09 Menyelesaikan Persamaan Matriks
MATRIKS
E. Menyelesaikan Persamaan Matriks
Salah satu diantara penggunaan invers matriks adalah untuk menyelesaikan
persamaan matriks. Ada dua macam rumus dasar menyelesaikan persamaan
matriks, yaitu :
(1) Jika A x B = C maka B = A 1 x C
(2) Jika A x B = C maka A = C x B 1
Bukti :
(1) Jika A x B = C maka
A 1 x A x B = A 1 x C
I x B = A 1 x C
B = A 1 x C
(2) Jika A x B = C maka
A x B x B 1 = C x B 1
A x I = C x B 1
A = C x B 1
Untuk lebih memahami rumus diatas, ikutilah contoh soal berikut ini :
2 3
01. Diketahui matriks A =
dan C =
3 4
jika B x A = C
Jawab
B x A = C
3 1
- 2 - 5 maka tentukanlah matriks B
B = C x A 1
3 1
1 4 3
B =
x
8 - 9 3 2
- 2 - 5
3 1
4 3
B =
x 1
- 2 - 5
3 2
3 1
4 3
B =
x
- 2 - 5
3 2
12 3 9 2
B =
8 15 6 10
9 7
B =
7 4
Matriks
1
2 - 1
02. Diketahui A =
, C =
4 - 3
jika A x C x B = D
Jawab
A x C x B = D
3 2
2 1 dan D =
0 4
- 2 2 maka tentukan matriks B
C x B = A 1 x D
B = C 1 x A 1 x D
B =
0 4
1 1 2
1 3 1
x
x
6 4 4 2
3 4 2 3
2 2
1 2
B = 1
2 3
B =
B =
B =
B =
B =
x
1 3 1
2 4 2
0 4
x
2 2
0 4
3 1
1 1 2
x
x
2 2 3
2 2
4 2
0 4
1 3 8 1 4
x
2 6 12 2 6
2 2
0 4
1 5 3
x
2 6 4
2 2
1 0 6 20 6
2 0 8 24 8
14
1 6
2 8 16
7
3
B =
4 8
- 2 3
03 Diketahui B =
dan C =
0 4
adalah …
Jawab
1 - 4
1
1
2 2 . Jika (A .B) = C maka matriks A
(A 1 .B) 1 = C
B 1 x (A 1 ) 1 = C
B 1 x A = C
B x B 1 x A = B x C
I x A = B x C
A = B x C
- 2 3
1 - 4
A =
x
0 4
2 2
Matriks
2
- 2 6 8 6
A =
0 8 0 8
4 14
A =
8 8
- 2 0
6 4
2 1
1
1
1
04. Jika A =
,C=
dan D =
, serta (A x .C) ( A x B) = D maka
1
4
2
8
5
3
tentukanlah matriks B
Jawab
(A 1x .C) 1 ( A 1 x B) = D
C 1 (A - 1 ) 1 A 1 B = D
C 1 A A 1 B = D
C 1 I B = D
C 1 B = D
B = C x D
2 1
B =
x
5 3
4 1
B =
10 3
- 2 0
1 4
04
0 12
3 4
B =
7 12
Kegunaan lain dari invers matriks adalah untuk menentukan penyelesaian sistim
persamaan linier. Tentu saja teknik penyelesaiannya dengan aturan persamaan
matriks, yaitu :
a1x + b1y = c1
Jika
a2x + b2y = c2
a
maka 1
a 2
b1 x
b 2 y
x
y
c
= 1
c 2
=
1 b2
ad bc a 2
b1 c 1
a 1 c 2
