Lineare Algebra Beispiele

$2 x - y = 4$ , $3 x + y - z = 10$ , $y + z = 3$

Schritt 1

Ermittle $A X = B$ aus dem Gleichungssystem.

$[\begin{array}{c} 2 & - 1 & 0 \\ 3 & 1 & - 1 \\ 0 & 1 & 1 \end{array}] \cdot [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 4 \\ 10 \\ 3 \end{array}]$

Schritt 2

Finde die Inverse der Koeffizientenmatrix.

Tippen, um mehr Schritte zu sehen ...

Schritt 2.1

Find the determinant.

Tippen, um mehr Schritte zu sehen ...

Schritt 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.

Tippen, um mehr Schritte zu sehen ...

Schritt 2.1.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Schritt 2.1.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Schritt 2.1.1.3

The minor for $a_{11}$ is the determinant with row $1$ and column $1$ deleted.

$| \begin{array}{c} 1 & - 1 \\ 1 & 1 \end{array} |$

Schritt 2.1.1.4

Multiply element $a_{11}$ by its cofactor.

$2 | \begin{array}{c} 1 & - 1 \\ 1 & 1 \end{array} |$

Schritt 2.1.1.5

The minor for $a_{12}$ is the determinant with row $1$ and column $2$ deleted.

$| \begin{array}{c} 3 & - 1 \\ 0 & 1 \end{array} |$

Schritt 2.1.1.6

Multiply element $a_{12}$ by its cofactor.

$1 | \begin{array}{c} 3 & - 1 \\ 0 & 1 \end{array} |$

Schritt 2.1.1.7

The minor for $a_{13}$ is the determinant with row $1$ and column $3$ deleted.

$| \begin{array}{c} 3 & 1 \\ 0 & 1 \end{array} |$

Schritt 2.1.1.8

Multiply element $a_{13}$ by its cofactor.

$0 | \begin{array}{c} 3 & 1 \\ 0 & 1 \end{array} |$

Schritt 2.1.1.9

Add the terms together.

$2 | \begin{array}{c} 1 & - 1 \\ 1 & 1 \end{array} | + 1 | \begin{array}{c} 3 & - 1 \\ 0 & 1 \end{array} | + 0 | \begin{array}{c} 3 & 1 \\ 0 & 1 \end{array} |$

Schritt 2.1.2

Mutltipliziere $0$ mit $| \begin{array}{c} 3 & 1 \\ 0 & 1 \end{array} |$ .

$2 | \begin{array}{c} 1 & - 1 \\ 1 & 1 \end{array} | + 1 | \begin{array}{c} 3 & - 1 \\ 0 & 1 \end{array} | + 0$

Schritt 2.1.3

Berechne $| \begin{array}{c} 1 & - 1 \\ 1 & 1 \end{array} |$ .

Tippen, um mehr Schritte zu sehen ...

Schritt 2.1.3.1

Die Determinante einer $2 \times 2$ -Matrix kann mithilfe der Formel $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ bestimmt werden.

$2 (1 \cdot 1 - 1 \cdot - 1) + 1 | \begin{array}{c} 3 & - 1 \\ 0 & 1 \end{array} | + 0$

Schritt 2.1.3.2

Vereinfache die Determinante.

Tippen, um mehr Schritte zu sehen ...

Schritt 2.1.3.2.1

Vereinfache jeden Term.

Tippen, um mehr Schritte zu sehen ...

Schritt 2.1.3.2.1.1

Mutltipliziere $1$ mit $1$ .

$2 (1 - 1 \cdot - 1) + 1 | \begin{array}{c} 3 & - 1 \\ 0 & 1 \end{array} | + 0$

Schritt 2.1.3.2.1.2

Mutltipliziere $- 1$ mit $- 1$ .

$2 (1 + 1) + 1 | \begin{array}{c} 3 & - 1 \\ 0 & 1 \end{array} | + 0$

Schritt 2.1.3.2.2

Addiere $1$ und $1$ .

