Lineare Algebra Beispiele

⎡ ⎢ ⎣ \begin{matrix} 1 & - 1 & 1 2 & 3 & 0 0 & 2 & - 1 \end{matrix} ⎤ ⎥ ⎦ X = ⎡ ⎢ ⎣ \begin{matrix} 6 & - 1 & 9 & 11 & 12 8 & 0 & - 7 & 0 & 1 7 & 9 & - 11 & 6 & 1 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & - 1 & 1 \\ 2 & 3 & 0 \\ 0 & 2 & - 1 \end{array}] X = [\begin{array}{c} 6 & - 1 & 9 & 11 & 12 \\ 8 & 0 & - 7 & 0 & 1 \\ 7 & 9 & - 11 & 6 & 1 \end{array}]$

Schritt 1

Find the inverse of

⎡ ⎢ ⎣ \begin{matrix} 1 & - 1 & 1 2 & 3 & 0 0 & 2 & - 1 \end{matrix} ⎤ ⎥ ⎦

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

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Schritt 1.1

Forme um.

∣ ∣ ∣ ∣ \begin{matrix} 1 & - 1 & 1 2 & 3 & 0 0 & 2 & - 1 \end{matrix} ∣ ∣ ∣ ∣

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

Schritt 1.2

Find the determinant.

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Schritt 1.2.1

Choose the row or column with the most

0

$0$ elements. If there are no

0

$0$ elements choose any row or column. Multiply every element in column

1

$1$ by its cofactor and add.

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Schritt 1.2.1.1

Consider the corresponding sign chart.

∣ ∣ ∣ ∣ \begin{matrix} + & - & + - & + & - + & - & + \end{matrix} ∣ ∣ ∣ ∣

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

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

a_{11}

$a_{11}$ is the determinant with row

1

$1$ and column

1

$1$ deleted.

∣ ∣ ∣ \begin{matrix} 3 & 0 2 & - 1 \end{matrix} ∣ ∣ ∣

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

Schritt 1.2.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

1 ∣ ∣ ∣ \begin{matrix} 3 & 0 2 & - 1 \end{matrix} ∣ ∣ ∣

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

Schritt 1.2.1.5

The minor for

a_{21}

$a_{21}$ is the determinant with row

2

$2$ and column

1

$1$ deleted.

∣ ∣ ∣ \begin{matrix} - 1 & 1 2 & - 1 \end{matrix} ∣ ∣ ∣

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

Schritt 1.2.1.6

Multiply element

a_{21}

$a_{21}$ by its cofactor.

- 2 ∣ ∣ ∣ \begin{matrix} - 1 & 1 2 & - 1 \end{matrix} ∣ ∣ ∣

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

Schritt 1.2.1.7

The minor for

a_{31}

$a_{31}$ is the determinant with row

3

$3$ and column

1

$1$ deleted.

∣ ∣ ∣ \begin{matrix} - 1 & 1 3 & 0 \end{matrix} ∣ ∣ ∣

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

Schritt 1.2.1.8

Multiply element

a_{31}

$a_{31}$ by its cofactor.

0 ∣ ∣ ∣ \begin{matrix} - 1 & 1 3 & 0 \end{matrix} ∣ ∣ ∣

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

Schritt 1.2.1.9

Add the terms together.

1 ∣ ∣ ∣ \begin{matrix} 3 & 0 2 & - 1 \end{matrix} ∣ ∣ ∣ - 2 ∣ ∣ ∣ \begin{matrix} - 1 & 1 2 & - 1 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} - 1 & 1 3 & 0 \end{matrix} ∣ ∣ ∣

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

1 ∣ ∣ ∣ \begin{matrix} 3 & 0 2 & - 1 \end{matrix} ∣ ∣ ∣ - 2 ∣ ∣ ∣ \begin{matrix} - 1 & 1 2 & - 1 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} - 1 & 1 3 & 0 \end{matrix} ∣ ∣ ∣