Selain dengan persamaan matriks, teknik menyelesaikan sistem persamaan linier
juga dapat dilakukan dengan determinan matriks. Aturan dengan cara ini adalah :
a b
a b
maka
det(A)
=
= ad – bc. sehingga
Jika matriks A =
c d
c d
Jika
Matriks
a1x + b1y = c1
a2x + b2y = c2
maka D =
a1
a2
b1
= a1b 2 b1a 2
b2
3
Dx =
c1
c2
b1
= c1b 2 b1c 2
b2
Dy =
a1
a2
c1
= a1c 2 c1a 2
c2
Dy
Dx
dan
y=
D
D
Untuk lebih jelanya, ikutolah contoh soal berikut ini:
Maka x =
05. Tentukan himpunan penyelesaian sistem persamaan 2x – 3y = 8 dan x + 2y = –3
dengan metoda:
(a) Invers matriks
(b) Determinan
Jawab
8
2 3 x
2x – 3y = 8
maka
x + 2y = –3
1 2 y
3
(a) Dengan metoda invers matriks diperoleh
8
2 3 x
1 2 y = 3
1
2 3 8
4 (3) 1 2 3
1 2 3 8
=
7 1 2 3
x
y
x
y
x
y
x
y
x
y
=
=
=
1 16 9
7 8 6
1 7
7 14
1
=
2
Jadi x = 1 dan y = –2
(b) Dengan metoda determinan matriks diperoleh
2 3
D =
= (2)(2) – (–3)(1) = 4 + 3 = 7
1
Dx =
Dy =
2
3
8
3
2
1
2
8
3
= (8)(2) – (–3)( –3) = 16 – 9 = 7
= (2)( –3) – (8)(1) = –6 – 8 = –14
Dx
7
=
= 1
D
7
Dy
14
y =
=
=2
D
7
Maka x =
Matriks
4
06. Tentukan himpunan penyelesaian sistem persamaan y =
dengan metoda:
(a) Invers matriks
Jawab
1
x + 5 (2)
2
2
x + 6 = y (3)
3
y=
1
2
x + 5 dan x + 6 = y
2
3
(b) Determinan
2y = x + 10
–x + 2y = 10
3x + 18 = 2y
3x – 2y = –18
–x + 2y = 10
3x – 2y = –18
Maka
D =
1
2
2
3
= (–1)(–2) – (2)(3) = 2 – 6 = –4
10
2
1
10
18 2
Dx =
Dy =
18
3
= (10)(–2) – (2)( –18) = –20 – (–36) = 16
= (–1)(–18) – (10)(3) = 18 – 30 = –12
Dx
16
=
= –4
D
4
Dy
12
y =
=
=3
D
4
(3) Sistem persamaan linier tiga variabel
a1x + b1y + c1z = d1
Jika
a2x + b2y + c2z = d2 diperoleh nilai determinan :
a3x + b3y + c3z = d3
Maka x =
Matriks
a1
b1
c1
D = a2
b2
c 2 = a1.b2.c3 + b1.c2.a3 + c1.a2.b3 – c1.b2.a3 – a1.c2.b3 – b1.a2.c3
a3
b3
c3
d1
b1
c1
Dx = d 2
b2
c 2 = d1.b2.c3 + b1.c2.d3 + c1.d2.b3 – c1.b2.d3 – d1.c2.b3 – b1.d2.c3
d3
b3
c3
a1
d1
c1
Dy = a 2
d2
c 2 = a1.d2.c3 + d1.c2.a3 + c1.a2.d3 – c1.d2.a3 – a1.c2.d3 – d1.a2.c3
a3
d3
c3
a1
b1
d1
Dz = a 2
b2
d 2 = a1.b2.d3 + b1.d2.a3 + d1.a2.b3 – d1.b2.a3 – a1.d2.b3 – b1.a2.d3
