ATURAN CRAMER

ATURAN CRAMER
MATA KULIAH ALJABAR
LINEAR
ANNISA
SALSABILA
19.1600.013
ATURAN CRAMER
◦ Misalkan kita mempunyai persamaan matriks seperti :
𝑎11 𝑎12 𝑎13 ⋯ 𝑎1 𝑛
𝑎21 𝑎22 𝑎23 ⋯ 𝑎2𝑛
𝑥1
𝑏1
⋮
⋮
⋮
⋯
⋮
𝑥2
𝑏2
◦ 𝑎
=
𝑎𝐼2 𝑎𝐼3 ⋯ 𝑎3𝑛
⋮
𝐼1
⋮
⋮
⋮
⋮
⋱
⋮
𝑥𝑛
𝑏𝑛
𝑎𝑛1 𝑎𝑛2 𝑎𝑛3 ⋯ 𝑎𝑛𝑛
𝑥1
𝑥2
◦ Ditanyakan : 𝑥3 =?
⋮
𝑥𝑛
𝑥1
𝑥2
◦ 𝑥3 = 𝐴−1
⋮
𝑥𝑛
𝑥1
𝑥2
1
◦ 𝑥3 = det 𝐴
⋮
𝑥𝑛
1
𝑏1
𝑏2
𝑏3
⋮
𝑏𝑛
𝐶11
𝐶12
𝐶13
⋮
𝐶1 𝑛
𝐶21
𝐶22
𝐶23
⋮
𝐶2 𝑛
⋯ 𝐶𝑛1
⋯ 𝐶𝑛2
⋯ 𝑐𝑛3
⋯
⋮
⋯ 𝐶𝑛𝑛
𝑏1
𝑏2
𝑏3
⋮
𝑏𝑛
◦ 𝑥𝐽 = det(𝐴) ∙ b1C1j+b2C2j+. . .+bnCnj
◦ 𝑥𝐽 =
det 𝐴𝐽
det 𝐴
◦ Ekspansi kofaktor sepanjang kolom ke-j
det(A) = a1jC1j+a2jC2j+. . .+anjCnj
𝑏1
𝑏
Aj ganti kolom ke-j dari A dengan 2
⋮
𝑏𝑛
◦ Contoh
2 0 3
◦ 0 3 2
−2 0 −4
𝑥1
◦ Dit . 𝑥2 ?
𝑥3
𝑥1
1
𝑥2 = 2
𝑥3
3
◦ Penyelesaian :
2 0 3
◦ 𝐴 = 0 3 2 = +3 −8 + 6 = −6
−2 0 −4
1 0 3
◦ 𝐴1 = 2 3 2 = +3 −4 − 9 = −39
3 0 −4
◦ 𝐴2
◦ 𝐴3
2
= 0
−2
2
= 0
−2
1
2
3
0
3
0
◦ 𝑥1 =
−39
−6
=
39
6
◦ 𝑥2 =
−20
−6
=
20
6
24
◦ 𝑥3 = −6 =
3
2 = +2 −8 + 6 − 2 6 + 2 = −20
−4
1
2 = +3 6 + 2 = 24
3
−24
6
◦ Kemudian kita cek :
◦
2 0
0 3
−2 0
3
2
−4
39/6
1
20/6
= 2
(−24/6
3
GUNAKAN ATURAN CRAMER UNTUK
MENCARI SOLUSI DARI SPL
◦ −𝑥 − 4𝑥 2 + 2𝑥 3 + 𝑥 4 = −32
◦ 2𝑥 1 − 𝑥 2 + 7𝑥 3 + 9𝑥 4 = 14
◦ −𝑥 1 + 𝑥 2 + 3𝑥 3 + 𝑥 4 = 11
◦ 𝑥 1 − 2𝑥 2 + 𝑥 3 − 4𝑥 4 = −32
◦ Ditanyakan
𝑥1
𝑥2
𝑥3 ?