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

Schritt 1.2.2

Mutltipliziere

0

$0$ mit

∣ ∣ ∣ \begin{matrix} - 1 & 1 3 & 0 \end{matrix} ∣ ∣ ∣

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

1 ∣ ∣ ∣ \begin{matrix} 3 & 0 2 & - 1 \end{matrix} ∣ ∣ ∣ - 2 ∣ ∣ ∣ \begin{matrix} - 1 & 1 2 & - 1 \end{matrix} ∣ ∣ ∣ + 0

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

Schritt 1.2.3

Berechne

∣ ∣ ∣ \begin{matrix} 3 & 0 2 & - 1 \end{matrix} ∣ ∣ ∣

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

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Schritt 1.2.3.1

Die Determinante einer

2 \times 2

$2 \times 2$ -Matrix kann mithilfe der Formel

∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ bestimmt werden.

1 (3 \cdot - 1 - 2 \cdot 0) - 2 ∣ ∣ ∣ \begin{matrix} - 1 & 1 2 & - 1 \end{matrix} ∣ ∣ ∣ + 0

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

Schritt 1.2.3.2

Vereinfache die Determinante.

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Schritt 1.2.3.2.1

Vereinfache jeden Term.

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Schritt 1.2.3.2.1.1

Mutltipliziere

3

$3$ mit

- 1

$- 1$ .

1 (- 3 - 2 \cdot 0) - 2 ∣ ∣ ∣ \begin{matrix} - 1 & 1 2 & - 1 \end{matrix} ∣ ∣ ∣ + 0

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

Schritt 1.2.3.2.1.2

Mutltipliziere

- 2

$- 2$ mit

0

$0$ .

1 (- 3 + 0) - 2 ∣ ∣ ∣ \begin{matrix} - 1 & 1 2 & - 1 \end{matrix} ∣ ∣ ∣ + 0

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

1 (- 3 + 0) - 2 ∣ ∣ ∣ \begin{matrix} - 1 & 1 2 & - 1 \end{matrix} ∣ ∣ ∣ + 0

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

Schritt 1.2.3.2.2

Addiere

- 3

$- 3$ und

0

$0$ .

1 \cdot - 3 - 2 ∣ ∣ ∣ \begin{matrix} - 1 & 1 2 & - 1 \end{matrix} ∣ ∣ ∣ + 0

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

Schritt 1.2.4

Berechne

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

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Schritt 1.2.4.1

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.

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

Schritt 1.2.4.2

Vereinfache die Determinante.

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Schritt 1.2.4.2.1

Vereinfache jeden Term.

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Schritt 1.2.4.2.1.1

Mutltipliziere

$- 1$ mit

$- 1$ .

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

Schritt 1.2.4.2.1.2

Mutltipliziere

$- 2$ mit

$1$ .

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

Schritt 1.2.4.2.2

Subtrahiere

$2$ von

$1$ .

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

Schritt 1.2.5

Vereinfache die Determinante.

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Schritt 1.2.5.1

Vereinfache jeden Term.

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Schritt 1.2.5.1.1

Mutltipliziere

$- 3$ mit

$1$ .

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

Schritt 1.2.5.1.2

Mutltipliziere

$- 2$ mit

$- 1$ .

$- 3 + 2 + 0$

Schritt 1.2.5.2

Addiere

$- 3$ und

$2$ .

$- 1 + 0$

Schritt 1.2.5.3

Addiere

$- 1$ und

$0$ .

$- 1$

Schritt 1.3

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

Schritt 1.4

Set up a

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

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

Schritt 1.5

Ermittele die normierte Zeilenstufenform.

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Schritt 1.5.1

Perform the row operation

$R_{2} = R_{2} - 2 R_{1}$ to make the entry at

$2, 1$ a

$0$ .

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Schritt 1.5.1.1

Perform the row operation

$R_{2} = R_{2} - 2 R_{1}$ to make the entry at

$2, 1$ a

$0$ .

