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Aljabar Linear Contoh
2x-5y+5z=282x−5y+5z=28 , -3x-2y+15z=35−3x−2y+15z=35 , -3x+6y-5z=33−3x+6y−5z=33
Langkah 1
Tentukan AX=BAX=B dari sistem persamaan tersebut.
[2-55-3-215-36-5]⋅[xyz]=[283533]⎡⎢⎣2−55−3−215−36−5⎤⎥⎦⋅⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣283533⎤⎥⎦
Langkah 2
Langkah 2.1
Find the determinant.
Langkah 2.1.1
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 row 11 by its cofactor and add.
Langkah 2.1.1.1
Consider the corresponding sign chart.
|+-+-+-+-+|∣∣
∣∣+−+−+−+−+∣∣
∣∣
Langkah 2.1.1.2
The cofactor is the minor with the sign changed if the indices match a -− position on the sign chart.
Langkah 2.1.1.3
The minor for a11a11 is the determinant with row 11 and column 11 deleted.
|-2156-5|∣∣∣−2156−5∣∣∣
Langkah 2.1.1.4
Multiply element a11a11 by its cofactor.
2|-2156-5|2∣∣∣−2156−5∣∣∣
Langkah 2.1.1.5
The minor for a12a12 is the determinant with row 11 and column 22 deleted.
|-315-3-5|∣∣∣−315−3−5∣∣∣
Langkah 2.1.1.6
Multiply element a12a12 by its cofactor.
5|-315-3-5|5∣∣∣−315−3−5∣∣∣
Langkah 2.1.1.7
The minor for a13a13 is the determinant with row 11 and column 33 deleted.
|-3-2-36|∣∣∣−3−2−36∣∣∣
Langkah 2.1.1.8
Multiply element a13a13 by its cofactor.
5|-3-2-36|5∣∣∣−3−2−36∣∣∣
Langkah 2.1.1.9
Add the terms together.
2|-2156-5|+5|-315-3-5|+5|-3-2-36|2∣∣∣−2156−5∣∣∣+5∣∣∣−315−3−5∣∣∣+5∣∣∣−3−2−36∣∣∣
2|-2156-5|+5|-315-3-5|+5|-3-2-36|2∣∣∣−2156−5∣∣∣+5∣∣∣−315−3−5∣∣∣+5∣∣∣−3−2−36∣∣∣
Langkah 2.1.2
Evaluasi |-2156-5|∣∣∣−2156−5∣∣∣.
Langkah 2.1.2.1
Determinan dari matriks 2×22×2 dapat dicari menggunakan rumus |abcd|=ad-cb∣∣∣abcd∣∣∣=ad−cb.
2(-2⋅-5-6⋅15)+5|-315-3-5|+5|-3-2-36|2(−2⋅−5−6⋅15)+5∣∣∣−315−3−5∣∣∣+5∣∣∣−3−2−36∣∣∣
Langkah 2.1.2.2
Sederhanakan determinannya.
Langkah 2.1.2.2.1
Sederhanakan setiap suku.
Langkah 2.1.2.2.1.1
Kalikan -2−2 dengan -5−5.
2(10-6⋅15)+5|-315-3-5|+5|-3-2-36|2(10−6⋅15)+5∣∣∣−315−3−5∣∣∣+5∣∣∣−3−2−36∣∣∣
Langkah 2.1.2.2.1.2
Kalikan -6−6 dengan 1515.
2(10-90)+5|-315-3-5|+5|-3-2-36|2(10−90)+5∣∣∣−315−3−5∣∣∣+5∣∣∣−3−2−36∣∣∣
2(10-90)+5|-315-3-5|+5|-3-2-36|2(10−90)+5∣∣∣−315−3−5∣∣∣+5∣∣∣−3−2−36∣∣∣
Langkah 2.1.2.2.2
Kurangi 9090 dengan 1010.
