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-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.
- 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 = (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 = 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 = (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.
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