# University Plagiarism Declaration And Math Questions

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 😡 = (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 😡 = 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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