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Aljabar Linear Contoh
14x-21y-7z=10 , -4x+2y-2z=4 , 56x-21y+7z=5
Langkah 1
Tentukan AX=B dari sistem persamaan tersebut.
[14-21-7-42-256-217]⋅[xyz]=[1045]
Langkah 2
Langkah 2.1
Find the determinant.
Langkah 2.1.1
Choose the row or column with the most 0 elements. If there are no 0 elements choose any row or column. Multiply every element in row 1 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 a11 is the determinant with row 1 and column 1 deleted.
|2-2-217|
Langkah 2.1.1.4
Multiply element a11 by its cofactor.
14|2-2-217|
Langkah 2.1.1.5
The minor for a12 is the determinant with row 1 and column 2 deleted.
|-4-2567|
Langkah 2.1.1.6
Multiply element a12 by its cofactor.
21|-4-2567|
Langkah 2.1.1.7
The minor for a13 is the determinant with row 1 and column 3 deleted.
|-4256-21|
Langkah 2.1.1.8
Multiply element a13 by its cofactor.
-7|-4256-21|
Langkah 2.1.1.9
Add the terms together.
14|2-2-217|+21|-4-2567|-7|-4256-21|
14|2-2-217|+21|-4-2567|-7|-4256-21|
Langkah 2.1.2
Evaluasi |2-2-217|.
Langkah 2.1.2.1
Determinan dari matriks 2×2 dapat dicari menggunakan rumus |abcd|=ad-cb.
14(2⋅7-(-21⋅-2))+21|-4-2567|-7|-4256-21|
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 dengan 7.
14(14-(-21⋅-2))+21|-4-2567|-7|-4256-21|
Langkah 2.1.2.2.1.2
Kalikan -(-21⋅-2).
Langkah 2.1.2.2.1.2.1
Kalikan -21 dengan -2.
14(14-1⋅42)+21|-4-2567|-7|-4256-21|
Langkah 2.1.2.2.1.2.2
Kalikan -1 dengan 42.
14(14-42)+21|-4-2567|-7|-4256-21|
14(14-42)+21|-4-2567|-7|-4256-21|
14(14-42)+21|-4-2567|-7|-4256-21|
Langkah 2.1.2.2.2
Kurangi 42 dengan 14.
14⋅-28+21|-4-2567|-7|-4256-21|
14⋅-28+21|-4-2567|-7|-4256-21|
14⋅-28+21|-4-2567|-7|-4256-21|
Langkah 2.1.3
Evaluasi |-4-2567|.
Langkah 2.1.3.1
Determinan dari matriks 2×2 dapat dicari menggunakan rumus |abcd|=ad-cb.
14⋅-28+21(-4⋅7-56⋅-2)-7|-4256-21|
Langkah 2.1.3.2
Sederhanakan determinannya.
Langkah 2.1.3.2.1
Sederhanakan setiap suku.
Langkah 2.1.3.2.1.1
Kalikan -4 dengan 7.
14⋅-28+21(-28-56⋅-2)-7|-4256-21|
Langkah 2.1.3.2.1.2
Kalikan -56 dengan -2.
14⋅-28+21(-28+112)-7|-4256-21|
14⋅-28+21(-28+112)-7|-4256-21|
Langkah 2.1.3.2.2
Tambahkan -28 dan 112.
14⋅-28+21⋅84-7|-4256-21|
14⋅-28+21⋅84-7|-4256-21|
14⋅-28+21⋅84-7|-4256-21|
Langkah 2.1.4
Evaluasi |-4256-21|.
Langkah 2.1.4.1
Determinan dari matriks 2×2 dapat dicari menggunakan rumus |abcd|=ad-cb.
14⋅-28+21⋅84-7(-4⋅-21-56⋅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 -4 dengan -21.
14⋅-28+21⋅84-7(84-56⋅2)
Langkah 2.1.4.2.1.2
Kalikan -56 dengan 2.
14⋅-28+21⋅84-7(84-112)
14⋅-28+21⋅84-7(84-112)
Langkah 2.1.4.2.2
Kurangi 112 dengan 84.
14⋅-28+21⋅84-7⋅-28
14⋅-28+21⋅84-7⋅-28
14⋅-28+21⋅84-7⋅-28
Langkah 2.1.5
Sederhanakan determinannya.
Langkah 2.1.5.1
Sederhanakan setiap suku.
Langkah 2.1.5.1.1
Kalikan 14 dengan -28.
-392+21⋅84-7⋅-28
Langkah 2.1.5.1.2
Kalikan 21 dengan 84.
-392+1764-7⋅-28
Langkah 2.1.5.1.3
Kalikan -7 dengan -28.
-392+1764+196
-392+1764+196
Langkah 2.1.5.2
Tambahkan -392 dan 1764.
1372+196
Langkah 2.1.5.3
Tambahkan 1372 dan 196.
1568
1568
1568
Langkah 2.2
Since the determinant is non-zero, the inverse exists.
Langkah 2.3
Set up a 3×6 matrix where the left half is the original matrix and the right half is its identity matrix.
[14-21-7100-42-201056-217001]
Langkah 2.4
Tentukan bentuk eselon baris yang dikurangi.
Langkah 2.4.1
Multiply each element of R1 by 114 to make the entry at 1,1 a 1.
Langkah 2.4.1.1
Multiply each element of R1 by 114 to make the entry at 1,1 a 1.
