Advanced Mathematics Assignment: Functions, Matrices, and Equations

This document presents a complete solution to an advanced mathematics assignment. It begins with an introduction and then addresses four main questions. Question 1 covers function analysis, including finding function values, inverse functions, domains, and ranges. Question 2 delves into function properties, linear functions, and composite functions. Question 3 focuses on matrices, including matrix operations, adjoints, and finding unknown matrices. Finally, Question 4 tackles matrix operations, solving simultaneous equations using inverse matrices, and expressing solutions in terms of variables. The assignment provides detailed step-by-step solutions and explanations for each problem, demonstrating various mathematical concepts.

Advance
Mathematics

Table of Contents
INTRODUCTION...........................................................................................................................1
Question-1........................................................................................................................................1
Question-2........................................................................................................................................2
Question -3.......................................................................................................................................3
Question 4 .......................................................................................................................................6

INTRODUCTION
Question-1
A) Give the function f(x) = 2√x+4
I) Find F(x)=0
f(0)=2√x+4
=2√0+4
=2*2
=4
II) f(x) = 9x^2
F(9x^2) = 2√x+4
=2√9x^2+4
=2√3x+2
III) Find domain and range for this function
f(x) = 2√x+4
Domain= x≥ -4
Range= f≥ 0 (all non-negative real numbers)
B) Show that f(x) =2x+3 and g(x)=x-3/2 are inverse
Given function where
f(x)=2x+3
let the y=2x+3
let the inverse function g(y)=x
Find the x in terms of y
Therefore,
2x=y-3
x=(y-3)/2
Therefore,
g(y)=(y-3)/2
y is the dummy variable of this function
Hence g(y) =(x-3)/2 which is an inverse function of f(x)=2x+3
1

c) The function f is defined by :
I) f(x)= 3(x+1)/(2x^2+7x-4) – (1/(x+4))
=3(x+1)/ (2x^2+8x-x-4) – (1/(x+4))
=3(x+1)/ (2x(x+4)-1(x+4)) – (1/(x+4))
=3(x+1)/ ((x+4) (2x+1))- (1/(x+4))
= (1/(x+4))( (3x+3-2x+1)/ (2x-1))
= (1/(x+4)) ((x+4)/ (2x+1))
=1/(2x+1)
Hence proved.
ii) Find f^-1 (x)
f^-1 (x)= 1/(2x+1)
= 2x-1
= x= 1/2
iii) Find the domain of f^-1
f^-1= (2x-1)
where x= ½ so the domain of this function are the all real numbers.
iv) Given that g(x) = In (x+1)
Find the solution of g(x) =1/7 give your answer in term of e.
=(1/7+1)
= 0.14+1=1.14
Question-2
The function f has domain -2≤x≤6 and is a linear from (-2,10) to (2,0) and from (2,0)
to (6,4). A sketch of the graph of y=f(x) is shown in figure 1.
I) write down the range of (f)
0≤F (X) ≤,0
II) Find ff(0)
The function is defined by g:x= (4-3x)/(5-x)
F(0)=5
F(5)=3
III) Find g^-1(x)
Y=4+3X/5-X
2

X= 4+3Y/5-Y
X(5-Y)=4+3Y
5X-XY=4+3Y
5X-4/3+X=Y
=G^-1(X)=5X-4/3+X
IV) Solve the equation g f(x) =16
G^-1=5(16)-4/3+(16)
=4
F(X)=4
X=6 OR X=2/5
Question -3
A) Given the matrices P= [3 -2] Q= [-4 2]
[7 -4] [-7 3]
I) PQ= P= [3 -2] Q= [-4 2]
[7 -4] [-7 3]
=(3*(-4)+ (-2)*(-7) 3*2+ (-2)*3)
(7*(-4) +(-4)*(-7) 7*2+(-4)*3)
=(2 0)
(0 2)
II) QP= Q= [-4 2] P= [3 -2]
[-7 3] [7 4]
=(-4*3+2*7 -4*(-2)+ 2*(-4))
(-7*3+3*7 -7*(-2)+ 3*(-4))
=(2 0)
(0 2)
III) Write a possible relationship between P and Q
Q= adj(P)
Q matrix is adjoint of P matrix2A+x=b
B) (I) Find A and B if 4[a 2] + [b -2] = [7 6]
[-1 3a] [5 -b] [1 1]
3

