Finite Mathematik Beispiele

$[\begin{array}{c} 0.8 & 0 & 0.4 \\ 0 & 0.9 & - 0.2 \\ - 0.4 & - 0.2 & 0.8 \end{array}]$

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 row $1$ 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.

$| \begin{array}{c} 0.9 & - 0.2 \\ - 0.2 & 0.8 \end{array} |$

Schritt 1.1.4

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

$0.8 | \begin{array}{c} 0.9 & - 0.2 \\ - 0.2 & 0.8 \end{array} |$

Schritt 1.1.5

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

$| \begin{array}{c} 0 & - 0.2 \\ - 0.4 & 0.8 \end{array} |$

Schritt 1.1.6

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

$0 | \begin{array}{c} 0 & - 0.2 \\ - 0.4 & 0.8 \end{array} |$

Schritt 1.1.7

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

$| \begin{array}{c} 0 & 0.9 \\ - 0.4 & - 0.2 \end{array} |$

Schritt 1.1.8

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

$0.4 | \begin{array}{c} 0 & 0.9 \\ - 0.4 & - 0.2 \end{array} |$

Schritt 1.1.9

Add the terms together.

$0.8 | \begin{array}{c} 0.9 & - 0.2 \\ - 0.2 & 0.8 \end{array} | + 0 | \begin{array}{c} 0 & - 0.2 \\ - 0.4 & 0.8 \end{array} | + 0.4 | \begin{array}{c} 0 & 0.9 \\ - 0.4 & - 0.2 \end{array} |$

Schritt 1.2

Mutltipliziere $0$ mit $| \begin{array}{c} 0 & - 0.2 \\ - 0.4 & 0.8 \end{array} |$ .

$0.8 | \begin{array}{c} 0.9 & - 0.2 \\ - 0.2 & 0.8 \end{array} | + 0 + 0.4 | \begin{array}{c} 0 & 0.9 \\ - 0.4 & - 0.2 \end{array} |$

Schritt 1.3

Berechne $| \begin{array}{c} 0.9 & - 0.2 \\ - 0.2 & 0.8 \end{array} |$ .

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Schritt 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.

$0.8 (0.9 \cdot 0.8 - (- 0.2 \cdot - 0.2)) + 0 + 0.4 | \begin{array}{c} 0 & 0.9 \\ - 0.4 & - 0.2 \end{array} |$

Schritt 1.3.2

Vereinfache die Determinante.

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Schritt 1.3.2.1

Vereinfache jeden Term.

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Schritt 1.3.2.1.1

Mutltipliziere $0.9$ mit $0.8$ .

$0.8 (0.72 - (- 0.2 \cdot - 0.2)) + 0 + 0.4 | \begin{array}{c} 0 & 0.9 \\ - 0.4 & - 0.2 \end{array} |$

Schritt 1.3.2.1.2

Multipliziere $- (- 0.2 \cdot - 0.2)$ .

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Schritt 1.3.2.1.2.1

Mutltipliziere $- 0.2$ mit $- 0.2$ .

$0.8 (0.72 - 1 \cdot 0.04) + 0 + 0.4 | \begin{array}{c} 0 & 0.9 \\ - 0.4 & - 0.2 \end{array} |$

Schritt 1.3.2.1.2.2

Mutltipliziere $- 1$ mit $0.04$ .

$0.8 (0.72 - 0.04) + 0 + 0.4 | \begin{array}{c} 0 & 0.9 \\ - 0.4 & - 0.2 \end{array} |$

Schritt 1.3.2.2

Subtrahiere $0.04$ von $0.72$ .

$0.8 \cdot 0.68 + 0 + 0.4 | \begin{array}{c} 0 & 0.9 \\ - 0.4 & - 0.2 \end{array} |$

Schritt 1.4

Berechne $| \begin{array}{c} 0 & 0.9 \\ - 0.4 & - 0.2 \end{array} |$ .

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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.

$0.8 \cdot 0.68 + 0 + 0.4 (0 \cdot - 0.2 - (- 0.4 \cdot 0.9))$

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 $0$ mit $- 0.2$ .

$0.8 \cdot 0.68 + 0 + 0.4 (0 - (- 0.4 \cdot 0.9))$

Schritt 1.4.2.1.2

Multipliziere $- (- 0.4 \cdot 0.9)$ .