2⋅-80+5|-315-3-5|+5|-3-2-36|2⋅−80+5∣∣∣−315−3−5∣∣∣+5∣∣∣−3−2−36∣∣∣
2⋅-80+5|-315-3-5|+5|-3-2-36|2⋅−80+5∣∣∣−315−3−5∣∣∣+5∣∣∣−3−2−36∣∣∣
2⋅-80+5|-315-3-5|+5|-3-2-36|2⋅−80+5∣∣∣−315−3−5∣∣∣+5∣∣∣−3−2−36∣∣∣
Langkah 2.1.3
Evaluasi |-315-3-5|∣∣∣−315−3−5∣∣∣.
Langkah 2.1.3.1
Determinan dari matriks 2×22×2 dapat dicari menggunakan rumus |abcd|=ad-cb∣∣∣abcd∣∣∣=ad−cb.
2⋅-80+5(-3⋅-5-(-3⋅15))+5|-3-2-36|2⋅−80+5(−3⋅−5−(−3⋅15))+5∣∣∣−3−2−36∣∣∣
Langkah 2.1.3.2
Sederhanakan determinannya.
Langkah 2.1.3.2.1
Sederhanakan setiap suku.
Langkah 2.1.3.2.1.1
Kalikan -3−3 dengan -5−5.
2⋅-80+5(15-(-3⋅15))+5|-3-2-36|2⋅−80+5(15−(−3⋅15))+5∣∣∣−3−2−36∣∣∣
Langkah 2.1.3.2.1.2
Kalikan -(-3⋅15)−(−3⋅15).
Langkah 2.1.3.2.1.2.1
Kalikan -3−3 dengan 1515.
2⋅-80+5(15--45)+5|-3-2-36|2⋅−80+5(15−−45)+5∣∣∣−3−2−36∣∣∣
Langkah 2.1.3.2.1.2.2
Kalikan -1−1 dengan -45−45.
2⋅-80+5(15+45)+5|-3-2-36|2⋅−80+5(15+45)+5∣∣∣−3−2−36∣∣∣
2⋅-80+5(15+45)+5|-3-2-36|2⋅−80+5(15+45)+5∣∣∣−3−2−36∣∣∣
2⋅-80+5(15+45)+5|-3-2-36|2⋅−80+5(15+45)+5∣∣∣−3−2−36∣∣∣
Langkah 2.1.3.2.2
Tambahkan 1515 dan 4545.
2⋅-80+5⋅60+5|-3-2-36|2⋅−80+5⋅60+5∣∣∣−3−2−36∣∣∣
2⋅-80+5⋅60+5|-3-2-36|2⋅−80+5⋅60+5∣∣∣−3−2−36∣∣∣
2⋅-80+5⋅60+5|-3-2-36|2⋅−80+5⋅60+5∣∣∣−3−2−36∣∣∣
Langkah 2.1.4
Evaluasi |-3-2-36|∣∣∣−3−2−36∣∣∣.
Langkah 2.1.4.1
Determinan dari matriks 2×22×2 dapat dicari menggunakan rumus |abcd|=ad-cb∣∣∣abcd∣∣∣=ad−cb.
2⋅-80+5⋅60+5(-3⋅6-(-3⋅-2))2⋅−80+5⋅60+5(−3⋅6−(−3⋅−2))
Langkah 2.1.4.2
Sederhanakan determinannya.
Langkah 2.1.4.2.1
Sederhanakan setiap suku.
Langkah 2.1.4.2.1.1
Kalikan -3−3 dengan 66.
2⋅-80+5⋅60+5(-18-(-3⋅-2))2⋅−80+5⋅60+5(−18−(−3⋅−2))
Langkah 2.1.4.2.1.2
Kalikan -(-3⋅-2)−(−3⋅−2).
Langkah 2.1.4.2.1.2.1
Kalikan -3−3 dengan -2−2.
2⋅-80+5⋅60+5(-18-1⋅6)2⋅−80+5⋅60+5(−18−1⋅6)
Langkah 2.1.4.2.1.2.2
Kalikan -1−1 dengan 66.
2⋅-80+5⋅60+5(-18-6)2⋅−80+5⋅60+5(−18−6)
2⋅-80+5⋅60+5(-18-6)2⋅−80+5⋅60+5(−18−6)
2⋅-80+5⋅60+5(-18-6)2⋅−80+5⋅60+5(−18−6)
Langkah 2.1.4.2.2
Kurangi 66 dengan -18−18.