a3
b3
d3
5
Dy
Dx
Dz
, y=
dan z =
D
D
D
Sehingga nilai x =
Untuk lebih jelanya, ikutolah contoh soal berikut ini:
07. Tentukanlah himpunan penyelesaian sistem persamaan linier
x + 2y + z = 2
x – y – 2z = –1
dengan menggunakan metoda determinan
x+ y– z = 3
Jawab
1 2
1 1 2
D = 1 1 2 = 1 1 2 1 1
1 1 1 1 1
1 1 1
1
2
1
D = (1)(–1)(–1) + (2)(–2)(1) + (1)(1)(1) – (1)(–1)(1) – (1)(–2)(1) – (2)(1)(–1)
D = 1–4+1+1+2+2
D = 3
2 2
1
2 2
Dx = 1 1 2 = 1 1 2 1 1
3 1 1 3 1
3
1 1
2
2
1
Dx = (2)(–1)(–1) + (2)(–2)(3) + (1)(–1)(1) – (1)( –1)(3) – (2)(–2)(1) – (2)(–1)(–1)
Dx = 2 – 12 – 1 + 3 + 4 – 2
Dx = –6
1 2
1
1 2
1 1 2
Dy = 1 1 2 = 1 1 2 1 1
1 3 1 1 3
1 3 1
Dy = (1)(–1)(–1) + (2)(–2)(1) + (1)(1)(3) – (1)(–1)(1) – (1)(–2)(3) – (2)(1)(–1)
Dy = 1 – 4 + 3 + 1 + 6 + 2
Dy = 9
1
2
2
1
2
2
1
2
Dz = 1 1 1 = 1 1 1 1 1
1 1
3
1 1 3 1 1
Dz = (1)(–1)(3) + (2)(–1)(1) + (2)(1)(1) – (2)(–1)(1) – (1)(–1)(1) – (2)(1)(3)
Dz = –3 – 2 + 2 + 2 + 1 – 6
Dz = –6
D
6
= –2
Jadi x = x =
3
D
Dy
9
y=
=
= 3
3
D
6
D
z= z =
= –2
3
D
Matriks
6
08. Tentukanlah himpunan penyelesaian sistem persamaan linier
x – 2y = –3
y+ z = 1
dengan menggunakan metoda determinan
2x + z = 1
Jawab
1 2 0
D = 0
1
2
0
D
D
D
D
=
=
=
=
(1)(1)(1) + (–2)(1)(2) + (0)(0)(0) – (0)(1)(2) – (1)(1)(0) – (–2)(0)(1)
(1) + (–4) + (0) – (0) – (0) – (0)
1–4+0+0+0+0
–3
3 2 0
3 2 0 3 2
Dx = 1
1
Dx =
Dx =
Dx =
Dx =
1
0
1 =
1
1
1
1
1
1
1
0
1
1
0
(–3)(1)(1) + (–2)(1)(1) + (0)(1)(0) – (0)(1)(1) – (–3)(1)(0) – (–2)(1)(1)
(–3) + (–2) + (0) – (0) – (0) – (–2)
–3 – 2 + 0 – 0 – 0 + 2
–3
1 3 0
1 3 0 1 3
Dy = 0
2
Dy =
Dy =
Dy =
Dy =
1 2 0 1 2
1 = 0 1 1 0 1
2 0 1 2 0
1
1
1
1 = 0
1
2
1
1 0
1
1
1 2
1
(1)(1)(1) + (–3)(1)(2) + (0)(0)(1) – (0)(1)(2) – (1)(1)(1) – (–3)(0)(1)
( 1) + (–6) + (0) – (0) – (1) – (0)
1–6+0+0–1–0
–6
1 2 3
1 2 3 1 2
Dz = 0
2
1
0
1 = 0
1
2
1
1
0
1
0
1
2
0
Dz = (1)(1)(1) + (–2)(1)(2) + (–3)(0)(0) – (–3)(1)(2) – (1)(1)(0) – (–2)(0)(1)
Dz = (1) + (–4) + (0) – (–6) – (0) – (0)
Dz = 1 – 4 + 0 + 6 – 0 – 0
Dz = 3
D
3
= 1
Jadi x = x =
3
D
Dy
6
y=
=
= 2
3
D
3
D
= –1