[1414-2114-714114014014-42-201056-217001]
Langkah 2.4.1.2
Sederhanakan R1.
[1-32-1211400-42-201056-217001]
[1-32-1211400-42-201056-217001]
Langkah 2.4.2
Perform the row operation R2=R2+4R1 to make the entry at 2,1 a 0.
Langkah 2.4.2.1
Perform the row operation R2=R2+4R1 to make the entry at 2,1 a 0.
[1-32-1211400-4+4⋅12+4(-32)-2+4(-12)0+4(114)1+4⋅00+4⋅056-217001]
Langkah 2.4.2.2
Sederhanakan R2.
[1-32-12114000-4-4271056-217001]
[1-32-12114000-4-4271056-217001]
Langkah 2.4.3
Perform the row operation R3=R3-56R1 to make the entry at 3,1 a 0.
Langkah 2.4.3.1
Perform the row operation R3=R3-56R1 to make the entry at 3,1 a 0.
[1-32-12114000-4-4271056-56⋅1-21-56(-32)7-56(-12)0-56(114)0-56⋅01-56⋅0]
Langkah 2.4.3.2
Sederhanakan R3.
[1-32-12114000-4-4271006335-401]
[1-32-12114000-4-4271006335-401]
Langkah 2.4.4
Multiply each element of R2 by -14 to make the entry at 2,2 a 1.
Langkah 2.4.4.1
Multiply each element of R2 by -14 to make the entry at 2,2 a 1.
[1-32-1211400-14⋅0-14⋅-4-14⋅-4-14⋅27-14⋅1-14⋅006335-401]
Langkah 2.4.4.2
Sederhanakan R2.
[1-32-1211400011-114-14006335-401]
[1-32-1211400011-114-14006335-401]
Langkah 2.4.5
Perform the row operation R3=R3-63R2 to make the entry at 3,2 a 0.
Langkah 2.4.5.1
Perform the row operation R3=R3-63R2 to make the entry at 3,2 a 0.
[1-32-1211400011-114-1400-63⋅063-63⋅135-63⋅1-4-63(-114)0-63(-14)1-63⋅0]
Langkah 2.4.5.2
Sederhanakan R3.
[1-32-1211400011-114-14000-28126341]
[1-32-1211400011-114-14000-28126341]
Langkah 2.4.6
Multiply each element of R3 by -128 to make the entry at 3,3 a 1.
Langkah 2.4.6.1
Multiply each element of R3 by -128 to make the entry at 3,3 a 1.
[1-32-1211400011-114-140-128⋅0-128⋅0-128⋅-28-128⋅12-128⋅634-128⋅1]
Langkah 2.4.6.2
Sederhanakan R3.
[1-32-1211400011-114-140001-156-916-128]
[1-32-1211400011-114-140001-156-916-128]
Langkah 2.4.7
Perform the row operation R2=R2-R3 to make the entry at 2,3 a 0.
Langkah 2.4.7.1
Perform the row operation R2=R2-R3 to make the entry at 2,3 a 0.
[1-32-12114000-01-01-1-114+156-14+9160+128001-156-916-128]
Langkah 2.4.7.2
Sederhanakan R2.
[1-32-1211400010-356516128001-156-916-128]
[1-32-1211400010-356516128001-156-916-128]
Langkah 2.4.8
Perform the row operation R1=R1+12R3 to make the entry at 1,3 a 0.
Langkah 2.4.8.1
Perform the row operation R1=R1+12R3 to make the entry at 1,3 a 0.
[1+12⋅0-32+12⋅0-12+12⋅1114+12(-156)0+12(-916)0+12(-128)010-356516128001-156-916-128]
Langkah 2.4.8.2
Sederhanakan R1.
[1-320116-932-156010-356516128001-156-916-128]
[1-320116-932-156010-356516128001-156-916-128]
Langkah 2.4.9
Perform the row operation R1=R1+32R2 to make the entry at 1,2 a 0.
Langkah 2.4.9.1
Perform the row operation R1=R1+32R2 to make the entry at 1,2 a 0.
[1+32⋅0-32+32⋅10+32⋅0116+32(-356)-932+32⋅516-156+32⋅128010-356516128001-156-916-128]
Langkah 2.4.9.2
Sederhanakan R1.
[100-156316128010-356516128001-156-916-128]
[100-156316128010-356516128001-156-916-128]
[100-156316128010-356516128001-156-916-128]
Langkah 2.5
The right half of the reduced row echelon form is the inverse.
[-156316128-356516128-156-916-128]
[-156316128-356516128-156-916-128]
Langkah 3
Kalikan kedua sisi persamaan matriks dengan matriks balikan.
([-156316128-356516128-156-916-128]⋅[14-21-7-42-256-217])⋅[xyz]=[-156316128-356516128-156-916-128]⋅[1045]
Langkah 4
Semua matriks akan selalu bernilai 1 jika dikalikan dengan balikannya. A⋅A-1=1.
[xyz]=[-156316128-356516128-156-916-128]⋅[1045]
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.
[-156⋅10+316⋅4+128⋅5-356⋅10+516⋅4+128⋅5-156⋅10-916⋅4-128⋅5]
Langkah 5.3
Sederhanakan setiap elemen dalam matriks dengan mengalikan semua pernyataannya.
[342528-7328]
[342528-7328]
Langkah 6
Sederhanakan sisi kiri dan kanan.
[xyz]=[342528-7328]
Langkah 7
Tentukan penyelesaiannya.
x=34
y=2528
z=-7328