Metode Numerik 1 - Universitas Indonesia
Metode Numerik 1 - Universitas Indonesia
Metode Numerik 1 - Universitas Indonesia
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Soal:Jawab:⎛ 2 -4 1 3⎞⎛x1⎞ ⎛2⎞⎜ ⎟⎜⎟ ⎜ ⎟⎜-1 2 3 -2⎟⎜x2⎟ ⎜2= ⎟⎜ 3 -4 1 2⎟⎜x ⎟ ⎜ ⎟32⎜ ⎟⎜⎟ ⎜ ⎟⎝ 1 -3 -1 5⎠⎝x4⎠ ⎝2⎠⎛ 2 -4 1 3 ⎞⎛x1⎞ ⎛ 2⎞⎜ ⎟⎜⎟ ⎜ ⎟⎜0 0 3.5 -0.5⎟⎜x2⎟ ⎜3= ⎟⎜0 2 -0.5 -2.5⎟⎜x ⎟ ⎜ ⎟3-1⎜ ⎟⎜⎟ ⎜ ⎟⎝0 -1 -1.5 3.5⎠⎝x4⎠ ⎝ 1⎠⎛ 2 -4 1 3 ⎞⎛x1⎞ ⎛ 2⎞⎜ ⎟⎜⎟ ⎜ ⎟⎜0 2 -0.5 -2.5⎟⎜x2⎟ ⎜-1= ⎟⎜0 0 3.5 -0.5⎟⎜x ⎟ ⎜ ⎟33⎜ ⎟⎜⎟ ⎜ ⎟⎝0 0 0 2 ⎠⎝x4⎠ ⎝ 2⎠baris 2 ditukardengan baris 3⎛ 2 -4 1 3 ⎞⎛x1⎞ ⎛ 2⎞⎜ ⎟⎜⎟ ⎜ ⎟⎜0 2 -0.5 -2.5⎟⎜x2⎟ ⎜-1= ⎟⎜0 0 3.5 -0.5⎟⎜x ⎟ ⎜ ⎟33⎜ ⎟⎜⎟ ⎜ ⎟⎝0 -1 -1.5 3.5⎠⎝x4⎠ ⎝ 1⎠⎛ 2 -4 1 3 ⎞⎛x1⎞ ⎛ 2 ⎞⎜ ⎟⎜⎟ ⎜ ⎟⎜0 2 -0.5 -2.5⎟⎜x2⎟ ⎜-1= ⎟⎜0 0 3.5 -0.5 ⎟⎜x ⎟ ⎜ ⎟33⎜ ⎟⎜⎟ ⎜ ⎟⎝0 0 -1.75 2.25⎠⎝x4⎠ ⎝ 0.5⎠27xx43= 13+0.5x=3.54= 1xx21−1+ 0.5x3+ 2.5x=22 + 4x2− x3−3x4=24= 1= 1
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