$2 \cdot 2 + 1 | \begin{array}{c} 3 & - 1 \\ 0 & 1 \end{array} | + 0$

Schritt 2.1.4

Berechne $| \begin{array}{c} 3 & - 1 \\ 0 & 1 \end{array} |$ .

Tippen, um mehr Schritte zu sehen ...

Schritt 2.1.4.1

Die Determinante einer $2 \times 2$ -Matrix kann mithilfe der Formel $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ bestimmt werden.

$2 \cdot 2 + 1 (3 \cdot 1 + 0 \cdot - 1) + 0$

Schritt 2.1.4.2

Vereinfache die Determinante.

Tippen, um mehr Schritte zu sehen ...

Schritt 2.1.4.2.1

Vereinfache jeden Term.

Tippen, um mehr Schritte zu sehen ...

Schritt 2.1.4.2.1.1

Mutltipliziere $3$ mit $1$ .

$2 \cdot 2 + 1 (3 + 0 \cdot - 1) + 0$

Schritt 2.1.4.2.1.2

Mutltipliziere $0$ mit $- 1$ .

$2 \cdot 2 + 1 (3 + 0) + 0$

Schritt 2.1.4.2.2

Addiere $3$ und $0$ .

$2 \cdot 2 + 1 \cdot 3 + 0$

Schritt 2.1.5

Vereinfache die Determinante.

Tippen, um mehr Schritte zu sehen ...

Schritt 2.1.5.1

Vereinfache jeden Term.

Tippen, um mehr Schritte zu sehen ...

Schritt 2.1.5.1.1

Mutltipliziere $2$ mit $2$ .

$4 + 1 \cdot 3 + 0$

Schritt 2.1.5.1.2

Mutltipliziere $3$ mit $1$ .

$4 + 3 + 0$

Schritt 2.1.5.2

Addiere $4$ und $3$ .

$7 + 0$

Schritt 2.1.5.3

Addiere $7$ und $0$ .

$7$

Schritt 2.2

Since the determinant is non-zero, the inverse exists.

Schritt 2.3

Set up a $3 \times 6$ matrix where the left half is the original matrix and the right half is its identity matrix.

$[\begin{array}{c} 2 & - 1 & 0 & 1 & 0 & 0 \\ 3 & 1 & - 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 & 0 & 1 \end{array}]$

Schritt 2.4

Ermittele die normierte Zeilenstufenform.

Tippen, um mehr Schritte zu sehen ...

Schritt 2.4.1

Multiply each element of $R_{1}$ by $\frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

Tippen, um mehr Schritte zu sehen ...

Schritt 2.4.1.1

Multiply each element of $R_{1}$ by $\frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{2}{2} & \frac{- 1}{2} & \frac{0}{2} & \frac{1}{2} & \frac{0}{2} & \frac{0}{2} \\ 3 & 1 & - 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 & 0 & 1 \end{array}]$

Schritt 2.4.1.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 3 & 1 & - 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 & 0 & 1 \end{array}]$

Schritt 2.4.2

Perform the row operation $R_{2} = R_{2} - 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

Tippen, um mehr Schritte zu sehen ...

Schritt 2.4.2.1

Perform the row operation $R_{2} = R_{2} - 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 3 - 3 \cdot 1 & 1 - 3 (- \frac{1}{2}) & - 1 - 3 \cdot 0 & 0 - 3 (\frac{1}{2}) & 1 - 3 \cdot 0 & 0 - 3 \cdot 0 \\ 0 & 1 & 1 & 0 & 0 & 1 \end{array}]$

Schritt 2.4.2.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & \frac{5}{2} & - 1 & - \frac{3}{2} & 1 & 0 \\ 0 & 1 & 1 & 0 & 0 & 1 \end{array}]$

Schritt 2.4.3

Multiply each element of $R_{2}$ by $\frac{2}{5}$ to make the entry at $2, 2$ a $1$ .

