Question1:
= = (Ans)
2A + x = B
=
= (Ans)
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 =
iv yA =
Question 2.
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.
Substituting the value of x in equation ii
3x + 2y = 30.
Putting y in equation i,
x + 3y = -11.
So, the solutions are x=16 and y = -9.
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
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:
2x + y = 5.x – y = 1.
x + 3y = 2 2x – y = 11.The equations are to be solved through inverse matrix method. x +3y = 2. 2x – y = 1.
=
Therefore, x = 14 and y = 7.
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:
x + 3y – z = -42x + z = 7X – 2y + 3z = 13
The matrix equation form is:- = ^(-1) * And the augmented matrix form is :
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.
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:
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.
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%.
5.11.17…..,599.
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 = n*(a1 +an)/2 = 100(5+599/2)100*(604/2) = 100*302 = 30200.Therefore, the sum is 30200.
Time or t = 5yrs.Rate of interest or r = 5% = 0.05.
A = P*[1 + (r/n)]nt = 2500 * [1+(0.05/1)]2 =2500*0.1025 = 2756.25.Therefore, the amount is RM 2756.25.
A = [p*{1+(r/n)}nt]= 2500[1+0.05]3 = 2894.063.Therefore, the amount is RM 2894.063
=500[{1+(5/100)5}/(5/100)] = 2762.815625. Therefore, the amount is RM 2762.815625.
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System of Equations |
Row operations |
Augmented matrix |
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x + y – z = 4 x – 2y – 2z = -5 2x – y + 2z = -2 |
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x + 3y – z = 4 -3y – z = -9 -3y + 4z = -10 |
L2 = L2 – (L1). L3 = L3 – 2*L1 |
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4x + y = 6 -15y = -46 -3y + 4z = -10 |
L1 = 4*L1 + L3 L2 = 4*(L2) + L3. |
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X = (11/15) y=(46/15) z = -(1/5) |
L1 = 15*L1 + L2 L3 = L3 – (L2/5) |
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System of equations |
Row operations |
Augmented matrix |
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x + 3y – z = -4 2x + z = 7 x – 2y + 3z = 13 |
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x + 3y – z = -4 6y – z = 15 y + 4z = 17 |
L2 = L2 – 2*(L1) L3 = L3 – L1 |
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x – 3y = -19 25y = 77 y + 4z = 17 |
L1 = L1 – L2 L2 = 4*L2 + L3 |
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x = -(244/25) y = (77/25) z = -(1700/25) |
L1 = L1 + (3/25)L2 L3 = L3 – (L2/25) |
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System of equations |
Row operations |
Augmented matrix |
|
x + y – z = 4 x – 2y + 2z = -5 2x – y + 2z = -2 |
||
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x + y – z = 4 -y + z = -3 -3y + 4z = -10 |
L2 = L2 – L1 L3 = L3 – 2*(L1) |
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x = 1 y = 2 z = -1 |
LI = LI + L2 L2 = 4*(L2) – L3 L3 = L3 – 3*(L2) |
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