= [4a 8] + [b -2] = [7 6]
[-4 12a] [5 -b] [1 1]
4a+b=7 -1equation
8-2=6 -2 equation
=6= 6
-4+5=1 -3 equation
=1=1
12a-b=1 -4 equation
Computing the equation 1 and 4
4a+b=7 -1equation
12a-b=1 -4 equation
multiplying equation 1 by 3
12a+3b=21
12a-b =1
= 2b=20
= b=10
Putting the value of b=10 in equation 4
12a-b=1
12a -10=1
12a=11
a=11/12
The values are (11/12 , 10)
(ii) If a = [3 2] and b= [0 -4]
[-1 1] [-2 8]
find matrix x such that 2A-X=B
2A-B=X
= 2[3 2] - [0 -4]
[-1 1] [-2 8]
= [6 4] - [0 -4]
4

[-2 2] [-2 8]
= [6 8]
[0 -6]
C) Given a= [0 1]
[2 3]
I) show that A is non-singular
A square matrix A, such that |A|≠0 is called non singular matrix
A= [0 1]
[2 3]
|A|=[0 1]
[2 3]
=0*3 – 1*2
= 0-2
= -2
Here |A|≠0 so A is non singular matrix−1s
II) Find B such that BA^2=A
B [0 1]^2 = [0 1]
[2 3]^2 [2 3]
B [0 1] = [0 1]
[4 9] [2 3]
let B matrix be a,b,c,d
[a b] * [0 1] = [0 1]
[c d] * [4 9] [2 3]
= [4b a+9b] = [0 1]
[4d c+9d] [2 3]
4b=0 a+9b=1
=b=0 =a+9(0)=1
=a+0=1
=a=1
4d=2 c+9d=3
d=2/4 c+9(1/2)=3
5

d=1/2 c= (6-9)/2
c= -3/2
D) The matrix A=[3 -5] is the inverse of B = [2 5]
[-1 2] [1 3]
Verify this statement.
B^-1=A
B^-1=1/1 [3 -5]
[-1 2]
Question 4
a) The matrices A and B are given by:
A= [2 3] B=[3 4]
[-1 1] [1 1]
i)Find ABA^-1 and AB^-1A^-1
A^-1= 1/2*1-3*1 [1 -3]
[1 2]
A^-1= 1/(2+3) [1 -3]
[1 2]
A^-1= [1/5 -3/5]
[1/5 2/5]
AA^-1 =I
[2 3] * [1/5 -3/5] = [1 0]
[-1 1] [1/5 2/5] [0 1]
AA^-1 B
[1 0] * [3 4]
[0 1] [1 1]
=[3+0] [4+0]
[0+1] [0+1]
=[3 4]
[1 1]
6

ii)AS^-1 A^-1
B= [3 4]
[1 1]
B^-1 = 1/(3-4) [1 -4]
[-1 3]
=1/-1 [1 -4]
[1 -3]
B^-1 = [-1 4]
[1 -3]
AA^-1 B^-1 =I
= [1 0] [-1 4]
[0 1] [1 -3]
= [-1 4]
[1 -3]
II) Verify that (ABA^-1)-1=AB^-1A^-1
= [3 4] = [-1 4]
[1 1] [1 -3]
=1/ (3-4) [1 -4]
[-1 3]
= 1/-1 [1 -4]
[-1 3]
= [-1 4]
[1 -3]
b) Three simultaneously equations are given below:
x+y =a
y+2z =b
x-y- 3z =c
I) Write the simultaneously equation above as matrix equation.
Solution: [1 1 0] [a]
[0 1 0] [x y z] = [b]
[1 -1 -3] [c]
7

II) Solve the simultaneous equation above by using inverse matrix method
A^-1 = [1 1 0]
[0 1 2]
[1 1 -3]
=[1*3-2*1 0*-3-2*1 0*1-1]
[1*-3-0 -3-0 1-1]
[2-0 2-0 1-0]
= [-5 -2 -1]
[-3 -3 0]
[2 2 1] (minor matrix)
Matrix of co factor:
[-5 2 -1]
[3 -3 0]
[2 -2 1]
Adjoint of matrix
[-5 2 2]
[3 -3 -2]
[-1 0 1]
Determinant= (1*-5)+(1*-2)+(0*-1) =-7
A^-1= 1/-7 [-5 2 2]
[3 -3 -2]
[-1 0 1]
A^-1= [-5/-7 2/-7 2/-7]
[3/-7 -3/-7 -2/-7]
[-1/-7 0/-7 1/-7]
x= [x] [0.7 -0.2 -0.2] [a]
[y] [-0.4 0.4 0.2] [b]
[z] [0.1 0 -0.1] [c]
8

III) Give the expression for x, y, z in term of a, b, c
[0.7a -0.2b -0.2c] [x]
[-0.4a 0.4b 0.2c] = [y]
[0.1a 0 -0.1c] [z]
x= 0.7a-0.2b-0.2c
y=-0.4a+0.4b+0.2c
z= 0.1a-0.1c
9