a.
Pengulangan Bintang
|
b.
Nilai Akhir Ujian
|
1.Matlab
a. Soal 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
>> B=5*A
B =
10 35 15 25
-10 10 5 20
0 5 45 10
15 20 10 5
>> invers=inv(A)
invers =
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
>> Determinan=det(A)
Determinan =
-7
>> Transpose=A'
Transpose =
2 -2 0 3
7 2 1 4
3 1 9 2
5 4 2 1
>> C=A^3
C =
55 383 614 378
6 71 144 78
42 248 853 292
40 254 407 247
>> D=A+B
D =
12 42 18 30
-12 12 6 24
0 6 54 12
18 24 12 6
b. Soal 2
i. Sistem Persamaan Linier 1
>> A=[2,-5,2,7;1,2,-4,3;3,-4,-6,5]
A =
2 -5 2 7
1 2 -4 3
3 -4 -6 5
>> k1=A(:,1)
k1 =
2
1
3
>> k2=A(:,2)
k2 =
-5
2
-4
>> k3=A(:,3)
k3 =
2
-4
-6
>> k4=A(:,4)
k4 =
7
3
5
>> A=[k1 k2 k3]
A =
2 -5 2
1 2 -4
3 -4 -6
>> det(A)
ans =
-46
>> A1=[k4 k2 k3]
A1 =
7 -5 2
3 2 -4
5 -4 -6
>> det(A1)
ans =
-230
>> A2=[k1 k4 k3]
A2 =
2 7 2
1 3 -4
3 5 -6
>> det(A2)
ans =
-46
>> A3=[k1 k2 k4]
A3 =
2 -5 7
1 2 3
3 -4 5
>> det(A3)
ans =
-46
>> x=det(A1)/det(A)
x =
5
>> y=det(A2)/det(A)
y =
1
>> z=det(A3)/det(A)
z =
1
Jadi solusi dari sistem persamaan linier tersebut adalah
X= 5
Y=1
Z=1
ii. Sistem Persamaan Linier 2
>> B=[2,3,1,6;1,1,2,4;3,4,3,9]
B =
2 3 1 6
1 1 2 4
3 4 3 9
>> k1=B(:,1)
k1 =
2
1
3
>> k2=B(:,2)
k2 =
3
1
4
>> k3=B(:,3)
k3 =
1
2
3
>> k4=B(:,4)
k4 =
6
4
9
>> B=[k1 k2 k3]
B =
2 3 1
1 1 2
3 4 3
>> det(B)
ans =
0
>> B1=[k4 k2 k3]
B1 =
6 3 1
4 1 2
9 4 3
>> det(B1)
ans =
-5
>> B2=[k1 k4 k3]
B2 =
2 6 1
1 4 2
3 9 3
>> det(B2)
ans =
3
>> B3=[k1 k2 k4]
B3 =
2 3 6
1 1 4
3 4 9
>> det(B3)
ans =
1
>> x=det(B1)/det(B)
x =
-Inf
>> y=det(B2)/det (B)
y =
Inf
>> z=det(B3)/det(B)
z =
Inf
Jadi sistem persamaan linier tersebut adalah tak hingga.
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