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Schritt 1.4.2.1.2.1

Mutltipliziere $- 0.4$ mit $0.9$ .

$0.8 \cdot 0.68 + 0 + 0.4 (0 - - 0.36)$

Schritt 1.4.2.1.2.2

Mutltipliziere $- 1$ mit $- 0.36$ .

$0.8 \cdot 0.68 + 0 + 0.4 (0 + 0.36)$

Schritt 1.4.2.2

Addiere $0$ und $0.36$ .

$0.8 \cdot 0.68 + 0 + 0.4 \cdot 0.36$

Schritt 1.5

Vereinfache die Determinante.

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Schritt 1.5.1

Vereinfache jeden Term.

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Schritt 1.5.1.1

Mutltipliziere $0.8$ mit $0.68$ .

$0.544 + 0 + 0.4 \cdot 0.36$

Schritt 1.5.1.2

Mutltipliziere $0.4$ mit $0.36$ .

$0.544 + 0 + 0.144$

Schritt 1.5.2

Addiere $0.544$ und $0$ .

$0.544 + 0.144$

Schritt 1.5.3

Addiere $0.544$ und $0.144$ .

$0.688$

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.

$[\begin{array}{c} 0.8 & 0 & 0.4 & 1 & 0 & 0 \\ 0 & 0.9 & - 0.2 & 0 & 1 & 0 \\ - 0.4 & - 0.2 & 0.8 & 0 & 0 & 1 \end{array}]$

Schritt 4

Ermittele die normierte Zeilenstufenform.

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Schritt 4.1

Multiply each element of $R_{1}$ by $\frac{1}{0.8}$ to make the entry at $1, 1$ a $1$ .

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Schritt 4.1.1

Multiply each element of $R_{1}$ by $\frac{1}{0.8}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{0.8}{0.8} & \frac{0}{0.8} & \frac{0.4}{0.8} & \frac{1}{0.8} & \frac{0}{0.8} & \frac{0}{0.8} \\ 0 & 0.9 & - 0.2 & 0 & 1 & 0 \\ - 0.4 & - 0.2 & 0.8 & 0 & 0 & 1 \end{array}]$

Schritt 4.1.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0.5 & 1.25 & 0 & 0 \\ 0 & 0.9 & - 0.2 & 0 & 1 & 0 \\ - 0.4 & - 0.2 & 0.8 & 0 & 0 & 1 \end{array}]$

Schritt 4.2

Perform the row operation $R_{3} = R_{3} + 0.4 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Schritt 4.2.1

Perform the row operation $R_{3} = R_{3} + 0.4 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 0 & 0.5 & 1.25 & 0 & 0 \\ 0 & 0.9 & - 0.2 & 0 & 1 & 0 \\ - 0.4 + 0.4 \cdot 1 & - 0.2 + 0.4 \cdot 0 & 0.8 + 0.4 \cdot 0.5 & 0 + 0.4 \cdot 1.25 & 0 + 0.4 \cdot 0 & 1 + 0.4 \cdot 0 \end{array}]$

Schritt 4.2.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & 0 & 0.5 & 1.25 & 0 & 0 \\ 0 & 0.9 & - 0.2 & 0 & 1 & 0 \\ 0 & - 0.2 & 1 & 0.5 & 0 & 1 \end{array}]$

Schritt 4.3

Multiply each element of $R_{2}$ by $\frac{1}{0.9}$ to make the entry at $2, 2$ a $1$ .

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Schritt 4.3.1

Multiply each element of $R_{2}$ by $\frac{1}{0.9}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 0 & 0.5 & 1.25 & 0 & 0 \\ \frac{0}{0.9} & \frac{0.9}{0.9} & \frac{- 0.2}{0.9} & \frac{0}{0.9} & \frac{1}{0.9} & \frac{0}{0.9} \\ 0 & - 0.2 & 1 & 0.5 & 0 & 1 \end{array}]$

Schritt 4.3.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 0.5 & 1.25 & 0 & 0 \\ 0 & 1 & - 0.\overline{2} & 0 & 1.\overline{1} & 0 \\ 0 & - 0.2 & 1 & 0.5 & 0 & 1 \end{array}]$

