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Q2)
The reflection on the line x2 = 2x1
Substituting a = 2, to obtain a standard matrix T2
Ta = 1
1+ a2 [1−a2 2 a
2 a a2−1 ]
= 1
1+ 22 [1−22 2∗2
2∗2 22−1 ]
= 1
5∗
[ −3 4
4 3 ] => A
When x2 = 1/3 x1
 A = 1/3
Standard matrix of the matrix T1/3
Ta = 1
1+ a2 [1−a2 2 a
2 a a2−1 ]
= 1
1+ 1
3
2
[1−1
3
2 2
3
2
3
1
3
2
−1 ]
= 1
10
9 [ 1− 1
9
2
3
2
3
1
9 −1 ]
= 9
10 [ 8
9
2
3
2
3
−8
9 ]
= 1
10 [8 6
6 −8 ] => B

Given T: |R2 →∨R2 corresponds to reflection through the line x2 = 2x1 followed by reflection
through line x2 = 1/3 x1.
Substituting the value of A and B, the standard matrix will be
T = B * A
A = 1
5 [−3 4
4 3 ]
B = 1
10 [8 6
6 −8 ]
Therefore,
T = 1
10
[8 6
6 −8 ]∗1
5 [−3 4
4 3 ]
= 1
50 [ 0 50
−50 0 ]
C = [ 0 1
−1 0 ]
For any (x, y)∈∨R2
T(x, y) = C( x
y )
Substituting for the value of C
T(x, y) = [ 0 1
−1 0 ] ( x
y )
T(x, y) = (y, x)
This is a reflection on line y = x

b)
T: ¿ R3 →∨R3
T corresponds to a rotation in anticlockwise by an angle of θ
Therefore,
T(a*,b*) = (a1, b1)
Shown on the sketch below
a = rcos ∝
b = rsin ∝
and
a1 = rcos (θ+ ∝ )
b2 = rsin ( θ+∝ )
Therefore,
cos (a+ b)=cos a cos b+sin a sin b
a1 = r cos θ cos ∝−rsin θ sin∝
a1 = a cos θ−bsin θ
b1 = rsin ( θ+∝ )

Therefore,
sin(a+b)=sin a cos b+ sin a cos b
= rsinθcos ∝+rcos ∝ sinθ
Where
a = rcos ∝
b = rsin ∝
b1 = bcosθ +asinθ
TƟ(a, b) = (a1, b1)
TƟ(a, b) =(acosθ −bsinθ ,bcosθ +asinθ ¿ ¿
a = 1, b = 0
substituting
T(1, 0) =(1∗cosθ−0∗sinθ, 0∗cosθ +1∗sinθ ¿¿ = ¿)
= cos θ ( 1,0 ) + sinθ (0,1)
Where, a = 0 and b = 1, when substituted
T(0, 1) =(0∗cosθ−1∗sinθ, 1∗cosθ+ 0∗sinθ ¿¿
= ¿)
= ¿)
Standard matrix for T
T = [cos θ −sin θ
sin θ cos θ ]
Matrix has the intended effect of leaving all points on the x – axis, are of the form (x1, 0)
T(x1, 0) = (cos θ −sin θ
sinθ cos θ )(x 1
0 )
= (x1cosθ+0∗−sin θ ,x1sinθ+0∗cos θ)
T(x1, 0) = (x1cosθ ,x1sinθ)