University regulations against plagiarism and collusion
Evidence of plagiarism or collusion will be taken seriously and the University regulations will be applied fully. You are advised to be familiar with the University’s definitions of plagiarism and collusion.
Instructions:
 This is an individual assignment. No duplication of work will be tolerated. Any plagiarism or collusion may result in disciplinary action. TMA 2 covers topics from Unit 3 and Unit 4. No need to submit your TMAs to Turnitin.
 Submit your TMA2 file through the Online Assignment Submission (OAS) system. The total marks for TMA 1 is 100 % and it contributes 25 % towards our total grade. Marks will be awarded for correct working steps and answers.TMA 1 contains FOUR (4) questions.
 Students are highly encouraged to passage their TMAs to the Turnitin system before submission, to encourage honest academic writing and it is not mandatory except for Project courses".
PLAGIARISM DECLARATION
I declare that the attached work is entirely my own (or when submitted to meet the requirements of an approved group assignment is the work of the group), except where materials quoted or paraphrased is acknowledged in the text. I also declare that this work / assignment has not been submitted for assessment in any other unit or course without due acknowledgement.
TMA Mark Deduction Penalties due to Plagiarism:
For first time offenders the following marks shall be deducted if Plagiarism is detected in the TMAs
 Assignments with plagiarized content of 10  30 %: 20% deduction from the total marks scored.
2, Assignments with plagiarized content of 31  50 %: 40% deduction from the total marks scored.
 Assignments with plagiarized content of more than 50%: Zero mark would be given.
Students shall face possible expulsion from the University (or any other penalties, as deemed appropriate) if they have been found to be repeat offenders of TMA plagiarism.
Question 1
(a) Given:
Compute the following:
 A+B [2 marks]
(ii) 2A3B [2 marks]
(iii) AB [2 marks]
(iv) AB+BA [2 marks]
(b) If and B , find matrices such that
[5 marks]
(c) For the matrix , find
 det (A)
[3 marks] [3 marks]
 if
[Hint: multiply both side by ]
[3 marks]
 if
[Hint: multiply both side by ]
[3 marks]
Question 2
(a) Solve the following system of linear equations by using the method of substitution. [4 marks]
(b) Solve the following system of linear equations by using the method of elimination. [5 marks]
(c) Solve the following equations by using Gaussian elimination. [8 marks]
Question 3
 Rewrite the following system of linear equations as matrix equations and solve them using inverse matrix method [4 marks](ii)
[4 marks]
 Write the following systems of linear equations as matrix equation and then as an augmented matrix:
[3 marks]
[3 marks]
 Use Gaussian elimination method to solve the system of 3 linear equations in (b)(i).
[5 marks]
(d) Use Cramer’s rule to solve the system of 2 linear equations in 3(a)(i) and 3(a)(ii) [6 marks]
question 4
 A sum of RM6 000 was deposited in a bank at a rate of 4.2% compounded semiannually. Calculate the total amount after 3 years. [3 marks]
(b) A sum of RM5 000 yielded an interest of RM978.10 after 3 years compounded quarterly. Find the annual compounded interest rate, correct to the nearest whole number. [4 marks]
(c) Mr.Gan decided to deposit RM2 500 at the end of every year for 5 years in an account with a bank. The annual interest is at 5.0% compounded annually. [This is an annuity question.]
Find the amount Mr.Gan has in the bank at the end of the
 second year; [3 marks]
(ii) third year; [3 marks]
(iii) 5^{th} year by using the formula given below:
Where is the future value,
is the deposit made every period,
is the interest rate at each period(in %),
is the number of periods involved in an annuity
[4 marks]
(c) An arithmetic sequence is given as 5, 11, 17, ………, 599.
(i) State the values of ‘a’, the first term of the sequence
[1 mark]
(ii) Find the value of ‘d’, the common difference of the sequence
[1 mark]
 Find T_{15}, the 15^{th} term of the sequence
[2 marks]
 Find the total number of terms, n, in the sequence, where 599 is the last term [2 marks]
 Find the sum of all the terms of the sequence [2 marks]
Question1:
 Given that, A = , B =
 A + B = + = (Ans)
 2A – 3B = 2 3 = (Ans)
 AB = = (Ans)
 AB + BA =
= = (Ans)
 A = , B =
2A + x = B
 x = B – 2A.
 x =  2
=
= (Ans)
 c. Given, A =
 det(A) = = 18 2 = 16 (Ans).
 A =
Matrix of minors of A is :
Matrix of cofactors of A is:
Adjugate of the last matrix is:
Determinant of the adjugate matrix is: = 16 = k (say)
Therefore, the inverse of A is :
A^{1} = (1/k) *A = (1/16) * = = (Ans).
iii. AX =
 A^{1}Ax = A^{1}[Premultiplying both sides by A^{1}]
 x = A^{1}
 x = [from ii]
 x = (Ans).
iv yA =
 yAA^{1} = [postmultiplying both sides by A]
 y = =
Question 2.
 Given the equations are:
x + 3y = 11 i
3x + 2y = 30 ii
The equations are to be solved by the method of substitution.
From equation i,
x + 3y = 11.
 x = 11 – 3y.
Substituting the value of x in equation ii
3x + 2y = 30.
 3 (113y) + 2y = 30.
 7y = 30+33 = 63
 y =  (63/7) = 9.
Putting y in equation i,
x + 3y = 11.
 x = 11 – 3y
 x = 11 + 27
 x = 16.
So, the solutions are x=16 and y = 9.
 Given the equations are:
3x + 2y + 9 = 0 i.
4x = 3y +5 ii
The equations are to be solved by method of elimination.
Multiplying i. by 4 and ii by 3
12x + 8y + 36 = 0
()12x –(+) 9y –(+)15 = 0
17y + 51 = 0
 y = (51/17) = 3.5.
Therefore, the solutions are:
x = 3.5 and y = 3.
Given the equations are:
x + y – z = 4.
x – 2y – 2z = 5.
2x – y + 2z = 2.
The equations are to be solved by Gaussian elimination method.
Therefore, the solution of the equations are :x = (11/15)y = (46*15)z = (1/5) Given the equations are: x + 3y – z = 42x + z = 7X – 2y + 3z = 13. The equations are to be solved by Gaussian elimination method. Therefore, the solutions of the equations are: x = (244/25)y = (77/25)z = (1700/25) Question 3: Given the equations are:
 [= []*1 = [].
 [] = []
 x = 5 and y = 1.
 Given the equations are:
 = .
 = *
 = 7*[ = [
 Given the linear equations are:
 * =
 Given the system of linear equations are:
 * =
 = ^(1) *
 Given the equation of b(i) are:
 Given the equations in 3.a.i. are:
 Given, principal or p = 6000.
 Given that,
 Given that the series is,
 The first term of the sequence is 5.
 The common difference of the sequence is 6.
 Sum of all the terms of the sequence is n*(a1 + an)/2, where, n= total number of terms, a1 = first term and an = last term.
 Given that the principal or p is 2500.
 Amount after 2 yrs is,
 Amount after 3 yrs or A is :
 A = p[{1 + (r/100)^{n} 1}/(R/100)]
System of Equations 
Row operations 
Augmented matrix 
x + y – z = 4 x – 2y – 2z = 5 2x – y + 2z = 2 