z= z =
3
D
Matriks
7
E. Menyelesaikan Persamaan Matriks
Salah satu diantara penggunaan invers matriks adalah untuk menyelesaikan
persamaan matriks. Ada dua macam rumus dasar menyelesaikan persamaan
matriks, yaitu :
(1) Jika A x B = C maka B = A 1 x C
(2) Jika A x B = C maka A = C x B 1
Bukti :
(1) Jika A x B = C maka
A 1 x A x B = A 1 x C
I x B = A 1 x C
B = A 1 x C
(2) Jika A x B = C maka
A x B x B 1 = C x B 1
A x I = C x B 1
A = C x B 1
Untuk lebih memahami rumus diatas, ikutilah contoh soal berikut ini :
2 3
01. Diketahui matriks A =
dan C =
3 4
jika B x A = C
Jawab
B x A = C
3 1
- 2 - 5 maka tentukanlah matriks B
B = C x A 1
3 1
1 4 3
B =
x
8 - 9 3 2
- 2 - 5
3 1
4 3
B =
x 1
- 2 - 5
3 2
3 1
4 3
B =
x
- 2 - 5
3 2
12 3 9 2
B =
8 15 6 10
9 7
B =
7 4
Matriks
1
2 - 1
02. Diketahui A =
, C =
4 - 3
jika A x C x B = D
Jawab
A x C x B = D
3 2
2 1 dan D =
0 4
- 2 2 maka tentukan matriks B
C x B = A 1 x D
B = C 1 x A 1 x D
B =
0 4
1 1 2
1 3 1
x
x
6 4 4 2
3 4 2 3
2 2
1 2
B = 1
2 3
B =
B =
B =
B =
B =
x
1 3 1
2 4 2
0 4
x
2 2
0 4
3 1
1 1 2
x
x
2 2 3
2 2
4 2
0 4
1 3 8 1 4
x
2 6 12 2 6
2 2
0 4
1 5 3
x
2 6 4
2 2
1 0 6 20 6
2 0 8 24 8
14
1 6
2 8 16
7
3
B =
4 8
- 2 3
03 Diketahui B =
dan C =
0 4
adalah …
Jawab
1 - 4
1
1
2 2 . Jika (A .B) = C maka matriks A
(A 1 .B) 1 = C
B 1 x (A 1 ) 1 = C
B 1 x A = C
B x B 1 x A = B x C
I x A = B x C
A = B x C
- 2 3
1 - 4
A =
x
0 4
2 2
Matriks
2
- 2 6 8 6
A =
0 8 0 8
4 14
A =
8 8
- 2 0
6 4
2 1
1
1
1
04. Jika A =
,C=
dan D =
, serta (A x .C) ( A x B) = D maka
1
4
2
8
5
3
tentukanlah matriks B
Jawab
(A 1x .C) 1 ( A 1 x B) = D
C 1 (A - 1 ) 1 A 1 B = D
C 1 A A 1 B = D
C 1 I B = D
C 1 B = D
B = C x D
2 1
B =
x
5 3
4 1
B =
10 3
- 2 0
1 4
04
0 12
3 4
B =
7 12
Kegunaan lain dari invers matriks adalah untuk menentukan penyelesaian sistim
persamaan linier. Tentu saja teknik penyelesaiannya dengan aturan persamaan
matriks, yaitu :
a1x + b1y = c1
Jika
a2x + b2y = c2
a
maka 1
a 2
b1 x
b 2 y
x
y
c
= 1
c 2
=
1 b2
ad bc a 2
b1 c 1
a 1 c 2
