PEMBAHASAN SOAL UJIAN KUIS APLIKASI KOMPUTER III MATERI : APLIKASI MATRIKS

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Sudirman Pranoto
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1 PEMBAHASAN SOAL UJIAN KUIS APLIKASI KOMPUTER III MATERI : APLIKASI MATRIKS JURUSAN PENDIDIKAN MATEMATIKA UNIVERSITAS MUHAMMADIYAH PAREPARE Solusi Kuis Aplikom Page

2 Kuis A Soal. Misalkan B Tunjukkan bahwa: (a) A+ (B+C) = (A+ B) +C (b) (AB)C = A(BC) (c) (a + b)c = ac + bc (d) a (B C) = ab bc. Tentukan semua nilai a, b sehingga A dan B keduanya tidak dapat dibalik > A:=matrix(,,[,-,,0,4,5,-,,4]): > B:=matrix(,,[8,-,-5,0,,,4,-7,6]): > C:=matrix(,,[,-,,0,4,5,-,,4]): > a:=4: > b:=7: K5 > evalm(a+(b+c))= evalm((a+b)+c); K5 4 > evalm((a.b).c)= evalm(a.(b.c)); > evalm((a+b).c)= evalm(a.c+b.c); > evalm(a.(b-c))= evalm(a.b-a.c); > A:=matrix(,,[a+b-,0,0,]): 44 K4 K K6 K06 57 K7 K48 9 K K 44 4 K8 K 0 K K 4 K 8 Solusi Kuis Aplikom Page

3 b 0 0 B b 7 > B:=matrix(,,[5,0,0,*a-*b-7]): > det(a); ac bk > det(b); 0 ak5 bk5 > x:=matrix(,,[,,0,-5]); > x:=matrix(,,[,,5,-5]); > x:=matrix(,,[,,0,5]); > a:=(det(x)/det(x)); > b:=(det(x)/det(x)); a := b :=K x := 0 K5 x := 5 K5 x := 0 5. Tentukan semua nilai λ dimana det (A) = 0 λ 5 λ 4 > A:=matrix(,,[lambda-,,-5,lambda+4]); A := lk K5 lc 4 > det(a); l C lk > solve(%);, K 4. Diketahui: > A:=matrix(,,[,0,0,8,,0,-5,,6]); Solusi Kuis Aplikom Page

4 Tentukan invers (a) Matriks Adjoin (b) Matriks elementer Soal 0 0 A := 8 0 K5 6 > det(a); > adj(a); > inva:=/det(a).adj(a); > inverse(a); K K6 > A:=matrix(,6,[,0,0,,0,0,8,,0,0,,0,- 5,,6,0,0,]); 0 0 K4 0 9 K 6 inva := K K6 Solusi Kuis Aplikom Page 4

5 A := K > gaussjord(a); > inverse(a); K4 0 9 K K4 0 9 K 6 Kuis II Soal. Misalkan B 0 Warning, the protected names norm and trace have been redefined and unprotected > A:=matrix(,,[,-,,0,4,5,-,,4]): > B:=matrix(,,[8,-,-5,0,,,4,-7,6]): 4 buktikan bahwa: > C:=matrix(,,[,-,,0,4,5,-,,4]): Solusi Kuis Aplikom Page 5

6 (a) (b) (c) (d) Soal a(bc) = (ab)c= B (ac) A(B C) = AB AC (B + C)A = BA + CA a(bc) = (ab)c > a:=4: > b:=7: > evalm(a.(b.c)); 04 K00 K44 K6 4 5 K6 K04 4 > evalm((a.b).c); 04 K00 K44 K6 4 5 K6 K04 4 > evalm(b.(a.c)); 04 K00 K44 K6 4 5 K6 K04 4 > evalm(a.(b-c)); 0 K5 K7 0 K5 K K > evalm(a.b-a.c); Solusi Kuis Aplikom Page 6

7 > evalm((b+c).a); > evalm(a.(b.c)); > evalm(a.(b.c)); 0 K5 K7 0 K5 K K 4 K8 K4 7 5 K6 K K K K K56 8. Misalkan 9 dan serta A dan B (a) (b) Tunjukkan bahwa p (A) = p (A).p (A) Tunjukkan bahwa p (B) = p (B).p (B) > A:=matrix(,,[,,,]); A := Solusi Kuis Aplikom Page 7

8 > p := x -> x^-9*linearalgebra:- IdentityMatrix(,); p := x/x K 9 (LinearAlgebra:-IdentityMatrix ) (, ) > > p := x -> x+*linearalgebra:-identitymatrix(,); p := x/xc (LinearAlgebra:-IdentityMatrix ) (, ) > p := x -> x-*linearalgebra:-identitymatrix(,); p := x/xk (LinearAlgebra:-IdentityMatrix ) (, ) > evalm(p(a)); > evalm(p(a).p(a)); 4 8 K6 4 8 K6. Tentukan semua nilai λ dimana det (A) = 0 λ λ 0 λ > A:=matrix(,,[lambda-4,0,0,0,lambda,,0,,lambda- ]); lk A := 0 l 0 lk > det(a); Solusi Kuis Aplikom Page 8

