Lineare Algebra Beispiele

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

Schritt 1

Find the determinant.

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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 column $1$ by its cofactor and add.

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Schritt 1.1.1

Consider the corresponding sign chart.

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

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

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

Schritt 1.1.4

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

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

Schritt 1.1.5

The minor for $a_{21}$ is the determinant with row $2$ and column $1$ deleted.

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

Schritt 1.1.6

Multiply element $a_{21}$ by its cofactor.

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

Schritt 1.1.7

The minor for $a_{31}$ is the determinant with row $3$ and column $1$ deleted.

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

Schritt 1.1.8

Multiply element $a_{31}$ by its cofactor.

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

Schritt 1.1.9

Add the terms together.

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

Schritt 1.2

Mutltipliziere $0$ mit $| \begin{array}{c} - 2 & 3 \\ 0 & 2 \end{array} |$ .

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

Schritt 1.3

Mutltipliziere $0$ mit $| \begin{array}{c} - 2 & 3 \\ 1 & 2 \end{array} |$ .

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

Schritt 1.4

Berechne $| \begin{array}{c} 1 & 2 \\ 0 & 2 \end{array} |$ .

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Schritt 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.

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

Schritt 1.4.2

Vereinfache die Determinante.

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Schritt 1.4.2.1

Vereinfache jeden Term.

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Schritt 1.4.2.1.1

Mutltipliziere $2$ mit $1$ .

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

Schritt 1.4.2.1.2

Mutltipliziere $0$ mit $2$ .

$1 (2 + 0) + 0 + 0$

Schritt 1.4.2.2

Addiere $2$ und $0$ .

$1 \cdot 2 + 0 + 0$

Schritt 1.5

Vereinfache die Determinante.

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Schritt 1.5.1

Mutltipliziere $2$ mit $1$ .

$2 + 0 + 0$

Schritt 1.5.2

Addiere $2$ und $0$ .

$2 + 0$

Schritt 1.5.3

Addiere $2$ und $0$ .

$2$

Schritt 2

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

Schritt 3

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

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

Schritt 4

Ermittele die normierte Zeilenstufenform.

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Schritt 4.1

Multiply each element of $R_{3}$ by $\frac{1}{2}$ to make the entry at $3, 3$ a $1$ .

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Schritt 4.1.1

Multiply each element of $R_{3}$ by $\frac{1}{2}$ to make the entry at $3, 3$ a $1$ .

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

Schritt 4.1.2

Vereinfache $R_{3}$ .

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

Schritt 4.2

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

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Schritt 4.2.1

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

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

Schritt 4.2.2

Vereinfache $R_{2}$ .

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

Schritt 4.3

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

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Schritt 4.3.1

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

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

Schritt 4.3.2

Vereinfache $R_{1}$ .

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

Schritt 4.4

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

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Schritt 4.4.1

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

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

Schritt 4.4.2

Vereinfache $R_{1}$ .

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

Schritt 5

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

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