Math1013, Mathematics and Applications 1: Assignment 2 Solution

This document contains the solutions for Math1013 Assignment 2 from the Australian National University, focusing on topics including linear algebra and calculus. The solutions cover reflection and rotation transformations, including finding standard matrices for combined reflections and rotations, and determining the resulting transformation. The assignment also includes problems related to linear transformations in vector spaces and their properties, such as finding the rank and nullity of a transformation. Furthermore, the solutions address optimization problems involving volume and surface area, utilizing calculus to minimize the surface area of a box with a fixed volume. Finally, the solutions include an analysis of a function defined by an integral, applying the fundamental theorem of calculus to find its derivative, and determining intervals of increasing/decreasing behavior and concavity.

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 , ∞ )