B'C' by transformation T1 = [[-1, 0], [0, 1]]. Triangle A'B'C' is then mapped to A''B''C'' by transformation T2 = [[2, 2], [0, 1

Step 1: Understand the transformations.
Triangle ABC with vertices $A(1, -1)$, $B(3, -1)$, $C(1, 3)$ is mapped to $A'B'C'$ by transformation $T_1 = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$.
Triangle $A'B'C'$ is then mapped to $A''B''C''$ by transformation $T_2 = \begin{pmatrix} 2 & 2 \\ 0 & 1 \end{pmatrix}$.

a) Find the coordinates of A'B'C'.
To find the coordinates of $A'B'C'$, we multiply the coordinates of $A, B, C$ by the matrix $T_1$.

For $A(1, -1)$:

$$A' = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} (-1)(1) + (0)(-1) \\ (0)(1) + (1)(-1) \end{pmatrix} = \begin{pmatrix} -1 \\ -1 \end{pmatrix}$$

So, $A'(-1, -1)$.

For $B(3, -1)$:

$$B' = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} (-1)(3) + (0)(-1) \\ (0)(3) + (1)(-1) \end{pmatrix} = \begin{pmatrix} -3 \\ -1 \end{pmatrix}$$

So, $B'(-3, -1)$.

For $C(1, 3)$:

$$C' = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} (-1)(1) + (0)(3) \\ (0)(1) + (1)(3) \end{pmatrix} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}$$

So, $C'(-1, 3)$.

The coordinates of $A'B'C'$ are $\boxed{A'(-1, -1), B'(-3, -1), C'(-1, 3)}$.

b) Find the matrix which maps A'B'C' onto A''B''C''.
The problem statement explicitly gives this matrix as the second transformation.
The matrix is $\boxed{\begin{pmatrix} 2 & 2 \\ 0 & 1 \end{pmatrix}}$.

c) Determine the area scale factor between A'B'C' and A''B''C''.
The transformation from $A'B'C'$ to $A''B''C''$ is given by the matrix $T_2 = \begin{pmatrix} 2 & 2 \\ 0 & 1 \end{pmatrix}$.
The area scale factor is the absolute value of the determinant of the transformation matrix.

$$det(T_2) = (2)(1) - (2)(0) = 2 - 0 = 2$$

The area scale factor is $|2| = \boxed{2}$.

d) Find the transformation matrix which maps A''B''C'' onto ABC.
This requires finding the inverse of the combined transformation from $ABC$ to $A''B''C''$.
First, find the combined transformation matrix $T_{total}$ that maps $ABC$ to $A''B''C''$. This is $T_2 \cdot T_1$.

$$T_{total} = T_2 \cdot T_1 = \begin{pmatrix} 2 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$$ $$T_{total} = \begin{pmatrix} (2)(-1) + (2)(0) & (2)(0) + (2)(1) \\ (0)(-1) + (1)(0) & (0)(0) + (1)(1) \end{pmatrix}$$ $$T_{total} = \begin{pmatrix} -2 & 2 \\ 0 & 1 \end{pmatrix}$$

Now, find the inverse of $T_{total}$. For a matrix $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, its inverse is $M^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$.
Here, $a=-2, b=2, c=0, d=1$.

$$det(T_{total}) = (-2)(1) - (2)(0) = -2 - 0 = -2$$ $$T_{total}^{-1} = \frac{1}{-2} \begin{pmatrix} 1 & -2 \\ 0 & -2 \end{pmatrix}$$ $$T_{total}^{-1} = \begin{pmatrix} \frac{1}{-2} & \frac{-2}{-2} \\ \frac{0}{-2} & \frac{-2}{-2} \end{pmatrix}$$ $$T_{total}^{-1} = \boxed{\begin{pmatrix} -\frac{1}{2} & 1 \\ 0 & 1 \end{pmatrix}}$$