Mapping a parallelogram into a square

It is apparent that finding a transformation from $\vec{y}$ to $\vec{x}$ is much easier than the reverse. Express this transformation as $\vec{x} = H\vec{y} + \vec{c}$ Notice that for the linear transformation $H\vec{y}$ , the image of $\vec{0}$ must be $\vec{0}$ . This is because $H$ 's columns represent the positions of the basis vectors $\hat{i},\hat{j}$ etc, and since a zero vector is $0\hat{i} + 0\hat{j}$ , the result must be $\vec{0}$ . However, we know that the image of $A' = \vec{0}$ is $A = \begin{pmatrix} 0\\3 \end{pmatrix}$ . Thus, $\begin{pmatrix} 0\\3 \end{pmatrix} = H\vec{0} + \vec{c} = \vec{c}$ To figure out the entries of $H$ , we find where the basis vectors $B' = \hat{i}$ and $D' = \hat{j}$ land following the transformation. $H\vec{y} = \vec{x} - \vec{c}$ $H\hat{i} = B - \begin{pmatrix} 0\\3 \end{pmatrix} = \begin{pmatrix} 10\\8 \end{pmatrix} - \begin{pmatrix} 0\\3 \end{pmatrix} = \begin{pmatrix} 10\\5 \end{pmatrix}$ $H\hat{j} = D - \begin{pmatrix} 0\\3 \end{pmatrix} = \begin{pmatrix} 2\\7 \end{pmatrix} - \begin{pmatrix} 0\\3 \end{pmatrix} = \begin{pmatrix} 2\\4 \end{pmatrix}$ Therefore, $H = \begin{pmatrix} 10 & 2\\ 5 & 4\end{pmatrix}$ Going back to the original equation, we can rearrange to see that $\vec{y} = H^{-1}(\vec{x} - \vec{c}) = H^{-1}\vec{x} - H^{-1} \vec{c}$ $H^{-1} = \frac{1}{\det(H)}\begin{pmatrix} 4 & -2\\ -5 & 10\end{pmatrix} = \frac{1}{30} \begin{pmatrix} 4 & -2\\ -5 & 10\end{pmatrix}$ Finally, $\vec{y} = \frac{1}{30} \begin{pmatrix} 4 & -2\\ -5 & 10\end{pmatrix}\vec{x} - \frac{1}{30} \begin{pmatrix} 4 & -2\\ -5 & 10\end{pmatrix}\begin{pmatrix} 0\\3 \end{pmatrix}$ $T = \frac{1}{30} \begin{pmatrix} 4 & -2\\ -5 & 10\end{pmatrix}$ $\vec{b} = -\frac{1}{30} \begin{pmatrix} -6\\ 30\end{pmatrix} = \begin{pmatrix} \frac{1}{5} \\-1\end{pmatrix}$ $S = \frac{4+2+5+10}{30} + \frac{1}{5} + 1 = \frac{57}{30} = \frac{19}{10} = \boxed{1.9}$

Let:

$T = \left[\begin{matrix} t_1&t_2\\t_3&t_4 \end{matrix} \right]$ $b = \left[\begin{matrix} b_1\\b_2 \end{matrix} \right]$

Now:

$\left[\begin{matrix} X_{A'}\\Y_{A'} \end{matrix} \right] = T \left[\begin{matrix} X_{A}\\Y_{A} \end{matrix} \right] + b$

Writing out the equations for each vector and its image gives rise to 8 equations in 6 unknowns ( $t_{1,2,3,4} \ , \ b_{1,2}$ ):

$X_{A'} = t_1 X_A + t_2Y_A + b_1$ $Y_{A'} = t_3 X_A + t_4Y_A + b_2$ $X_{B'} = t_1 X_B + t_2Y_B + b_1$ $Y_{B'} = t_3 X_B + t_4Y_B + b_2$ $X_{C'} = t_1 X_C + t_2Y_C + b_1$ $Y_{C'} = t_3 X_C + t_4Y_C + b_2$ $X_{D'} = t_1 X_D + t_2Y_D + b_1$ $Y_{D'} = t_3 X_D + t_4Y_D + b_2$

Using the first six equations to simultaneously solve for the unknown elements of $T$ and $b$ gives the required answer. The last two equations can be used for a sanity check.

$\left[ \begin{matrix} X_{A'}\\Y_{A'}\\X_{B'}\\Y_{B'}\\X_{C'}\\Y_{C'}\end{matrix} \right] = \left[ \begin{matrix} X_A&Y_A&0&0&1&0\\0&0&X_A&Y_A&0&1\\X_B&Y_B&0&0&1&0\\0&0&X_B&Y_B&0&1\\X_C&Y_C&0&0&1&0\\0&0&X_C&Y_C&0&1 \end{matrix} \right] \left[\begin{matrix} t_1\\t_2\\t_3\\t_4\\t_5\\t_6 \end{matrix} \right]$ $\implies \left[\begin{matrix} t_1\\t_2\\t_3\\t_4\\t_5\\t_6 \end{matrix} \right] = \left[ \begin{matrix} X_A&Y_A&0&0&1&0\\0&0&X_A&Y_A&0&1\\X_B&Y_B&0&0&1&0\\0&0&X_B&Y_B&0&1\\X_C&Y_C&0&0&1&0\\0&0&X_C&Y_C&0&1 \end{matrix} \right]^{-1}\left[ \begin{matrix} X_{A'}\\Y_{A'}\\X_{B'}\\Y_{B'}\\X_{C'}\\Y_{C'}\end{matrix} \right]$

The required answer is:

$\boxed{\lvert t_1 \rvert + \lvert t_2 \rvert + \lvert t_3 \rvert + \lvert t_4 \rvert + \lvert b_1 \rvert + \lvert b_2 \rvert = 1.9}$