Tippen, um mehr Schritte zu sehen ...

Schritt 2.4.3.1

Multiply each element of $R_{2}$ by $\frac{2}{5}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ \frac{2}{5} \cdot 0 & \frac{2}{5} \cdot \frac{5}{2} & \frac{2}{5} \cdot - 1 & \frac{2}{5} (- \frac{3}{2}) & \frac{2}{5} \cdot 1 & \frac{2}{5} \cdot 0 \\ 0 & 1 & 1 & 0 & 0 & 1 \end{array}]$

Schritt 2.4.3.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & - \frac{2}{5} & - \frac{3}{5} & \frac{2}{5} & 0 \\ 0 & 1 & 1 & 0 & 0 & 1 \end{array}]$

Schritt 2.4.4

Perform the row operation $R_{3} = R_{3} - R_{2}$ to make the entry at $3, 2$ a $0$ .

Tippen, um mehr Schritte zu sehen ...

Schritt 2.4.4.1

Perform the row operation $R_{3} = R_{3} - R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & - \frac{2}{5} & - \frac{3}{5} & \frac{2}{5} & 0 \\ 0 - 0 & 1 - 1 & 1 + \frac{2}{5} & 0 + \frac{3}{5} & 0 - \frac{2}{5} & 1 - 0 \end{array}]$

Schritt 2.4.4.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & - \frac{2}{5} & - \frac{3}{5} & \frac{2}{5} & 0 \\ 0 & 0 & \frac{7}{5} & \frac{3}{5} & - \frac{2}{5} & 1 \end{array}]$

Schritt 2.4.5

Multiply each element of $R_{3}$ by $\frac{5}{7}$ to make the entry at $3, 3$ a $1$ .

Tippen, um mehr Schritte zu sehen ...

Schritt 2.4.5.1

Multiply each element of $R_{3}$ by $\frac{5}{7}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & - \frac{2}{5} & - \frac{3}{5} & \frac{2}{5} & 0 \\ \frac{5}{7} \cdot 0 & \frac{5}{7} \cdot 0 & \frac{5}{7} \cdot \frac{7}{5} & \frac{5}{7} \cdot \frac{3}{5} & \frac{5}{7} (- \frac{2}{5}) & \frac{5}{7} \cdot 1 \end{array}]$

Schritt 2.4.5.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & - \frac{2}{5} & - \frac{3}{5} & \frac{2}{5} & 0 \\ 0 & 0 & 1 & \frac{3}{7} & - \frac{2}{7} & \frac{5}{7} \end{array}]$

Schritt 2.4.6

Perform the row operation $R_{2} = R_{2} + \frac{2}{5} R_{3}$ to make the entry at $2, 3$ a $0$ .

Tippen, um mehr Schritte zu sehen ...

Schritt 2.4.6.1

Perform the row operation $R_{2} = R_{2} + \frac{2}{5} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 + \frac{2}{5} \cdot 0 & 1 + \frac{2}{5} \cdot 0 & - \frac{2}{5} + \frac{2}{5} \cdot 1 & - \frac{3}{5} + \frac{2}{5} \cdot \frac{3}{7} & \frac{2}{5} + \frac{2}{5} (- \frac{2}{7}) & 0 + \frac{2}{5} \cdot \frac{5}{7} \\ 0 & 0 & 1 & \frac{3}{7} & - \frac{2}{7} & \frac{5}{7} \end{array}]$

Schritt 2.4.6.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 & - \frac{3}{7} & \frac{2}{7} & \frac{2}{7} \\ 0 & 0 & 1 & \frac{3}{7} & - \frac{2}{7} & \frac{5}{7} \end{array}]$

Schritt 2.4.7

Perform the row operation $R_{1} = R_{1} + \frac{1}{2} R_{2}$ to make the entry at $1, 2$ a $0$ .