Schritt 4.4

Perform the row operation $R_{3} = R_{3} + 0.2 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Schritt 4.4.1

Perform the row operation $R_{3} = R_{3} + 0.2 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 0 & 0.5 & 1.25 & 0 & 0 \\ 0 & 1 & - 0.\overline{2} & 0 & 1.\overline{1} & 0 \\ 0 + 0.2 \cdot 0 & - 0.2 + 0.2 \cdot 1 & 1 + 0.2 \cdot - 0.\overline{2} & 0.5 + 0.2 \cdot 0 & 0 + 0.2 \cdot 1.\overline{1} & 1 + 0.2 \cdot 0 \end{array}]$

Schritt 4.4.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & 0 & 0.5 & 1.25 & 0 & 0 \\ 0 & 1 & - 0.\overline{2} & 0 & 1.\overline{1} & 0 \\ 0 & 0 & 0.9\overline{5} & 0.5 & 0.\overline{2} & 1 \end{array}]$

Schritt 4.5

Multiply each element of $R_{3}$ by $\frac{1}{0.9\overline{5}}$ to make the entry at $3, 3$ a $1$ .

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Schritt 4.5.1

Multiply each element of $R_{3}$ by $\frac{1}{0.9\overline{5}}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 0 & 0.5 & 1.25 & 0 & 0 \\ 0 & 1 & - 0.\overline{2} & 0 & 1.\overline{1} & 0 \\ \frac{0}{0.9\overline{5}} & \frac{0}{0.9\overline{5}} & \frac{0.9\overline{5}}{0.9\overline{5}} & \frac{0.5}{0.9\overline{5}} & \frac{0.\overline{2}}{0.9\overline{5}} & \frac{1}{0.9\overline{5}} \end{array}]$

Schritt 4.5.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & 0 & 0.5 & 1.25 & 0 & 0 \\ 0 & 1 & - 0.\overline{2} & 0 & 1.\overline{1} & 0 \\ 0 & 0 & 1 & 0.52325581 & 0.23255813 & 1.04651162 \end{array}]$

Schritt 4.6

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

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Schritt 4.6.1

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

$[\begin{array}{c} 1 & 0 & 0.5 & 1.25 & 0 & 0 \\ 0 + 0.\overline{2} \cdot 0 & 1 + 0.\overline{2} \cdot 0 & - 0.\overline{2} + 0.\overline{2} \cdot 1 & 0 + 0.\overline{2} \cdot 0.52325581 & 1.\overline{1} + 0.\overline{2} \cdot 0.23255813 & 0 + 0.\overline{2} \cdot 1.04651162 \\ 0 & 0 & 1 & 0.52325581 & 0.23255813 & 1.04651162 \end{array}]$

Schritt 4.6.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 0.5 & 1.25 & 0 & 0 \\ 0 & 1 & 0 & 0.11627906 & 1.16279069 & 0.23255813 \\ 0 & 0 & 1 & 0.52325581 & 0.23255813 & 1.04651162 \end{array}]$

Schritt 4.7

Perform the row operation $R_{1} = R_{1} - 0.5 R_{3}$ to make the entry at $1, 3$ a $0$ .

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Schritt 4.7.1

Perform the row operation $R_{1} = R_{1} - 0.5 R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 0.5 \cdot 0 & 0 - 0.5 \cdot 0 & 0.5 - 0.5 \cdot 1 & 1.25 - 0.5 \cdot 0.52325581 & 0 - 0.5 \cdot 0.23255813 & 0 - 0.5 \cdot 1.04651162 \\ 0 & 1 & 0 & 0.11627906 & 1.16279069 & 0.23255813 \\ 0 & 0 & 1 & 0.52325581 & 0.23255813 & 1.04651162 \end{array}]$

Schritt 4.7.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0.98837209 & - 0.11627906 & - 0.52325581 \\ 0 & 1 & 0 & 0.11627906 & 1.16279069 & 0.23255813 \\ 0 & 0 & 1 & 0.52325581 & 0.23255813 & 1.04651162 \end{array}]$

Schritt 5

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

$[\begin{array}{c} 0.98837209 & - 0.11627906 & - 0.52325581 \\ 0.11627906 & 1.16279069 & 0.23255813 \\ 0.52325581 & 0.23255813 & 1.04651162 \end{array}]$