Lineare Algebra Beispiele

Löse die Matrixgleichung [[1,0,0],[1,1,0],[1,1,1]]y=[[1,2],[3,3],[2,1]]
[100110111]y=[123321]100110111y=123321
Schritt 1
Find the inverse of [100110111]100110111.
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Schritt 1.1
Forme um.
|100110111|∣ ∣100110111∣ ∣
Schritt 1.2
Find the determinant.
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Schritt 1.2.1
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.
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Schritt 1.2.1.1
Consider the corresponding sign chart.
|+-+-+-+-+|∣ ∣+++++∣ ∣
Schritt 1.2.1.2
The cofactor is the minor with the sign changed if the indices match a - position on the sign chart.
Schritt 1.2.1.3
The minor for a11a11 is the determinant with row 11 and column 11 deleted.
|1011|1011
Schritt 1.2.1.4
Multiply element a11a11 by its cofactor.
1|1011|11011
Schritt 1.2.1.5
The minor for a12a12 is the determinant with row 11 and column 22 deleted.
|1011|1011
Schritt 1.2.1.6
Multiply element a12a12 by its cofactor.
0|1011|01011
Schritt 1.2.1.7
The minor for a13a13 is the determinant with row 11 and column 33 deleted.
|1111|1111
Schritt 1.2.1.8
Multiply element a13a13 by its cofactor.
0|1111|01111
Schritt 1.2.1.9
Add the terms together.
1|1011|+0|1011|+0|1111|11011+01011+01111
1|1011|+0|1011|+0|1111|11011+01011+01111
Schritt 1.2.2
Mutltipliziere 00 mit |1011|1011.
1|1011|+0+0|1111|
Schritt 1.2.3
Mutltipliziere 0 mit |1111|.
1|1011|+0+0
Schritt 1.2.4
Berechne |1011|.
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Schritt 1.2.4.1
Die Determinante einer 2×2-Matrix kann mithilfe der Formel |abcd|=ad-cb bestimmt werden.
1(11-10)+0+0
Schritt 1.2.4.2
Vereinfache die Determinante.
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Schritt 1.2.4.2.1
Mutltipliziere 1 mit 1.
1(1-10)+0+0
Schritt 1.2.4.2.2
Subtrahiere 0 von 1.
11+0+0
11+0+0
11+0+0
Schritt 1.2.5
Vereinfache die Determinante.
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Schritt 1.2.5.1
Mutltipliziere 1 mit 1.
1+0+0
Schritt 1.2.5.2
Addiere 1 und 0.
1+0
Schritt 1.2.5.3
Addiere 1 und 0.
1
1
1
Schritt 1.3
Since the determinant is non-zero, the inverse exists.
Schritt 1.4
Set up a 3×6 matrix where the left half is the original matrix and the right half is its identity matrix.
[100100110010111001]
Schritt 1.5
Ermittele die normierte Zeilenstufenform.
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Schritt 1.5.1
Perform the row operation R2=R2-R1 to make the entry at 2,1 a 0.
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Schritt 1.5.1.1
Perform the row operation R2=R2-R1 to make the entry at 2,1 a 0.
[1001001-11-00-00-11-00-0111001]
Schritt 1.5.1.2
Vereinfache R2.
[100100010-110111001]
[100100010-110111001]
Schritt 1.5.2
Perform the row operation R3=R3-R1 to make the entry at 3,1 a 0.
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Schritt 1.5.2.1
Perform the row operation R3=R3-R1 to make the entry at 3,1 a 0.
[100100010-1101-11-01-00-10-01-0]
Schritt 1.5.2.2
Vereinfache R3.
[100100010-110011-101]
[100100010-110011-101]
Schritt 1.5.3
Perform the row operation R3=R3-R2 to make the entry at 3,2 a 0.
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Schritt 1.5.3.1
Perform the row operation R3=R3-R2 to make the entry at 3,2 a 0.
[100100010-1100-01-11-0-1+10-11-0]
Schritt 1.5.3.2
Vereinfache R3.
[100100010-1100010-11]
[100100010-1100010-11]
[100100010-1100010-11]
Schritt 1.6
The right half of the reduced row echelon form is the inverse.
[100-1100-11]
[100-1100-11]
Schritt 2
Multiply both sides by the inverse of [100110111].
[100-1100-11][100110111]y=[100-1100-11][123321]
Schritt 3
Vereinfache die Gleichung.
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Schritt 3.1
Multipliziere [100-1100-11][100110111].
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Schritt 3.1.1
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 3×3.
Schritt 3.1.2
Multipliziere jede Zeile in der ersten Matrix mit jeder Spalte in der zweiten Matrix.
[11+01+0110+01+0110+00+01-11+11+01-0+11+01-0+10+0101-11+1100-11+1100-0+11]y=[100-1100-11][123321]
Schritt 3.1.3
Vereinfache jedes Element der Matrix durch Ausmultiplizieren aller Ausdrücke.
[100010001]y=[100-1100-11][123321]
[100010001]y=[100-1100-11][123321]
Schritt 3.2
Multiplying the identity matrix by any matrix A is the matrix A itself.
y=[100-1100-11][123321]
Schritt 3.3
Multipliziere [100-1100-11][123321].
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Schritt 3.3.1
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 3×2.
Schritt 3.3.2
Multipliziere jede Zeile in der ersten Matrix mit jeder Spalte in der zweiten Matrix.
y=[11+03+0212+03+01-11+13+02-12+13+0101-13+1202-13+11]
Schritt 3.3.3
Vereinfache jedes Element der Matrix durch Ausmultiplizieren aller Ausdrücke.
y=[1221-1-2]
y=[1221-1-2]
y=[1221-1-2]
 [x2  12  π  xdx ]