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Lineare Algebra Beispiele
[4-10291-25-37-19]⎡⎢⎣4−10291−25−37−19⎤⎥⎦
Schritt 1
Schritt 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.
Schritt 1.1.1
Consider the corresponding sign chart.
|+-+-+-+-+|
Schritt 1.1.2
The cofactor is the minor with the sign changed if the indices match a - position on the sign chart.
Schritt 1.1.3
The minor for a11 is the determinant with row 1 and column 1 deleted.
|-257-19|
Schritt 1.1.4
Multiply element a11 by its cofactor.
4|-257-19|
Schritt 1.1.5
The minor for a12 is the determinant with row 1 and column 2 deleted.
|15-3-19|
Schritt 1.1.6
Multiply element a12 by its cofactor.
10|15-3-19|
Schritt 1.1.7
The minor for a13 is the determinant with row 1 and column 3 deleted.
|1-2-37|
Schritt 1.1.8
Multiply element a13 by its cofactor.
29|1-2-37|
Schritt 1.1.9
Add the terms together.
4|-257-19|+10|15-3-19|+29|1-2-37|
4|-257-19|+10|15-3-19|+29|1-2-37|
Schritt 1.2
Berechne |-257-19|.
Schritt 1.2.1
Die Determinante einer 2×2-Matrix kann mithilfe der Formel |abcd|=ad-cb bestimmt werden.
4(-2⋅-19-7⋅5)+10|15-3-19|+29|1-2-37|
Schritt 1.2.2
Vereinfache die Determinante.
Schritt 1.2.2.1
Vereinfache jeden Term.
Schritt 1.2.2.1.1
Mutltipliziere -2 mit -19.
4(38-7⋅5)+10|15-3-19|+29|1-2-37|
Schritt 1.2.2.1.2
Mutltipliziere -7 mit 5.
4(38-35)+10|15-3-19|+29|1-2-37|
4(38-35)+10|15-3-19|+29|1-2-37|
Schritt 1.2.2.2
Subtrahiere 35 von 38.
4⋅3+10|15-3-19|+29|1-2-37|
4⋅3+10|15-3-19|+29|1-2-37|
4⋅3+10|15-3-19|+29|1-2-37|
Schritt 1.3
Berechne |15-3-19|.
Schritt 1.3.1
Die Determinante einer 2×2-Matrix kann mithilfe der Formel |abcd|=ad-cb bestimmt werden.
4⋅3+10(1⋅-19-(-3⋅5))+29|1-2-37|
Schritt 1.3.2
Vereinfache die Determinante.
Schritt 1.3.2.1
Vereinfache jeden Term.
Schritt 1.3.2.1.1
Mutltipliziere -19 mit 1.
4⋅3+10(-19-(-3⋅5))+29|1-2-37|
Schritt 1.3.2.1.2
Multipliziere -(-3⋅5).
Schritt 1.3.2.1.2.1
Mutltipliziere -3 mit 5.
4⋅3+10(-19--15)+29|1-2-37|
Schritt 1.3.2.1.2.2
Mutltipliziere -1 mit -15.
4⋅3+10(-19+15)+29|1-2-37|
4⋅3+10(-19+15)+29|1-2-37|
4⋅3+10(-19+15)+29|1-2-37|
Schritt 1.3.2.2
Addiere -19 und 15.
4⋅3+10⋅-4+29|1-2-37|
4⋅3+10⋅-4+29|1-2-37|
4⋅3+10⋅-4+29|1-2-37|
Schritt 1.4
Berechne |1-2-37|.
Schritt 1.4.1
Die Determinante einer 2×2-Matrix kann mithilfe der Formel |abcd|=ad-cb bestimmt werden.
4⋅3+10⋅-4+29(1⋅7-(-3⋅-2))
Schritt 1.4.2
Vereinfache die Determinante.
Schritt 1.4.2.1
Vereinfache jeden Term.
Schritt 1.4.2.1.1
Mutltipliziere 7 mit 1.
4⋅3+10⋅-4+29(7-(-3⋅-2))
Schritt 1.4.2.1.2
Multipliziere -(-3⋅-2).
Schritt 1.4.2.1.2.1
Mutltipliziere -3 mit -2.
4⋅3+10⋅-4+29(7-1⋅6)
Schritt 1.4.2.1.2.2
Mutltipliziere -1 mit 6.
4⋅3+10⋅-4+29(7-6)
4⋅3+10⋅-4+29(7-6)
4⋅3+10⋅-4+29(7-6)
Schritt 1.4.2.2
Subtrahiere 6 von 7.
4⋅3+10⋅-4+29⋅1
4⋅3+10⋅-4+29⋅1
4⋅3+10⋅-4+29⋅1
Schritt 1.5
Vereinfache die Determinante.