9 (lk 4) (l KlK 6) > solve(%); 4,, K 4. Diketahui: Tentukan invers (a) Matriks Adjoin (b) Matriks elementer > A:=matrix(,,[,-,5,0,,-,0,0,]); K 5 A := 0 K 0 0 > A:=matrix(,6,[,-,5,,0,0,0,,-,0,,0,0,0,,0,0,]); K A := 0 K > > gaussjord(a); Solusi Kuis Aplikom Page 9

10 > inverse(a); Kuis III Soal 5. Misalkan B buktikan bahwa: Warning, the protected names norm and trace have been redefined and unprotected > A:=matrix(,,[,-,,0,4,5,-,,4]): > B:=matrix(,,[8,-,-5,0,,,4,-7,6]): > C:=matrix(,,[,-,,0,4,5,-,,4]): > a:=4: Solusi Kuis Aplikom Page 0

11 (a) (A T ) T = A (b) (A+B) T = A T + B T (c) (ac) T = ac T (d) (AB) T = B T A T Soal > b:=7: > evalm(transpose((transpose(a)))); K K 4 > evalm(transpose(a+b)); 0 0 K4 5 K6 K 7 0 > evalm(transpose(a)+transpose(b)); 0 0 K4 5 K6 K 7 0 > evalm(transpose(a.c)); 8 0 K8 K > evalm(transpose(a.b)); Solusi Kuis Aplikom Page

12 8 0 0 K8 K K > evalm(transpose(b).transpose(a)); K8 K K Misalkan A adalah matriks pada setiap bagian, tentukan P(A) (a) P(x) = x (b) P(x) = x x + (c) P(x) = x x + 4 > A:=matrix(,,[,,,]); A := > p := x -> x-*linearalgebra:-identitymatrix(,); p := x/x K (LinearAlgebra:-IdentityMatrix ) (, ) > p := x -> *x^-x+*linearalgebra:- IdentityMatrix(,); p := x/ x K xc (LinearAlgebra:-IdentityMatrix ) (, ) > p := x -> x^-*x+4*linearalgebra:- IdentityMatrix(,); p := x/x K xc 4 (LinearAlgebra:-IdentityMatrix ) (, ) > evalm(p(a)); Solusi Kuis Aplikom Page

13 > evalm(p(a)); > evalm(p(a)); 9 6 K Selesaikan x pada: > A:=matrix(,,[x,-,,-x]); A := x K Kx > B:=matrix(,,[,0,-,,x,-6,,,x-5]); 0 K B := x K6 xk 5 > det(a); xk x C > det(b); x K x Solusi Kuis Aplikom Page

14 > det(a)-det(b); xk x C 8. Diketahui: Tentukan invers (a) Matriks Adjoin (b) Matriks elementer > solve(%); 4 K, 4 4 C 4 > A:=matrix(,,[,0,,0,,,-,0,-4]); 0 A := 0 K 0 K4 > A:=matrix(,6,[,0,,,0,0,0,,,0,,0,0,0,,0,0,]) ; A := > gaussjord(a); K K Solusi Kuis Aplikom Page 4

15 > inverse(a); 0 K 0 K Kuis IV 9. Misalkan Soal 5 B buktikan bahwa: (a) (A - ) - = A (b) (B T ) - = (B - ) T (c) (AB) - = B - A - (d) (ABC) - = C - B - A - 0 Warning, the protected names norm and trace have been redefined and unprotected > A:=matrix(,,[,,5,]); A := 5 > B:=matrix(,,[,-,4,4]); B := K 4 4 > C:=matrix(,,[,0,0,]); C := 0 0 Solusi Kuis Aplikom Page 5

16 > inverse(inverse(a)); > inverse(transpose(b)); > transpose(inverse(b)); > inverse(a.b); 5 > evalm(inverse(b).inverse(a)); K7 0 K9 0 K7 0 K9 0 K 5 0 K Solusi Kuis Aplikom Page 6

17 > inverse(a.b.c); > evalm(inverse(c).inverse(b).inverse(a)); K7 40 K 0 K7 40 K Misalkan A adalah matriks 0 4 Hitunglah A, A - dan A A + I > A:=matrix(,,[,0,4,]); > evalm(a^); > evalm((inverse(a))^); A := Solusi Kuis Aplikom Page 7

18 8 K7 > evalm(a^-*a+linearalgebra:-identitymatrix(,)); Selesaikan x pada: 0 9 > A:=matrix(,,[,x,x^,,,,,-,9]); x x A := K 9 > det(a); K8 xk 4 x > solve(%); K,. Diketahui: Tentukan invers > A:=matrix(,,[,5,5,-,-,0,,4,]); Solusi Kuis Aplikom Page 8

19 (a) (b) Matriks Adjoin Matriks elementer Soal 5 5 A := K K 0 4 > adj(a); K 5 5 K4 K5 K > det(a); K > inv_a:=evalm(/det(a)*adj(a)); K5 K5 inv_a := K 4 5 K K > inverse(a); K5 K5 K 4 5 K K > A:=matrix(,6,[,5,5,,0,0,-,-,0,0,,0,,4,,0,0,]); Solusi Kuis Aplikom Page 9

20 > gaussjord(a); A := K K K5 K5 0 0 K K K Solusi Kuis Aplikom Page 0

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