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Matematika Berhingga Contoh
[1012-2-1300]⎡⎢⎣1012−2−1300⎤⎥⎦
Langkah 1
Langkah 1.1
Choose the row or column with the most 00 elements. If there are no 00 elements choose any row or column. Multiply every element in column 22 by its cofactor and add.
Langkah 1.1.1
Consider the corresponding sign chart.
|+-+-+-+-+|∣∣
∣∣+−+−+−+−+∣∣
∣∣
Langkah 1.1.2
The cofactor is the minor with the sign changed if the indices match a -− position on the sign chart.
Langkah 1.1.3
The minor for a12a12 is the determinant with row 11 and column 22 deleted.
|2-130|∣∣∣2−130∣∣∣
Langkah 1.1.4
Multiply element a12a12 by its cofactor.
0|2-130|0∣∣∣2−130∣∣∣
Langkah 1.1.5
The minor for a22a22 is the determinant with row 22 and column 22 deleted.
|1130|∣∣∣1130∣∣∣
Langkah 1.1.6
Multiply element a22a22 by its cofactor.
-2|1130|−2∣∣∣1130∣∣∣
Langkah 1.1.7
The minor for a32a32 is the determinant with row 33 and column 22 deleted.
|112-1|∣∣∣112−1∣∣∣
Langkah 1.1.8
Multiply element a32a32 by its cofactor.
0|112-1|0∣∣∣112−1∣∣∣
Langkah 1.1.9
Add the terms together.
0|2-130|-2|1130|+0|112-1|0∣∣∣2−130∣∣∣−2∣∣∣1130∣∣∣+0∣∣∣112−1∣∣∣
0|2-130|-2|1130|+0|112-1|0∣∣∣2−130∣∣∣−2∣∣∣1130∣∣∣+0∣∣∣112−1∣∣∣
Langkah 1.2
Kalikan 00 dengan |2-130|∣∣∣2−130∣∣∣.
0-2|1130|+0|112-1|0−2∣∣∣1130∣∣∣+0∣∣∣112−1∣∣∣
Langkah 1.3
Kalikan 00 dengan |112-1|∣∣∣112−1∣∣∣.
0-2|1130|+00−2∣∣∣1130∣∣∣+0
Langkah 1.4
Evaluasi |1130|∣∣∣1130∣∣∣.
Langkah 1.4.1
Determinan dari matriks 2×22×2 dapat dicari menggunakan rumus |abcd|=ad-cb∣∣∣abcd∣∣∣=ad−cb.
0-2(1⋅0-3⋅1)+00−2(1⋅0−3⋅1)+0
Langkah 1.4.2
Sederhanakan determinannya.
Langkah 1.4.2.1
Sederhanakan setiap suku.
Langkah 1.4.2.1.1
Kalikan 00 dengan 11.
0-2(0-3⋅1)+00−2(0−3⋅1)+0
Langkah 1.4.2.1.2
Kalikan -3−3 dengan 11.
0-2(0-3)+00−2(0−3)+0
0-2(0-3)+00−2(0−3)+0
Langkah 1.4.2.2
Kurangi 33 dengan 00.
0-2⋅-3+00−2⋅−3+0
0-2⋅-3+00−2⋅−3+0
0-2⋅-3+00−2⋅−3+0
Langkah 1.5
Sederhanakan determinannya.
Langkah 1.5.1
Kalikan -2−2 dengan -3−3.
0+6+00+6+0
Langkah 1.5.2
Tambahkan 00 dan 66.
6+06+0
Langkah 1.5.3
Tambahkan 66 dan 00.
66
66
66
Langkah 2
Since the determinant is non-zero, the inverse exists.
Langkah 3
Set up a 3×63×6 matrix where the left half is the original matrix and the right half is its identity matrix.
[1011002-2-1010300001]⎡⎢⎣1011002−2−1010300001⎤⎥⎦
Langkah 4
Langkah 4.1
Perform the row operation R2=R2-2R1R2=R2−2R1 to make the entry at 2,12,1 a 00.
Langkah 4.1.1
Perform the row operation R2=R2-2R1R2=R2−2R1 to make the entry at 2,12,1 a 00.
[1011002-2⋅1-2-2⋅0-1-2⋅10-2⋅11-2⋅00-2⋅0300001]⎡⎢⎣1011002−2⋅1−2−2⋅0−1−2⋅10−2⋅11−2⋅00−2⋅0300001⎤⎥⎦
Langkah 4.1.2
Sederhanakan R2R2.
[1011000-2-3-210300001]⎡⎢⎣1011000−2−3−210300001⎤⎥⎦
[1011000-2-3-210300001]⎡⎢⎣1011000−2−3−210300001⎤⎥⎦
Langkah 4.2
Perform the row operation R3=R3-3R1R3=R3−3R1 to make the entry at 3,13,1 a 00.
Langkah 4.2.1
Perform the row operation R3=R3-3R1R3=R3−3R1 to make the entry at 3,13,1 a 00.
[1011000-2-3-2103-3⋅10-3⋅00-3⋅10-3⋅10-3⋅01-3⋅0]⎡⎢⎣1011000−2−3−2103−3⋅10−3⋅00−3⋅10−3⋅10−3⋅01−3⋅0⎤⎥⎦
Langkah 4.2.2
Sederhanakan R3R3.
[1011000-2-3-21000-3-301]⎡⎢⎣1011000−2−3−21000−3−301⎤⎥⎦
[1011000-2-3-21000-3-301]⎡⎢⎣1011000−2−3−21000−3−301⎤⎥⎦
Langkah 4.3
Multiply each element of R2R2 by -12−12 to make the entry at 2,22,2 a 11.
Langkah 4.3.1
Multiply each element of R2R2 by -12−12 to make the entry at 2,22,2 a 11.
[101100-12⋅0-12⋅-2-12⋅-3-12⋅-2-12⋅1-12⋅000-3-301]⎡⎢
⎢⎣101100−12⋅0−12⋅−2−12⋅−3−12⋅−2−12⋅1−12⋅000−3−301⎤⎥
⎥⎦
Langkah 4.3.2
Sederhanakan R2.
[10110001321-12000-3-301]
[10110001321-12000-3-301]
Langkah 4.4
Multiply each element of R3 by -13 to make the entry at 3,3 a 1.
Langkah 4.4.1
Multiply each element of R3 by -13 to make the entry at 3,3 a 1.
[10110001321-120-13⋅0-13⋅0-13⋅-3-13⋅-3-13⋅0-13⋅1]
Langkah 4.4.2
Sederhanakan R3.
[10110001321-12000110-13]
[10110001321-12000110-13]
Langkah 4.5
Perform the row operation R2=R2-32R3 to make the entry at 2,3 a 0.
Langkah 4.5.1
Perform the row operation R2=R2-32R3 to make the entry at 2,3 a 0.
[1011000-32⋅01-32⋅032-32⋅11-32⋅1-12-32⋅00-32(-13)00110-13]
Langkah 4.5.2
Sederhanakan R2.
[101100010-12-121200110-13]
[101100010-12-121200110-13]
Langkah 4.6
Perform the row operation R1=R1-R3 to make the entry at 1,3 a 0.
Langkah 4.6.1
Perform the row operation R1=R1-R3 to make the entry at 1,3 a 0.
[1-00-01-11-10-00+13010-12-121200110-13]
Langkah 4.6.2
Sederhanakan R1.
[1000013010-12-121200110-13]
[1000013010-12-121200110-13]
[1000013010-12-121200110-13]
Langkah 5
The right half of the reduced row echelon form is the inverse.
[0013-12-121210-13]