## University’s Definition of Plagiarism and Collusion

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 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
^{-1}Ax = 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.

- 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.

- 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

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:

- Given the equations are:

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 = 4

x – 2y + 2z = -5

2x – 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 = -4

2x + z = 7

X – 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 = 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.

- 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.

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

iii. The 15^{th} term of the sequence is :

T_{15} = (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.