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 (x′y′)=(2x−yx+2y).
This corresponds to the matrix A=[21−12].
Step 2: Apply matrix A to each vertex of S.
For P(0,0):
P′=[21−12][00]=[00]
For Q(2,0):
Q′=[21−12][20]=[2(2)+(−1)(0)1(2)+2(0)]=[42]
For R(2,2):
R′=[21−12][22]=[2(2)+(−1)(2)1(2)+2(2)]=[4−22+4]=[26]
For S(0,2):
S′=[21−12][02]=[2(0)+(−1)(2)1(0)+2(2)]=[−24]
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=[2−112].
Step 2: Apply matrix B to each vertex of S.
For P(0,0):
P′′=[2−112][00]=[00]
For Q(2,0):
Q′′=[2−112][20]=[2(2)+1(0)−1(2)+2(0)]=[4−2]
For R(2,2):
R′′=[2−112][22]=[2(2)+1(2)−1(2)+2(2)]=[4+2−2+4]=[62]
For S(0,2):
S′′=[2−112][02]=[2(0)+1(2)−1(0)+2(2)]=[24]
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=[21−12][2−112]
AB=[(2)(2)+(−1)(−1)(1)(2)+(2)(−1)(2)(1)+(−1)(2)(1)(1)+(2)(2)]
AB=[4+12−22−21+4]
AB = \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′′′=[5005][00]=[00]
For Q(2,0):