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

Schritt 1.5.1.2

Vereinfache

$R_{2}$ .

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

Schritt 1.5.2

Multiply each element of

$R_{2}$ by

$\frac{1}{5}$ to make the entry at

$2, 2$ a

$1$ .

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Schritt 1.5.2.1

Multiply each element of

$R_{2}$ by

$\frac{1}{5}$ to make the entry at

$2, 2$ a

$1$ .

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

Schritt 1.5.2.2

Vereinfache

$R_{2}$ .

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

Schritt 1.5.3

Perform the row operation

$R_{3} = R_{3} - 2 R_{2}$ to make the entry at

$3, 2$ a

$0$ .

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Schritt 1.5.3.1

Perform the row operation

$R_{3} = R_{3} - 2 R_{2}$ to make the entry at

$3, 2$ a

$0$ .

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

Schritt 1.5.3.2

Vereinfache

$R_{3}$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & - \frac{2}{5} & - \frac{2}{5} & \frac{1}{5} & 0 \\ 0 & 0 & - \frac{1}{5} & \frac{4}{5} & - \frac{2}{5} & 1 \end{array}]$

Schritt 1.5.4

Multiply each element of

$R_{3}$ by

$- 5$ to make the entry at

$3, 3$ a

$1$ .

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Schritt 1.5.4.1

Multiply each element of

$R_{3}$ by

$- 5$ to make the entry at

$3, 3$ a

$1$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & - \frac{2}{5} & - \frac{2}{5} & \frac{1}{5} & 0 \\ - 5 \cdot 0 & - 5 \cdot 0 & - 5 (- \frac{1}{5}) & - 5 (\frac{4}{5}) & - 5 (- \frac{2}{5}) & - 5 \cdot 1 \end{array}]$

Schritt 1.5.4.2

Vereinfache

$R_{3}$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & - \frac{2}{5} & - \frac{2}{5} & \frac{1}{5} & 0 \\ 0 & 0 & 1 & - 4 & 2 & - 5 \end{array}]$

Schritt 1.5.5

Perform the row operation

$R_{2} = R_{2} + \frac{2}{5} R_{3}$ to make the entry at

$2, 3$ a

$0$ .

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Schritt 1.5.5.1

Perform the row operation

$R_{2} = R_{2} + \frac{2}{5} R_{3}$ to make the entry at

$2, 3$ a

$0$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 \\ 0 + \frac{2}{5} \cdot 0 & 1 + \frac{2}{5} \cdot 0 & - \frac{2}{5} + \frac{2}{5} \cdot 1 & - \frac{2}{5} + \frac{2}{5} \cdot - 4 & \frac{1}{5} + \frac{2}{5} \cdot 2 & 0 + \frac{2}{5} \cdot - 5 \\ 0 & 0 & 1 & - 4 & 2 & - 5 \end{array}]$

Schritt 1.5.5.2

Vereinfache

$R_{2}$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & - 2 & 1 & - 2 \\ 0 & 0 & 1 & - 4 & 2 & - 5 \end{array}]$

Schritt 1.5.6

Perform the row operation

$R_{1} = R_{1} - R_{3}$ to make the entry at

$1, 3$ a

$0$ .

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Schritt 1.5.6.1

Perform the row operation

$R_{1} = R_{1} - R_{3}$ to make the entry at

$1, 3$ a

$0$ .

$[\begin{array}{c} 1 - 0 & - 1 - 0 & 1 - 1 & 1 + 4 & 0 - 2 & 0 + 5 \\ 0 & 1 & 0 & - 2 & 1 & - 2 \\ 0 & 0 & 1 & - 4 & 2 & - 5 \end{array}]$

Schritt 1.5.6.2

Vereinfache

$R_{1}$ .

$[\begin{array}{c} 1 & - 1 & 0 & 5 & - 2 & 5 \\ 0 & 1 & 0 & - 2 & 1 & - 2 \\ 0 & 0 & 1 & - 4 & 2 & - 5 \end{array}]$

Schritt 1.5.7

Perform the row operation

$R_{1} = R_{1} + R_{2}$ to make the entry at

$1, 2$ a

$0$ .

