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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 ∝−rsinθ 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 )+sin θ (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 ,x1 sinθ+0∗cos θ)
T(x1, 0) = (x1cosθ ,x1sinθ)

When θ=0
Substituting to T(x1, 0)
Therefore, cos θ=1∧sin θ=0
So,
T(x1, 0) =(x1, 0)
= (x1x1, x1x0)
T(x1, 0) =(x1, 0)
That unvariant under T
c)
let T :∨R5 →∨R4 be the linear transformation
T(x1, x2, x3, x4, x5) = (x1 + x4 + 5x5 – x1 – x3, x2 + 2x3, 2x3, 2x1 – 2x2-2x3)
We know that
B = {e1, e2, e3, e4, e5} is the standard order for |R5
B1 = {e1, e2, e3, e4} is the standard order for |R4
T(e1) = T(1, 0, 0, 0, 0) = (1, -1, 0, 2)
= 1*e1 – 1*e2 + 0*e3 + 2*e4
T(e2) = T(0, 0,, 1, -2) = (1, -1, 0, 2)
= 0*e1 – 0*e2 + 1*e3 - 2*e4
T(e3) = T(0, 0, 1, 0, 0) = (0, -1, 2, 3)
= 0*e1 – 1*e2 + 2*e3 + 3*e4
T(e4) = T(0, 0, 0, 1, 0) = (4, 0, 0, 0)
= 4*e1 +0*e2 + 0*e3 + 0*e4
T(e5) = T(0, 0, 0, 0, 1) = (0, 5, 0, 0)
= 0*e1 +5*e2 + 0*e3 + 0*e4

Presenting the information in matrix form
{ T } B1
B =
[ 1 0 0
−1 0 −1
0 1 2
4 0
0 5
0 0
2 −2 −2 0 0 ]
{ T } B1
B has four linearly independent rows
Rank T = 4 =dim= ¿ R4
T is onto
From rank – Nullity
Rank T + Nullity T = dim¿ R5=5
Nullity T = 1 ≠ 0
T is therefore not one to one
Q4)
a) Let A = [0 0 0
1 0 0
0 1 0 ]
A2 = A * A = [ 0 0 0
1 0 0
0 1 0 ]∗
[ 0 0 0
1 0 0
0 1 0 ]
A2 = [0 0 0
0 0 0
1 0 0 ]
A3 = [ 0 0 0
0 0 0
1 0 0 ]∗
[ 0 0 0
1 0 0
0 1 0 ] = [ 0 0 0
0 0 0
0 0 0 ]
i) A3 = 0
ii) I – A = [1 0 0
0 1 0
0 0 1 ]− [0 0 0
1 0 0
0 1 0 ]=
[ 1 0 0
−1 1 0
0 −1 1 ]

I + A + A2 = [1 0 0
0 1 0
0 0 1 ]+
[0 0 0
1 0 0
0 1 0 ]+ [0 0 0
0 0 0
1 0 0 ]=
[1 0 0
1 1 0
1 1 1 ]
(I-A)(I + A+ A2) = [ 1 0 0
−1 1 0
0 −1 1 ]∗
[1 0 0
1 1 0
1 1 1 ]= [1 0 0
0 1 0
0 0 1 ] = I
I + A + A2 = I
(1− A)= ( I− A )−1
Hence
( I − A ) −1=I + A+ A2= [ 1 0 0
0 1 0
0 0 1 ]
b) if we take A4 = 0
then
I2 – A4 = I
(I – A2) (I – A2) = I
(I – A2) (I – A2) = I
I + A = ( I − A ) −1
( I − A )−1=(I + A2 )(I+ A2)
( I− A) ( I −A )−1=(I − A)( I+ A)(I + A2 )
= ( I− A2 )( I + A2) = I 2− A4
Since I2 = I and we took A4 = 0
( I − A ) ( I− A )−1 =I
Q6)
a) From the figure below volume of box is equivalent:

V = s*s * h
Therefore,
4 = s2h
h = 4
s2
So
Surface area of box is:
A = 4( h*s) + 2(s*s)
A = 2(s2 + 2s*h)
Writing the value of h in terms of s
A = 2(s2 + 2s * 4
s2 )
A = 2(s2 + 8
s ¿
b) Minimizing the value of A
da
ds =0
d
ds (2 (s2+ 8
s ))=0
2* d
ds ( s2 + 8
s ) =0

2*(2s - 8
s2 ¿=0
2s - 8
s2 =0
2s = 8
s2 =0
S3 = 8/2 = 4
S = 3
√4
d2 A
d s2 = d
ds (2 s− 8
s2 )
= 2 + 16
s3
At, s= 3
√4
d2 A
d s2 =2+ 16
( 3
√ 4 )
3
= 2 + 4 = 6
Second derivative at s = 3
√4 is positive
The minimum value of A at s = 3
√4
The value of h that minimizes A is
h = 4
s2 = 4
( 3
√ 4 )
2
=
4
4
2
3
= 41/3

Q7)
F(x) = ∫
0
x
t−3
t2+7 dt
Using fundamental theorem of calculus
a) F1(x) = x−3
x2 +7 , where, 1
x2 +7 >0 , ⍱ x ∈(−∞ , ∞)
F1(x) < 0, for x < 3
F1(x) > 0, for x > 3
Therefore,
F(x) is increasing in (3 , ∞)
F(x) is decreasing in (−∞, 3)
b)
F11(x) = x2 +7− ( x−3 ) ∗2 x
( x2 +7 ) 2 = x2+7−2 x2 +6 x
( x2 +7 ) 2
F11(x) = −x2+6 x +7
¿ ¿
therefore 1
( x2+7 ) 2 >0, for all x ∈(−∞, ∞)
concave upward F11(x) > 0, -x2 + 6x + 7 > 0
x2 - 6x - 7 < 0
(x-7)(x+1) < 0
x ∈(−1 , 7)
concave downward F11(x) < 0, -x2 + 6x + 7 < 0
x2 - 6x - 7 > 0
(x-7)(x+1) < 0

Therefore,
x ∈(−∞,−1) ( 7 , ∞ )