MATRICES AND INTEREST RATE.

Running Head: MATRICES AND INTEREST RATES
MATRICES AND INTEREST RATES
Name
Institute of Affiliation
Date

MATRICES AND INTEREST RATE 2
Question 1
i. ,
A=
[1 0
3 −5 ] , B= [−2 1
4 −3 ]
A +B = [ −2 1
4 −3 ] + [ 1 0
3 −5 ]
= [−1 1
7 −8 ]
ii. ,
2A - 3B =2 [ −2 1
4 −3 ] -3 [ 1 0
3 −5 ]
= [−4 2
8 −6 ] - [−3 0
−9 −15 ]
= [ −1 2
17 9 ]
iii. ,
AB= [−2 1
4 −3 ] . [1 0
3 −5 ]
= [ −2 1
−26 18 ]
iv. ,
AB+BA = [−2 1
4 −3 ] . [1 0
3 −5 ] +. [1 0
3 −5 ] .[−2 1
4 −3 ]
= [ −1 −4
−31 33 ]
b. ,
A= [ 3 2
−1 1 ]

MATRICES AND INTEREST RATE 3
B= [ 0 4
−2 8 ]
X=2A -B
= 2 [ 3 2
−1 1 ] - [ 0 4
−2 8 ]
= [ −2 1
−56 39 ]
c. ,
i.
A= [ 3 2
1 6 ]
Det A= (3 x 6) – (1 x 2)
=16
ii.
A−1 = 1
16 [ 6 −2
−1 3 ]
= [ 0.3750 −0.1250
−0.0625 0.1875 ]
iii. ,
AX=[ 5 6
7 2 ]
X= A−1
[ 5 6
7 2 ]
= [ 0.3750 −0.1250
−0.0625 0.1875 ]. [5 6
7 2 ]
= 1.5000 0.5000
2.5000 0.5000

MATRICES AND INTEREST RATE 4
v. ,
YA=[ 5 6
7 2 ]
Y= A−1
[ 5 6
7 2 ]
= 0 1
0 .5000 −0.5000
Question 2
a. ,
X+3y = -10
3x+2y = 5
.x = -3y-10
3(-3y-10) + 2y = 5
-9y -30 + 2y = 5
-7y = 35
Y=-5
X= -3(-5) -10
= 5
b. ,
3x +2y = 9
4x -3y=12
12x + 8y = 36
12x -9y = 36
17y =0
Y=0

MATRICES AND INTEREST RATE 5
3x +0 = 9
X=3
c. ,
x1−4 x2 + x3=0
2 x1 −3 x2 +7 x3=0
x1−2 x2 =0
[1 −4 1
2 −3 7
1 −2 0 ] { x1
x2
x3
} = {
0
0
0 }
Gaussian elimination
[1 −4 1 ¿ 0
2 −3 7 ¿ 0
1 −2 0 ¿ 0 ]
d. 2x+y = 5
x-y= 1
[2 1
1 −1 ] { x
y } = [5
1 ]
Det a = (-1*2)-(1*1)
= -3
Inv a = 1
−3 [ −1 −1
−1 2 ]
= [ 0.3750 −0.1250
−0.0625 0.1875 ]
{ x
y }=¿ [ 0.3750 −0.1250
−0.0625 0.1875 ][ 5
1 ]
= [ 2
1 ]

MATRICES AND INTEREST RATE 6
e. ,
f. 2x+y = 5
x-y= 1
[ 2 1
1 −1 ] { x
y } = [ 5
1 ]
Augmented matrix is as follows
[ 2 1 ¿5
1 −1 ¿ 1 ]
g. , crammers rule
For y y= [ 2 5
1 1 ]
Det y = ( 2∗1 )−(5∗1)
= -3
For x [5 1
1 −1 ]
Det x = 5*-1 – 1*1
= -6
X= detx/det
= -6/-2
= 2
Y= dety/det
= -3/-3
= 1
Question 3

MATRICES AND INTEREST RATE 7
a. ,
A=P(1+ 2.1
100)6
= 6000(1+ 2.1
100)6
= 6796.82
b. ,
5978.1 = 5000(1+ r
400 ) 12
1.1956 = (1+ r
400 ) 12
Log 1.1956 = 12 log (1+0.0025 r)
0.0064654 = log 1 + log 0.0025 r
1.015 = r
Rate = 1.5 %
c. ,
F=A
( (1+ r
100 )n
−1
r
100 )
i. F = 2500
( (1+ 5
100 )2
−1
5
100 )
= 5125

MATRICES AND INTEREST RATE 8
ii. F =(5125+ 2500)
( (1+ 5
100 )1
−1
5
100 )
= 7625
iii. F =(7625+ 2500)
( (1+ 5
100 )1
−1
5
100 )
= 10125
F =(10125+ 2500)
( (1+ 5
100 )1
−1
5
100 )
= 12626
Q4
a. ,
A =P(1+ R
100)N
= 500(1+
9
12
100
)18
= 571.98
b. ,
Compound interest
10000= 5000(1.06)N
Log 2 = Nlog 1.06

MATRICES AND INTEREST RATE 9
N= log 2
log 1.06
= 11.89 years
c. ,
i. ,
a= 5
ii. ,
d= 11-5 = 17-11
= 6
iii. ,
The 15th term = a + d(15-1)
= 5 + ( 6 *14)
= 89