Schritt 1.5.1
Vereinfache jeden Term.
Schritt 1.5.1.1
Mutltipliziere 4 mit 3.
12+10⋅-4+29⋅1
Schritt 1.5.1.2
Mutltipliziere 10 mit -4.
12-40+29⋅1
Schritt 1.5.1.3
Mutltipliziere 29 mit 1.
12-40+29
12-40+29
Schritt 1.5.2
Subtrahiere 40 von 12.
-28+29
Schritt 1.5.3
Addiere -28 und 29.
1
1
1
Schritt 2
Since the determinant is non-zero, the inverse exists.
Schritt 3
Set up a 3×6 matrix where the left half is the original matrix and the right half is its identity matrix.
[4-10291001-25010-37-19001]
Schritt 4
Schritt 4.1
Multiply each element of R1 by 14 to make the entry at 1,1 a 1.
Schritt 4.1.1
Multiply each element of R1 by 14 to make the entry at 1,1 a 1.
[44-1042941404041-25010-37-19001]
Schritt 4.1.2
Vereinfache R1.
[1-5229414001-25010-37-19001]
[1-5229414001-25010-37-19001]
Schritt 4.2
Perform the row operation R2=R2-R1 to make the entry at 2,1 a 0.
Schritt 4.2.1
Perform the row operation R2=R2-R1 to make the entry at 2,1 a 0.
[1-5229414001-1-2+525-2940-141-00-0-37-19001]
Schritt 4.2.2
Vereinfache R2.
[1-522941400012-94-1410-37-19001]
[1-522941400012-94-1410-37-19001]
Schritt 4.3
Perform the row operation R3=R3+3R1 to make the entry at 3,1 a 0.
Schritt 4.3.1
Perform the row operation R3=R3+3R1 to make the entry at 3,1 a 0.
[1-522941400012-94-1410-3+3⋅17+3(-52)-19+3(294)0+3(14)0+3⋅01+3⋅0]
Schritt 4.3.2
Vereinfache R3.
[1-522941400012-94-14100-121143401]
[1-522941400012-94-14100-121143401]
Schritt 4.4
Multiply each element of R2 by 2 to make the entry at 2,2 a 1.
Schritt 4.4.1
Multiply each element of R2 by 2 to make the entry at 2,2 a 1.
[1-5229414002⋅02(12)2(-94)2(-14)2⋅12⋅00-121143401]
Schritt 4.4.2
Vereinfache R2.
[1-52294140001-92-12200-121143401]
[1-52294140001-92-12200-121143401]
Schritt 4.5
Perform the row operation R3=R3+12R2 to make the entry at 3,2 a 0.
Schritt 4.5.1
Perform the row operation R3=R3+12R2 to make the entry at 3,2 a 0.
[1-52294140001-92-12200+12⋅0-12+12⋅1114+12(-92)34+12(-12)0+12⋅21+12⋅0]
Schritt 4.5.2
Vereinfache R3.
[1-52294140001-92-122000121211]
[1-52294140001-92-122000121211]
Schritt 4.6
Multiply each element of R3 by 2 to make the entry at 3,3 a 1.
Schritt 4.6.1
Multiply each element of R3 by 2 to make the entry at 3,3 a 1.
[1-52294140001-92-12202⋅02⋅02(12)2(12)2⋅12⋅1]
Schritt 4.6.2
Vereinfache R3.
[1-52294140001-92-1220001122]
[1-52294140001-92-1220001122]
Schritt 4.7
Perform the row operation R2=R2+92R3 to make the entry at 2,3 a 0.
Schritt 4.7.1
Perform the row operation R2=R2+92R3 to make the entry at 2,3 a 0.
[1-5229414000+92⋅01+92⋅0-92+92⋅1-12+92⋅12+92⋅20+92⋅2001122]
Schritt 4.7.2
Vereinfache R2.
[1-5229414000104119001122]
[1-5229414000104119001122]
Schritt 4.8
Perform the row operation R1=R1-294R3 to make the entry at 1,3 a 0.
Schritt 4.8.1
Perform the row operation R1=R1-294R3 to make the entry at 1,3 a 0.
[1-294⋅0-52-294⋅0294-294⋅114-294⋅10-294⋅20-294⋅20104119001122]
Schritt 4.8.2
Vereinfache R1.
[1-520-7-292-2920104119001122]
[1-520-7-292-2920104119001122]
Schritt 4.9
Perform the row operation R1=R1+52R2 to make the entry at 1,2 a 0.
Schritt 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⋅0-7+52⋅4-292+52⋅11-292+52⋅90104119001122]
Schritt 4.9.2
Vereinfache R1.
[10031380104119001122]
[10031380104119001122]
[10031380104119001122]
Schritt 5
The right half of the reduced row echelon form is the inverse.
[31384119122]