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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

  1. 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.   
  1. 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.
  1. 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

  1. 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.

  1. 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)    2A-3B                                                                                                  [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 semi-annually. 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)      5th 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 T15, the 15th 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:

  1. Given that, A =  , B = 
  1. A + B =  +  =  (Ans)
  2. 2A – 3B = 2- 3 =  (Ans)
  • AB = =  (Ans)
  1. AB + BA =   

=  =  (Ans)

  1. A = , B = 

2A + x = B

  • x = B – 2A.
  • x = -  2

     =

     = (Ans)

  1. c. Given, A =
  2. det(A) = = 18 -2 = 16 (Ans).
  3. A =

Matrix of minors of A is :-   

Matrix of co-factors 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-1Ax = A-1[Pre-multiplying both sides by A-1]
  • x = A-1
  • x = [from ii]
  • x = (Ans).

iv      yA =

  • yAA-1 = [post-multiplying both sides by A]
  • y = =  

Question 2. 

  1. 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 (-11-3y) + 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. 

  1. 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:
2x + y = 5.x – y = 1.
  • [= []*-1 = [].
  • [] = []
  • x = -5 and y = -1.
             
  • Given the equations are:
            x + 3y = 2            2x – y = 11.The equations are to be solved through inverse matrix method.            x +3y = 2.            2x – y = 1.
  • = .
  • = *
      =
  • = 7*[ = [
Therefore, x = 14 and y = 7.
  • Given the linear equations are:
      x + y – z = 4.x – 2y + 2z = -5.2x – y + 2z = -2.            The system of equations are to be written in matrix equation form or in the form of augmented matrix.Therefore,x + y – z = 4x – 2y + 2z = -52x – y + 2z = -2
  • *      = 
Therefore, the matrix equation form is:   = and the augmented matrix is of the form:
  • Given the system of linear equations are:
x + 3y – z = -42x + z = 7X – 2y + 3z = 13
  • * =
  •  =  ^(-1) * 
The matrix equation form is:- = ^(-1) *            And the augmented matrix form is :                                                                        
  • Given the equation of b(i) are:
x + y –z = 4x – 2y + 2z = -52x – y + 2z = -2The equations are to be solved using Gaussian elimination method. Therefore, the solutions of the equations are:x = 1,y = 2,z = -1.
  • Given the equations in 3.a.i. are:
x + y – z = 4x – 2y + 2z = -52x – y + 2z = -2The equations are to be solved by Cramer’s rule.By the rule:x = (Dx/D), y = (Dy/D), z = (Dz/D).where,D =Dx =Dy =Dz =x = (Dx/D) =      = [4*(-2) -1*(-6)  - 1*1]/[1*(-2) – 1*(-2) – 1*3] =  (-8+6-1)/(-2+2-3) =-(3/3)= 1.y = (Dy/D) =        = [{1*(-6)} –{4*(-2)} –(1*8)]/[{1*(-2)}-{1*(-2)} – (1*3)] = (-6+8-8)/(-2+2-3) = 6/3 = 2.z = (Dz/D)  =           = [{1*(-1)}-{1*8} + {4*3}]/[{1*(-2)} – {1*(-2)} – (1*3)] = [-1-8+12]/[-2+2-3] = -(3/3) = -1.   Therefore, the equation solves at:x = -(3/5)y = 2z = -1.Again, the given the equations in 3.a.ii. are:-x + 3y – z = -4.2x + z = 7.x – 2y + 3z = 13.The equations are to be solved by Cramer’s rule:x = (Dx/D) =      = [{(-4)*2} – {3*8) – {(-14)*1}] / [{1*2}-{3*5} – {1*(-4)}] = (-8-24+14)/(2-15+4) = -(18/9) = -2.y = (Dy/D) =      = [{1*8} + {4*5} –{1*19}] / [{1*2} – {3*5}+ {1*4}] = [8+20-19] / [2-15+4] = -(9/9) = -1.z = (Dz/D) =       = [14-(3*19)+(4*4)] / [(1*2) – (3*5) + (1*4)] = [14-57+16] / [2-15+4] = -3.Therefore, the system of equation solves into :x = 1;,y= -1,z = -3 Question 4:
  • Given, principal or p = 6000.
Rate of interest or r = 4.2% = 0.042.Interest is compounded semi-annually.Therefore, amount received after 3 years or A = P*[1+ (r/n)]nt = 6000[1+(0.042/2)]2*3 = 6796.81898.Therefore, the total amount is RM 6796.81898.
  • Given that,
Principal or P = 5000.Amount or A = 5000+978.10 = 5978.10.Time or t = 3.The interest rate has compounded quarterly.Therefore, the rate of interest or r = [(A/P)^(1/3) – 1] = [(5978.10/5000)^(1/3) – 1] =6%Therefore, the required rate of interest is 6%.
  • Given that the series is,
   5.11.17.....,599.
  • The first term of the sequence is 5.
  • The common difference of the sequence is 6.
iii. The 15th term of the sequence is :T15 = (d*n) + (a-d), where, d = common difference.                                              a = first term                                               n = required number of terms       = 6*15 + (5-6) = 89.   Iv  Total number of terms in the sequence, that is n = [(last term – first term)/common difference] + 1 = (549/6) +1 =99+1=100.Therefore, there are 100 terms in the sequence.
  • 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.
Sum = n*(a1 +an)/2        = 100(5+599/2)100*(604/2) = 100*302 = 30200.Therefore, the sum is 30200.
  • Given that the principal or p is 2500.
Time or t = 5yrs.Rate of interest or r = 5% = 0.05.
  • Amount after 2 yrs is,
A = P*[1 + (r/n)]nt    =  2500 * [1+(0.05/1)]2     =2500*0.1025 = 2756.25.Therefore, the amount is RM 2756.25.
  • Amount after 3 yrs or A is :
A =  [p*{1+(r/n)}nt]= 2500[1+0.05]3  = 2894.063.Therefore, the amount is RM 2894.063
  • A = p[{1 + (r/100)n -1}/(R/100)]
     =500[{1+(5/100)5}/(5/100)]      = 2762.815625. Therefore, the amount is RM 2762.815625.

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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My Assignment Help (2020) University Plagiarism Declaration And Math Questions [Online]. Available from: https://myassignmenthelp.com/free-samples/wuc115-university-mathematics-b
[Accessed 14 July 2024].

My Assignment Help. 'University Plagiarism Declaration And Math Questions' (My Assignment Help, 2020) <https://myassignmenthelp.com/free-samples/wuc115-university-mathematics-b> accessed 14 July 2024.

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