University Plagiarism Declaration And Math Questions
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 😡 = (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 = (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+61)/(2+23) =(3/3)= 1.y = (D_{y}/D) = = [{1*(6)} –{4*(2)} –(1*8)]/[{1*(2)}{1*(2)} – (1*3)] = (6+88)/(2+23) = 6/3 = 2.z = (D_{z}/D) = = [{1*(1)}{1*8} + {4*3}]/[{1*(2)} – {1*(2)} – (1*3)] = [18+12]/[2+23] = (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 = (D_{x}/D) = = [{(4)*2} – {3*8) – {(14)*1}] / [{1*2}{3*5} – {1*(4)}] = (824+14)/(215+4) = (18/9) = 2.y = (D_{y}/D) = = [{1*8} + {4*5} –{1*19}] / [{1*2} – {3*5}+ {1*4}] = [8+2019] / [215+4] = (9/9) = 1.z = (D_{z}/D) = = [14(3*19)+(4*4)] / [(1*2) – (3*5) + (1*4)] = [1457+16] / [215+4] = 3.Therefore, the system of equation solves into 😡 = 1;,y= 1,z = 3 Question 4:
 Given, principal or p = 6000.
Rate of interest or r = 4.2% = 0.042.Interest is compounded semiannually.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) + (ad), where, d = common difference. a = first term n = required number of terms = 6*15 + (56) = 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) 