Selain dengan persamaan matriks, teknik menyelesaikan sistem persamaan linier
juga dapat dilakukan dengan determinan matriks. Aturan dengan cara ini adalah :
a b
a b
maka
det(A)
=
= ad – bc. sehingga
Jika matriks A =
c d
c d
Jika
Matriks
a1x + b1y = c1
a2x + b2y = c2
maka D =
a1
a2
b1
= a1b 2 b1a 2
b2
3
Dx =
c1
c2
b1
= c1b 2 b1c 2
b2
Dy =
a1
a2
c1
= a1c 2 c1a 2
c2
Dy
Dx
dan
y=
D
D
Untuk lebih jelanya, ikutolah contoh soal berikut ini:
Maka x =
05. Tentukan himpunan penyelesaian sistem persamaan 2x – 3y = 8 dan x + 2y = –3
dengan metoda:
(a) Invers matriks
(b) Determinan
Jawab
8
2 3 x
2x – 3y = 8
maka
x + 2y = –3
1 2 y
3
(a) Dengan metoda invers matriks diperoleh
8
2 3 x
1 2 y = 3
1
2 3 8
4 (3) 1 2 3
1 2 3 8
=
7 1 2 3
x
y
x
y
x
y
x
y
x
y
=
=
=
1 16 9
7 8 6
1 7
7 14
1
=
2
Jadi x = 1 dan y = –2
(b) Dengan metoda determinan matriks diperoleh
2 3
D =
= (2)(2) – (–3)(1) = 4 + 3 = 7
1
Dx =
Dy =
2
3
8
3
2
1
2
8
3
= (8)(2) – (–3)( –3) = 16 – 9 = 7
= (2)( –3) – (8)(1) = –6 – 8 = –14
Dx
7
=
= 1
D
7
Dy
14
y =
=
=2
D
7
Maka x =
Matriks
4
06. Tentukan himpunan penyelesaian sistem persamaan y =
dengan metoda:
(a) Invers matriks
Jawab
1
x + 5 (2)
2
2
x + 6 = y (3)
3
y=
1
2
x + 5 dan x + 6 = y
2
3
(b) Determinan
2y = x + 10
–x + 2y = 10
3x + 18 = 2y
3x – 2y = –18
–x + 2y = 10
3x – 2y = –18
Maka
D =
1
2
2
3
= (–1)(–2) – (2)(3) = 2 – 6 = –4
10
2
1
10
18 2
Dx =
Dy =
18
3
= (10)(–2) – (2)( –18) = –20 – (–36) = 16
= (–1)(–18) – (10)(3) = 18 – 30 = –12
Dx
16
=
= –4
D
4
Dy
12
y =
=
=3
D
4
(3) Sistem persamaan linier tiga variabel
a1x + b1y + c1z = d1
Jika
a2x + b2y + c2z = d2 diperoleh nilai determinan :
a3x + b3y + c3z = d3
Maka x =
Matriks
a1
b1
c1
D = a2
b2
c 2 = a1.b2.c3 + b1.c2.a3 + c1.a2.b3 – c1.b2.a3 – a1.c2.b3 – b1.a2.c3
a3
b3
c3
d1
b1
c1
Dx = d 2
b2
c 2 = d1.b2.c3 + b1.c2.d3 + c1.d2.b3 – c1.b2.d3 – d1.c2.b3 – b1.d2.c3
d3
b3
c3
a1
d1
c1
Dy = a 2
d2
c 2 = a1.d2.c3 + d1.c2.a3 + c1.a2.d3 – c1.d2.a3 – a1.c2.d3 – d1.a2.c3
a3
d3
c3
a1
b1
d1
Dz = a 2
b2
d 2 = a1.b2.d3 + b1.d2.a3 + d1.a2.b3 – d1.b2.a3 – a1.d2.b3 – b1.a2.d3
a3
b3
d3
5
Dy