2⋅-80+5⋅60+5⋅-242⋅−80+5⋅60+5⋅−24
2⋅-80+5⋅60+5⋅-242⋅−80+5⋅60+5⋅−24
2⋅-80+5⋅60+5⋅-242⋅−80+5⋅60+5⋅−24
Langkah 2.1.5
Sederhanakan determinannya.
Langkah 2.1.5.1
Sederhanakan setiap suku.
Langkah 2.1.5.1.1
Kalikan 22 dengan -80−80.
-160+5⋅60+5⋅-24−160+5⋅60+5⋅−24
Langkah 2.1.5.1.2
Kalikan 55 dengan 6060.
-160+300+5⋅-24−160+300+5⋅−24
Langkah 2.1.5.1.3
Kalikan 55 dengan -24−24.
-160+300-120−160+300−120
-160+300-120−160+300−120
Langkah 2.1.5.2
Tambahkan -160−160 dan 300300.
140-120140−120
Langkah 2.1.5.3
Kurangi 120120 dengan 140140.
2020
2020
2020
Langkah 2.2
Since the determinant is non-zero, the inverse exists.
Langkah 2.3
Set up a 3×63×6 matrix where the left half is the original matrix and the right half is its identity matrix.
[2-55100-3-215010-36-5001]⎡⎢⎣2−55100−3−215010−36−5001⎤⎥⎦
Langkah 2.4
Tentukan bentuk eselon baris yang dikurangi.
Langkah 2.4.1
Multiply each element of R1R1 by 1212 to make the entry at 1,11,1 a 11.
Langkah 2.4.1.1
Multiply each element of R1R1 by 1212 to make the entry at 1,11,1 a 11.
[22-5252120202-3-215010-36-5001]⎡⎢
⎢⎣22−5252120202−3−215010−36−5001⎤⎥
⎥⎦
Langkah 2.4.1.2
Sederhanakan R1R1.
[1-52521200-3-215010-36-5001]⎡⎢
⎢⎣1−52521200−3−215010−36−5001⎤⎥
⎥⎦
[1-52521200-3-215010-36-5001]⎡⎢
⎢⎣1−52521200−3−215010−36−5001⎤⎥
⎥⎦
Langkah 2.4.2
Perform the row operation R2=R2+3R1R2=R2+3R1 to make the entry at 2,12,1 a 00.
Langkah 2.4.2.1
Perform the row operation R2=R2+3R1R2=R2+3R1 to make the entry at 2,12,1 a 00.
[1-52521200-3+3⋅1-2+3(-52)15+3(52)0+3(12)1+3⋅00+3⋅0-36-5001]⎡⎢
⎢
⎢⎣1−52521200−3+3⋅1−2+3(−52)15+3(52)0+3(12)1+3⋅00+3⋅0−36−5001⎤⎥
⎥
⎥⎦
Langkah 2.4.2.2
Sederhanakan R2.
[1-525212000-1924523210-36-5001]
[1-525212000-1924523210-36-5001]
Langkah 2.4.3
Perform the row operation R3=R3+3R1 to make the entry at 3,1 a 0.
Langkah 2.4.3.1
Perform the row operation R3=R3+3R1 to make the entry at 3,1 a 0.
[1-525212000-1924523210-3+3⋅16+3(-52)-5+3(52)0+3(12)0+3⋅01+3⋅0]
Langkah 2.4.3.2
Sederhanakan R3.
[1-525212000-19245232100-32523201]
[1-525212000-19245232100-32523201]
Langkah 2.4.4
Multiply each element of R2 by -219 to make the entry at 2,2 a 1.
Langkah 2.4.4.1
Multiply each element of R2 by -219 to make the entry at 2,2 a 1.
[1-52521200-219⋅0-219(-192)-219⋅452-219⋅32-219⋅1-219⋅00-32523201]
Langkah 2.4.4.2
Sederhanakan R2.
[1-5252120001-4519-319-21900-32523201]
[1-5252120001-4519-319-21900-32523201]
Langkah 2.4.5
Perform the row operation R3=R3+32R2 to make the entry at 3,2 a 0.
