1.
>> A=[2,7,3,5;-2,2,1,4;0,1,9,2;3,4,2,1]
A =
2 7 3 5
-2 2 1 4
0 1 9 2
3 4 2 1
a)
>> B=5*A
B =
10 35 15 25
-10 10 5 20
0 5 45 10
15 20 10 5
b)
>> inv(A)
ans =
17.0000 -16.2857 1.0000 -21.8571
-16.0000 15.2857 -1.0000 20.8571
-2.0000 1.8571 -0.0000 2.5714
17.0000 -16.0000 1.0000 -22.0000
c)
>> det(A)
ans =
-7
d)
>> A'
ans =
2 -2 0 3
7 2 1 4
3 1 9 2
5 4 2 1
e)
>> C=A^3
C =
55 383 614 378
6 71 144 78
42 248 853 292
40 254 407 247
f)
>> D=A+B
D =
12 42 18 30
-12 12 6 24
0 6 54 12
18 24 12 6
2.a)
>> a=[2;1;3]
a =
2
1
3
>> a1=[-5;2;-4]
a1 =
-5
2
-4
>> a2=[2;-4;-6]
a2 =
2
-4
-6
>> a3=[7;3;5]
a3 =
7
3
5
>> A=[a,a1,a2]
A =
2 -5 2
1 2 -4
3 -4 -6
>> det(A)
ans =
-46
>> A1=[a3,a1,a2]
A1 =
7 -5 2
3 2 -4
5 -4 -6
>> det(A1)
ans =
-230
>> nilai_x=(det(A1))/(det(A))
nilai_x =
5
>> B1=[a,a3,a2]
B1 =
2 7 2
1 3 -4
3 5 -6
>> det(B1)
ans =
-46
>> nilai_y=(det(B1))/(det(A))
nilai_y =
1
>> C1=[a,a1,a3]
C1 =
2 -5 7
1 2 3
3 -4 5
>> det(C1)
ans =
-46
>> nilai_z=(det(C1))/(det(A))
nilai_z =
1
-->> Jadi solusi dari sistem persamaan linier tersebut adalah
x= 5
y=1
z=1
-->> Jadi solusi dari sistem persamaan linier tersebut adalah
x= 5
y=1
z=1
b)
>> b=[2;1;3]
b =
2
1
3
>> b1=[3;1;4]
b1 =
3
1
4
>> b2=[1;2;3]
b2 =
1
2
3
>> b3=[6;4;9]
b3 =
6
4
9
>> B=[b,b1,b2]
B =
2 3 1
1 1 2
3 4 3
>> det(B)
ans =
0
>> B1=[b3,b1,b2]
B1 =
6 3 1
4 1 2
9 4 3
>> det(B1)
ans =
-5
>> nilai_x=(det(B1))/(det(B))
Warning: Divide by zero.
nilai_x =
-Inf
>> B2=[b,b3,b2]
B2 =
2 6 1
1 4 2
3 9 3
>> det(B2)
ans =
3
>> nilai_y=(det(B2))/(det(B))
Warning: Divide by zero.
nilai_y =
Inf
>> B3=[b,b1,b3]
B3 =
2 3 6
1 1 4
3 4 9
>> det(B3)
ans =
1
>> nilai_z=(det(B3))/(det(B))
Warning: Divide by zero.
nilai_z =
Inf
-->> Jadi sistem persamaan linier tersebut adalah tak hingga.
-->> Jadi sistem persamaan linier tersebut adalah tak hingga.
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