Dx
Dz
, y=
dan z =
D
D
D
Sehingga nilai x =
Untuk lebih jelanya, ikutolah contoh soal berikut ini:
07. Tentukanlah himpunan penyelesaian sistem persamaan linier
x + 2y + z = 2
x – y – 2z = –1
dengan menggunakan metoda determinan
x+ y– z = 3
Jawab
1 2
1 1 2
D = 1 1 2 = 1 1 2 1 1
1 1 1 1 1
1 1 1
1
2
1
D = (1)(–1)(–1) + (2)(–2)(1) + (1)(1)(1) – (1)(–1)(1) – (1)(–2)(1) – (2)(1)(–1)
D = 1–4+1+1+2+2
D = 3
2 2
1
2 2
Dx = 1 1 2 = 1 1 2 1 1
3 1 1 3 1
3
1 1
2
2
1
Dx = (2)(–1)(–1) + (2)(–2)(3) + (1)(–1)(1) – (1)( –1)(3) – (2)(–2)(1) – (2)(–1)(–1)
Dx = 2 – 12 – 1 + 3 + 4 – 2
Dx = –6
1 2
1
1 2
1 1 2
Dy = 1 1 2 = 1 1 2 1 1
1 3 1 1 3
1 3 1
Dy = (1)(–1)(–1) + (2)(–2)(1) + (1)(1)(3) – (1)(–1)(1) – (1)(–2)(3) – (2)(1)(–1)
Dy = 1 – 4 + 3 + 1 + 6 + 2
Dy = 9
1
2
2
1
2
2
1
2
Dz = 1 1 1 = 1 1 1 1 1
1 1
3
1 1 3 1 1
Dz = (1)(–1)(3) + (2)(–1)(1) + (2)(1)(1) – (2)(–1)(1) – (1)(–1)(1) – (2)(1)(3)
Dz = –3 – 2 + 2 + 2 + 1 – 6
Dz = –6
D
6
= –2
Jadi x = x =
3
D
Dy
9
y=
=
= 3
3
D
6
D
z= z =
= –2
3
D
Matriks
6
08. Tentukanlah himpunan penyelesaian sistem persamaan linier
x – 2y = –3
y+ z = 1
dengan menggunakan metoda determinan
2x + z = 1
Jawab
1 2 0
D = 0
1
2
0
D
D
D
D
=
=
=
=
(1)(1)(1) + (–2)(1)(2) + (0)(0)(0) – (0)(1)(2) – (1)(1)(0) – (–2)(0)(1)
(1) + (–4) + (0) – (0) – (0) – (0)
1–4+0+0+0+0
–3
3 2 0
3 2 0 3 2
Dx = 1
1
Dx =
Dx =
Dx =
Dx =
1
0
1 =
1
1
1
1
1
1
1
0
1
1
0
(–3)(1)(1) + (–2)(1)(1) + (0)(1)(0) – (0)(1)(1) – (–3)(1)(0) – (–2)(1)(1)
(–3) + (–2) + (0) – (0) – (0) – (–2)
–3 – 2 + 0 – 0 – 0 + 2
–3
1 3 0
1 3 0 1 3
Dy = 0
2
Dy =
Dy =
Dy =
Dy =
1 2 0 1 2
1 = 0 1 1 0 1
2 0 1 2 0
1
1
1
1 = 0
1
2
1
1 0
1
1
1 2
1
(1)(1)(1) + (–3)(1)(2) + (0)(0)(1) – (0)(1)(2) – (1)(1)(1) – (–3)(0)(1)
( 1) + (–6) + (0) – (0) – (1) – (0)
1–6+0+0–1–0
–6
1 2 3
1 2 3 1 2
Dz = 0
2
1
0
1 = 0
1
2
1
1
0
1
0
1
2
0
Dz = (1)(1)(1) + (–2)(1)(2) + (–3)(0)(0) – (–3)(1)(2) – (1)(1)(0) – (–2)(0)(1)
Dz = (1) + (–4) + (0) – (–6) – (0) – (0)
Dz = 1 – 4 + 0 + 6 – 0 – 0
Dz = 3
D
3
= 1
Jadi x = x =
3
D
Dy
6
y=
=
= 2
3
D
3
D
= –1
z= z =
3
D
Matriks
7