A square S has vertices (0,0), (2,0), (2,2) and (0,2). It is mapped onto a shape T by the transformation represented by the matrix A = [[2,-1],[1,2]]. Find the coordinates of the vertices of T.

Here is the solution to question 1:

a)

i) The vertices of square S are (0, 0), (2, 0), (2, 2), and (0, 2).
(Drawing on a grid is not possible in this format).

ii)

Step 1: Identify the transformation matrix A from the given mapping.
The mapping is $\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 2x-y \\ x+2y \end{pmatrix}$.
This corresponds to the matrix $A = \begin{bmatrix} 2 & -1 \\ 1 & 2 \end{bmatrix}$.

Step 2: Apply matrix A to each vertex of S.
For P(0,0):

$$P' = \begin{bmatrix} 2 & -1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

For Q(2,0):

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

For R(2,2):

$$R' = \begin{bmatrix} 2 & -1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 2 \\ 2 \end{bmatrix} = \begin{bmatrix} 2(2) + (-1)(2) \\ 1(2) + 2(2) \end{bmatrix} = \begin{bmatrix} 4 - 2 \\ 2 + 4 \end{bmatrix} = \begin{bmatrix} 2 \\ 6 \end{bmatrix}$$

For S(0,2):

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

The coordinates of the image T are P'(0,0), Q'(4,2), R'(2,6), S'(-2,4).
(Drawing on a grid is not possible in this format).

iii)

Step 1: Identify the transformation matrix B.
The matrix is $B = \begin{bmatrix} 2 & 1 \\ -1 & 2 \end{bmatrix}$.

Step 2: Apply matrix B to each vertex of S.
For P(0,0):

$$P'' = \begin{bmatrix} 2 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

For Q(2,0):

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

For R(2,2):

$$R'' = \begin{bmatrix} 2 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 \\ 2 \end{bmatrix} = \begin{bmatrix} 2(2) + 1(2) \\ -1(2) + 2(2) \end{bmatrix} = \begin{bmatrix} 4 + 2 \\ -2 + 4 \end{bmatrix} = \begin{bmatrix} 6 \\ 2 \end{bmatrix}$$

For S(0,2):

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

The coordinates of the image U are P''(0,0), Q''(4,-2), R''(6,2), S''(2,4).
(Drawing on a grid is not possible in this format).

b)

i)

Step 1: Find the product AB.

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

Step 2: Apply the matrix AB to each vertex of S to find the image V.
For P(0,0):

$$P''' = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

For Q(2,0):

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