𝑥4
−1 −4 2
2 −1 7
−1 1 3
1 −2 1
1
9
1
−4
−1 −4 2
2 −1 7
𝐴 =
−1 1 3
1 −2 1
= -1 0 −22
0 −21
1 −7
Ekspansi b3
𝑥1
𝑥2
𝑥3
𝑥4
=
−32
14
11
−4
1 −𝑏3 + 𝑏1 0 −5 −1 0 −5𝐶 + 𝐶 0
0
3
2
2 −1 7
9
9
2 −36
−1 1
3
1
1
−1 −14
1 −2 1 −4
−4
1
−7
17
−22
−3 = −1 + 1
−21
−4
−1 0 −2𝑏4 + 𝑏2
7
9
3
1
1 −4 𝑏4 + 𝑏3
0
0
0 −22
0 −21
1 −7
−1 0
5 17
4 −3
1 −4
17
−3
= −1 1 ∙ 66 + 357
= -1 (423)
= -234
Ekspansi sepanjang
𝐴1
−32
14
=
11
−4
0
−287
−118
−47
−1 -519
−4
−1
1
−2
2 1
7 9
3 1
−𝑏3 + 𝑏1
1 −4
0
−1 0
−36 7
9
−14 3
1
−7
1 −4
−36 9
−287
+ 63
−14 1
−118
−43 −5 −1 0 −43𝑏3 + 𝑏1 0
14 −1 7
9
−287
11
1
3
1
−118
−5𝑏
+
𝑏
3
2 −47
−4 −2 1 −4
−287
−1
−118
−47
−36
−14
−7
9 4𝑏2 + 𝑏3 −287
−118
1
−4
−1 −519
9
1
-1
−519 −36 + 126 + 63(−287 + 1062)
−1
−519 90 + 63(775) = −1 −46 ∙ 710 + 48 ∙ 825 = −2115
0
−1 0
−36 7
9
−14 3
1
−7
1 −4
−36
−14
−63
9
1
0
−1
2
𝐴2 =
−1
1
2
−1 −1
0
−32 2
19 7
11 3
−4 1
−287
−118
−165
1 −𝑏 + 𝑏 0 −43 −1 0 −43𝑏 + 𝑏 0
0
3
2
3
1
9
2 −287
2
14
7
9
−1 −188
1
−1 11
3
1
1
−47
−4
1
−4
1 −4
9
2 9
2 −287
−3
−1 165
1
−1 1
−1 −118
−3
−1 165 2 + 9 − 3(−236 − 287)
−1 165 11 − 3(−523)
−1 1815 + 1569
−1 3384 = −3384
−1
7
3
1
0 𝑏 +𝑏
3
4
9
1
−4
0
0
2 −287
−1 −118
0 −165
−1
7
3
4
0
9
1
−3
𝐴3
−1 −4
2 −1
=
−1 1
1 −2
2
5 −1
0
5 −7
14
11
7
2
−1
−32
14
11
−4
1
0 −1
−𝑏3 + 𝑏1
9
2 −1
1
−1 1
−4
1 −2
9
2
1 −43 −1
−3
0
9
2 19
−3
1
−1 11
−1
1
−1
−43
14
11
−4
0
𝑏 + 𝑏4
9 3
1
−4
9
1
−3
−43 +1
2
−1
9
2 −1
−3
1
−1 1
5 −7 (2 + 9) −3 (22 + 14) −43 +1 2 + 9 − 3(2 − 1)
5
5
−7 (11) −3 (36) −43 11 − 3(1)
−77 − 108 −43 11 − 3
5 −185 −43 8 = −925 − 344 = −1269
0 −5
2 −1
−1 1
1 −1
−43
14
11
7
0
9
1
−3
𝐴4
−1 −4 2
2 −1 7
=
−1 1 3
1 −2 1
2
−1 −1
0
−36
−14
−21
−32
−𝑏3 + 𝑏1
14
11
−4
−287
−118
−165
0 −5
2 −1
−1 1
1 −2
−1 −10.9831 + 10.560
−1 (−423) = 423
−43 −5𝑏3 + 𝑏2
14
11
−4 −43𝑏3 + 𝑏4
−1 21 2 −287 −165 2
−1 −118
−1
−1 21 −236 − 287 − 165(−28 − 36)
−1 21 −523 − 165(−64)
−1
7
3
1
−36
−14
0
2
−1
1
0
−36
1
−2
−1 0
0
𝑏3 + 𝑏4 0
7 14
2 −36
3 11
−1 −14
1 −4
0 −21
−1
0
7 −287
3 −118
4 −165
◦ 𝐴 = −423
◦ 𝐴1 = −2115
◦ 𝐴2 = −3384
◦ 𝐴3 = −1269
◦ 𝐴4 = 423
𝑥1 =
−2115
=5
−423
𝑥2 =
−3384
=8
−423
𝑥3 =
−1269
=3
−423
𝑥4 =
423
= −1
−423
◦ Pembuktian :
−1
2
◦
−1
1
−4
−1
1
−2
2 1
−32
5
7 9
14
8
=
3 1
11
3
1 −4 −1
−4