x + 3y – z = 4 3y – z = 9 3y + 4z = 10 
L2 = L2 – (L1). L3 = L3 – 2*L1 

4x + y = 6 15y = 46 3y + 4z = 10 
L1 = 4*L1 + L3 L2 = 4*(L2) + L3. 

X = (11/15) y=(46/15) z = (1/5) 
L1 = 15*L1 + L2 L3 = L3 – (L2/5) 

System of equations 
Row operations 
Augmented matrix 
x + 3y – z = 4 2x + z = 7 x – 2y + 3z = 13 

x + 3y – z = 4 6y – z = 15 y + 4z = 17 
L2 = L2 – 2*(L1) L3 = L3 – L1 

x – 3y = 19 25y = 77 y + 4z = 17 
L1 = L1 – L2 L2 = 4*L2 + L3 

x = (244/25) y = (77/25) z = (1700/25) 
L1 = L1 + (3/25)L2 L3 = L3 – (L2/25) 

System of equations 
Row operations 
Augmented matrix 
x + y – z = 4 x – 2y + 2z = 5 2x – y + 2z = 2 

x + y – z = 4 y + z = 3 3y + 4z = 10 
L2 = L2 – L1 L3 = L3 – 2*(L1) 

x = 1 y = 2 z = 1 
LI = LI + L2 L2 = 4*(L2) – L3 L3 = L3 – 3*(L2) 
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