A transformation matrix $T1 = eginpmatrix 1 & 0 \ 0 & 2 ext maps a triangle ABC onto triangle A'B'C'. Another transformation matrix T2 = eginpmatrix 3 & -2 \ 2 & -1 ext maps triangle A'B'C' into tri

18.

a) A transformation matrix $T_1 = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$ maps a triangle ABC onto triangle A'B'C'. Another transformation matrix $T_2 = \begin{pmatrix} 3 & -2 \\ 2 & -1 \end{pmatrix}$ maps triangle A'B'C' into triangle A''B''C''. The coordinates of point C'' is (10, 8) and the area of triangle A''B''C'' is 15 square units.

i) Determine the coordinates of C.

Step 1: Find the coordinates of C' using the transformation $C'' = T_2 \cdot C'$.
Let $C' = \begin{pmatrix} x' \\ y' \end{pmatrix}$.

$$\begin{pmatrix} 10 \\ 8 \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} x' \\ y' \end{pmatrix}$$

This gives the system of equations:

$$3x' - 2y' = 10 \quad \text{(1)}$$ $$2x' - y' = 8 \quad \text{(2)}$$

From (2), $y' = 2x' - 8$.
Substitute this into (1):

$$3x' - 2(2x' - 8) = 10$$ $$3x' - 4x' + 16 = 10$$ $$-x' = 10 - 16$$ $$-x' = -6 \implies x' = 6$$

Substitute $x' = 6$ back into $y' = 2x' - 8$:

$$y' = 2(6) - 8 = 12 - 8 = 4$$

So, the coordinates of C' are $(6, 4)$.

Step 2: Find the coordinates of C using the transformation $C' = T_1 \cdot C$.
Let $C = \begin{pmatrix} x \\ y \end{pmatrix}$.

$$\begin{pmatrix} 6 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$

This gives the system of equations:

$$1x + 0y = 6 \implies x = 6$$ $$0x + 2y = 4 \implies 2y = 4 \implies y = 2$$

The coordinates of C are:

$$\boxed{(6, 2)}$$

ii) Determine the area of triangle ABC.
The area of an image after a transformation is given by Area_image = |determinant of matrix| $\times$ Area_object.
Let Area(ABC) be $A$.
Let Area(A'B'C') be $A'$.
Let Area(A''B''C'') be $A''$. We are given $A'' = 15$ square units.

Step 1: Calculate the determinant of $T_1$.

$$\det(T_1) = \det \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} = (1)(2) - (0)(0) = 2$$

Step 2: Calculate the determinant of $T_2$.

$$\det(T_2) = \det \begin{pmatrix} 3 & -2 \\ 2 & -1 \end{pmatrix} = (3)(-1) - (-2)(2) = -3 - (-4) = -3 + 4 = 1$$

Step 3: Relate the areas.

$$A'' = |\det(T_2)| \times A'$$ $$A' = |\det(T_1)| \times A$$

Combining these, we get:

$$A'' = |\det(T_2)| \times |\det(T_1)| \times A$$ $$15 = |1| \times |2| \times A$$ $$15 = 1 \times 2 \times A$$ $$15 = 2A$$ $$A = \frac{15}{2} = 7.5$$

The area of triangle ABC is:

$$\boxed{7.5 \text{ square units}}$$

b) The coordinates of points B and B'' are $(x, y)$ and $(6x + 1, 8)$ respectively. Determine the value of y.

Step 1: Find the composite transformation matrix $T = T_2 T_1$.

$$T = T_2 T_1 = \begin{pmatrix} 3 & -2 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$$ $$T = \begin{pmatrix} (3)(1) + (-2)(0) & (3)(0) + (-2)(2) \\ (2)(1) + (-1)(0) & (2)(0) + (-1)(2) \end{pmatrix}$$ $$T = \begin{pmatrix} 3+0 & 0-4 \\ 2+0 & 0-2 \end{pmatrix} = \begin{pmatrix} 3 & -4 \\ 2 & -2 \end{pmatrix}$$

Step 2: Apply the composite matrix to point B to get B''.
Let $B = \begin{pmatrix} x \\ y \end{pmatrix}$ and $B'' = \begin{pmatrix} 6x+1 \\ 8 \end{pmatrix}$.

$$\begin{pmatrix} 6x+1 \\ 8 \end{pmatrix} = \begin{pmatrix} 3 & -4 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$

This gives the system of equations:

$$3x - 4y = 6x + 1 \quad \text{(1)}$$ $$2x - 2y = 8 \quad \text{(2)}$$

Step 3: Solve the system of equations for $y$.
From (2), divide by 2:

$$x - y = 4 \implies x = y + 4$$

Substitute $x = y + 4$ into (1):

$$3(y+4) - 4y = 6(y+4) + 1$$ $$$$