Welcome back Sewercydaa — missed you this week.
Here is the solution to question 16:
a)
Step 1: Write the coordinates of triangle XYZ as column vectors.
X=[2−1],Y=[4−1],Z=[42]
Step 2: Apply transformation T1=[10−31] to find the coordinates of X'Y'Z'.
X′=[10−31][2−1]=[1(2)+(−3)(−1)0(2)+1(−1)]=[2+30−1]=[5−1]
Y′=[10−31][4−1]=[1(4)+(−3)(−1)0(4)+1(−1)]=[4+30−1]=[7−1]
Z′=[10−31][42]=[1(4)+(−3)(2)0(4)+1(2)]=[4−60+2]=[−22]
The coordinates of triangle X'Y'Z' are X'(5, -1), Y'(7, -1), Z'(-2, 2).
(Drawing on a grid is not possible in this format, but these are the coordinates you would plot).
b)
Step 1: Apply transformation T2=[0110] to the coordinates of X'Y'Z' to find X''Y''Z''.
X′′=[0110][5−1]=[0(5)+1(−1)1(5)+0(−1)]=[−15]
Y′′=[0110][7−1]=[0(7)+1(−1)1(7)+0(−1)]=[−17]
Z′′=[0110][−22]=[0(−2)+1(2)1(−2)+0(2)]=[2−2]
The coordinates of triangle X''Y''Z'' are X''(-1, 5), Y''(-1, 7), Z''(2, -2).
(Drawing on a grid is not possible in this format, but these are the coordinates you would plot).
c)
Step 1: The single transformation matrix T that maps XYZ to X''Y''Z'' is the product of T2 and T1, in that order (T2 first, then T1 applied to the original object, so T=T2T1).
T=T2T1=[0110][10−31]
Step 2: Perform the matrix multiplication.
T=[(0)(1)+(1)(0)(1)(1)+(0)(0)(0)(−3)+(1)(1)(1)(−3)+(0)(1)]
T=[0+01+00+1−3+0]
T = \begin{bmatrix 0 & 1 \\ 1 & -3 \end{bmatrix}}
d)
Step 1: The area of the image is given by the formula: Area of image = ∣det(T)∣× Area of object.
Step 2: Calculate the determinant of the transformation matrix T.
T=[011−3]
det(T)=(0)(−3)−(1)(1)=0−1=−1
Step 3: Calculate the area of triangle X''Y''Z''.
Area(X”Y”Z”)=∣det(T)∣×Area(XYZ)
Area(X”Y”Z”)=∣−1∣×15cm2
Area(X”Y”Z”)=1×15cm2
\text{Area(X''Y''Z'')} = 15 \text{ cm^2}
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