Langkah 2.4.5.1
Perform the row operation R3=R3+32R2 to make the entry at 3,2 a 0.
[1-5252120001-4519-319-21900+32⋅0-32+32⋅152+32(-4519)32+32(-319)0+32(-219)1+32⋅0]
Langkah 2.4.5.2
Sederhanakan R3.
[1-5252120001-4519-319-219000-20192419-3191]
[1-5252120001-4519-319-219000-20192419-3191]
Langkah 2.4.6
Multiply each element of R3 by -1920 to make the entry at 3,3 a 1.
Langkah 2.4.6.1
Multiply each element of R3 by -1920 to make the entry at 3,3 a 1.
[1-5252120001-4519-319-2190-1920⋅0-1920⋅0-1920(-2019)-1920⋅2419-1920(-319)-1920⋅1]
Langkah 2.4.6.2
Sederhanakan R3.
[1-5252120001-4519-319-2190001-65320-1920]
[1-5252120001-4519-319-2190001-65320-1920]
Langkah 2.4.7
Perform the row operation R2=R2+4519R3 to make the entry at 2,3 a 0.
Langkah 2.4.7.1
Perform the row operation R2=R2+4519R3 to make the entry at 2,3 a 0.
[1-525212000+4519⋅01+4519⋅0-4519+4519⋅1-319+4519(-65)-219+4519⋅3200+4519(-1920)001-65320-1920]
Langkah 2.4.7.2
Sederhanakan R2.
[1-52521200010-314-94001-65320-1920]
[1-52521200010-314-94001-65320-1920]
Langkah 2.4.8
Perform the row operation R1=R1-52R3 to make the entry at 1,3 a 0.
Langkah 2.4.8.1
Perform the row operation R1=R1-52R3 to make the entry at 1,3 a 0.
[1-52⋅0-52-52⋅052-52⋅112-52(-65)0-52⋅3200-52(-1920)010-314-94001-65320-1920]
Langkah 2.4.8.2
Sederhanakan R1.
[1-52072-38198010-314-94001-65320-1920]
[1-52072-38198010-314-94001-65320-1920]
Langkah 2.4.9
Perform the row operation R1=R1+52R2 to make the entry at 1,2 a 0.
Langkah 2.4.9.1
Perform the row operation R1=R1+52R2 to make the entry at 1,2 a 0.
[1+52⋅0-52+52⋅10+52⋅072+52⋅-3-38+52⋅14198+52(-94)010-314-94001-65320-1920]
Langkah 2.4.9.2
Sederhanakan R1.
[100-414-134010-314-94001-65320-1920]
[100-414-134010-314-94001-65320-1920]
[100-414-134010-314-94001-65320-1920]
Langkah 2.5
The right half of the reduced row echelon form is the inverse.
[-414-134-314-94-65320-1920]
[-414-134-314-94-65320-1920]
Langkah 3
Kalikan kedua sisi persamaan matriks dengan matriks balikan.
([-414-134-314-94-65320-1920]⋅[2-55-3-215-36-5])⋅[xyz]=[-414-134-314-94-65320-1920]⋅[283533]
Langkah 4
Semua matriks akan selalu bernilai 1 jika dikalikan dengan balikannya. A⋅A-1=1.
[xyz]=[-414-134-314-94-65320-1920]⋅[283533]
Langkah 5
Langkah 5.1
Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. In this case, the first matrix is 3×3 and the second matrix is 3×1.
Langkah 5.2
Kalikan setiap baris pada matriks pertama dengan setiap kolom pada matriks kedua.
[-4⋅28+14⋅35-134⋅33-3⋅28+14⋅35-94⋅33-65⋅28+320⋅35-1920⋅33]
Langkah 5.3
Sederhanakan setiap elemen dalam matriks dengan mengalikan semua pernyataannya.
[-4212-2992-59710]
[-4212-2992-59710]
Langkah 6
Sederhanakan sisi kiri dan kanan.
[xyz]=[-4212-2992-59710]
Langkah 7
Tentukan penyelesaiannya.
x=-4212
y=-2992
z=-59710