University’s Definition of Plagiarism and Collusion

Question1:

1. Given that, A =  , B = A + B =  +  =  (Ans)

1. 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.

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) 

Therefore, the solution of the equations are :

x = (11/15)

y = (46*15)

z = -(1/5) 

Given the equations are: 

x + 3y – z = -4

2x + z = 7

X – 2y + 3z = 13. 

 The equations are to be solved by Gaussian elimination method.

System of equations

Row operations

Augmented matrix

x + 3y – z = -4

2x + z = 7

x – 2y + 3z = 13 

TMA Mark Deduction Penalties due to Plagiarism

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) 

Therefore, the solutions of the equations are: 

x = -(244/25)

y = (77/25)

z = -(1700/25) 

Question 3:

1. Given the equations are:

2x + y = 5.

x – y = 1.

• [= []*-1 = [].
• [] = []
• x = -5 and y = -1.             

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.

1. 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 = 4

x – 2y + 2z = -5

2x – y + 2z = -2

• *      = 

Therefore, the matrix equation form is:

   = 

and the augmented matrix is of the form:

1. Given the system of linear equations are:

x + 3y – z = -4

2x + z = 7

X – 2y + 3z = 13

• * =
•  =  ^(-1) * 

The matrix equation form is:-

 = ^(-1) *

            And the augmented matrix form is :                                                         

1. Given the equation of b(i) are:

x + y –z = 4

x – 2y + 2z = -5

2x – y + 2z = -2

The equations are to be solved using Gaussian elimination method.

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) 

Therefore, the solutions of the equations are:

x = 1,

Question 1

y = 2,

z = -1.

1. Given the equations in 3.a.i. are:

x + y – z = 4

x – 2y + 2z = -5

2x – y + 2z = -2

The equations are to be solved by Cramer’s rule.

By the rule:

x = (D_x/D), y = (D_y/D), z = (D_z/D).

where,

D =

D_x =

D_y =

D_z =

x = (D_x/D) =      = [4*(-2) -1*(-6)  – 1*1]/[1*(-2) – 1*(-2) – 1*3] =  (-8+6-1)/(-2+2-3) =-(3/3)= 1.

y = (D_y/D) =        = [{1*(-6)} –{4*(-2)} –(1*8)]/[{1*(-2)}-{1*(-2)} – (1*3)] = (-6+8-8)/(-2+2-3) = 6/3 = 2.

z = (D_z/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 = 2

z = -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 = (D_x/D) =      = [{(-4)*2} – {3*8) – {(-14)*1}] / [{1*2}-{3*5} – {1*(-4)}] = (-8-24+14)/(2-15+4) = -(18/9) = -2.

y = (D_y/D) =      = [{1*8} + {4*5} –{1*19}] / [{1*2} – {3*5}+ {1*4}] = [8+20-19] / [2-15+4] = -(9/9) = -1.

z = (D_z/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.

1. The first term of the sequence is 5.
2. The common difference of the sequence is 6.

iii. The 15^th 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.

1. 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.

1. Given that the principal or p is 2500.

Time or t = 5yrs.

Rate of interest or r = 5% = 0.05.

1. 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.

1. 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.