Tippen, um mehr Schritte zu sehen ...

Schritt 2.4.7.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{2} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & 0 + \frac{1}{2} \cdot 0 & \frac{1}{2} + \frac{1}{2} (- \frac{3}{7}) & 0 + \frac{1}{2} \cdot \frac{2}{7} & 0 + \frac{1}{2} \cdot \frac{2}{7} \\ 0 & 1 & 0 & - \frac{3}{7} & \frac{2}{7} & \frac{2}{7} \\ 0 & 0 & 1 & \frac{3}{7} & - \frac{2}{7} & \frac{5}{7} \end{array}]$

Schritt 2.4.7.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{2}{7} & \frac{1}{7} & \frac{1}{7} \\ 0 & 1 & 0 & - \frac{3}{7} & \frac{2}{7} & \frac{2}{7} \\ 0 & 0 & 1 & \frac{3}{7} & - \frac{2}{7} & \frac{5}{7} \end{array}]$

Schritt 2.5

The right half of the reduced row echelon form is the inverse.

$[\begin{array}{c} \frac{2}{7} & \frac{1}{7} & \frac{1}{7} \\ - \frac{3}{7} & \frac{2}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{2}{7} & \frac{5}{7} \end{array}]$

Schritt 3

Multipliziere beide Seiten der Matrizengleichung von links mit der inversen Matrix.

$([\begin{array}{c} \frac{2}{7} & \frac{1}{7} & \frac{1}{7} \\ - \frac{3}{7} & \frac{2}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{2}{7} & \frac{5}{7} \end{array}] \cdot [\begin{array}{c} 2 & - 1 & 0 \\ 3 & 1 & - 1 \\ 0 & 1 & 1 \end{array}]) \cdot [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{2}{7} & \frac{1}{7} & \frac{1}{7} \\ - \frac{3}{7} & \frac{2}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{2}{7} & \frac{5}{7} \end{array}] \cdot [\begin{array}{c} 4 \\ 10 \\ 3 \end{array}]$

Schritt 4

Jede Matrix multipliziert mit ihrer Inversen ist immer gleich $1$ . $A \cdot A^{- 1} = 1$ .

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{2}{7} & \frac{1}{7} & \frac{1}{7} \\ - \frac{3}{7} & \frac{2}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{2}{7} & \frac{5}{7} \end{array}] \cdot [\begin{array}{c} 4 \\ 10 \\ 3 \end{array}]$

Schritt 5

Multipliziere $[\begin{array}{c} \frac{2}{7} & \frac{1}{7} & \frac{1}{7} \\ - \frac{3}{7} & \frac{2}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{2}{7} & \frac{5}{7} \end{array}] [\begin{array}{c} 4 \\ 10 \\ 3 \end{array}]$ .

Tippen, um mehr Schritte zu sehen ...

Schritt 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 $3 \times 3$ and the second matrix is $3 \times 1$ .

Schritt 5.2

Multipliziere jede Zeile in der ersten Matrix mit jeder Spalte in der zweiten Matrix.

$[\begin{array}{c} \frac{2}{7} \cdot 4 + \frac{1}{7} \cdot 10 + \frac{1}{7} \cdot 3 \\ - \frac{3}{7} \cdot 4 + \frac{2}{7} \cdot 10 + \frac{2}{7} \cdot 3 \\ \frac{3}{7} \cdot 4 - \frac{2}{7} \cdot 10 + \frac{5}{7} \cdot 3 \end{array}]$

Schritt 5.3

Vereinfache jedes Element der Matrix durch Ausmultiplizieren aller Ausdrücke.

$[\begin{array}{c} 3 \\ 2 \\ 1 \end{array}]$

Schritt 6

Vereinfache die linke und rechte Seite.

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 3 \\ 2 \\ 1 \end{array}]$

Schritt 7

Ermittle die Lösung.

$x = 3$

$y = 2$

$z = 1$