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Schritt 1.5.7.1

Perform the row operation

$R_{1} = R_{1} + R_{2}$ to make the entry at

$1, 2$ a

$0$ .

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 0 + 0 & 5 - 2 & - 2 + 1 \cdot 1 & 5 - 2 \\ 0 & 1 & 0 & - 2 & 1 & - 2 \\ 0 & 0 & 1 & - 4 & 2 & - 5 \end{array}]$

Schritt 1.5.7.2

Vereinfache

$R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 3 & - 1 & 3 \\ 0 & 1 & 0 & - 2 & 1 & - 2 \\ 0 & 0 & 1 & - 4 & 2 & - 5 \end{array}]$

Schritt 1.6

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

$[\begin{array}{c} 3 & - 1 & 3 \\ - 2 & 1 & - 2 \\ - 4 & 2 & - 5 \end{array}]$

Schritt 2

Multiply both sides by the inverse of

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

$[\begin{array}{c} 3 & - 1 & 3 \\ - 2 & 1 & - 2 \\ - 4 & 2 & - 5 \end{array}] [\begin{array}{c} 1 & - 1 & 1 \\ 2 & 3 & 0 \\ 0 & 2 & - 1 \end{array}] X = [\begin{array}{c} 3 & - 1 & 3 \\ - 2 & 1 & - 2 \\ - 4 & 2 & - 5 \end{array}] [\begin{array}{c} 6 & - 1 & 9 & 11 & 12 \\ 8 & 0 & - 7 & 0 & 1 \\ 7 & 9 & - 11 & 6 & 1 \end{array}]$

Schritt 3

Vereinfache die Gleichung.

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Schritt 3.1

Multipliziere

$[\begin{array}{c} 3 & - 1 & 3 \\ - 2 & 1 & - 2 \\ - 4 & 2 & - 5 \end{array}] [\begin{array}{c} 1 & - 1 & 1 \\ 2 & 3 & 0 \\ 0 & 2 & - 1 \end{array}]$ .

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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

$3 \times 3$ and the second matrix is

$3 \times 3$ .

Schritt 3.1.2

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

$[\begin{array}{c} 3 \cdot 1 - 1 \cdot 2 + 3 \cdot 0 & 3 \cdot - 1 - 1 \cdot 3 + 3 \cdot 2 & 3 \cdot 1 - 0 + 3 \cdot - 1 \\ - 2 \cdot 1 + 1 \cdot 2 - 2 \cdot 0 & - 2 \cdot - 1 + 1 \cdot 3 - 2 \cdot 2 & - 2 \cdot 1 + 1 \cdot 0 - 2 \cdot - 1 \\ - 4 \cdot 1 + 2 \cdot 2 - 5 \cdot 0 & - 4 \cdot - 1 + 2 \cdot 3 - 5 \cdot 2 & - 4 \cdot 1 + 2 \cdot 0 - 5 \cdot - 1 \end{array}] X = [\begin{array}{c} 3 & - 1 & 3 \\ - 2 & 1 & - 2 \\ - 4 & 2 & - 5 \end{array}] [\begin{array}{c} 6 & - 1 & 9 & 11 & 12 \\ 8 & 0 & - 7 & 0 & 1 \\ 7 & 9 & - 11 & 6 & 1 \end{array}]$

Schritt 3.1.3

Vereinfache jedes Element der Matrix durch Ausmultiplizieren aller Ausdrücke.

$[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] X = [\begin{array}{c} 3 & - 1 & 3 \\ - 2 & 1 & - 2 \\ - 4 & 2 & - 5 \end{array}] [\begin{array}{c} 6 & - 1 & 9 & 11 & 12 \\ 8 & 0 & - 7 & 0 & 1 \\ 7 & 9 & - 11 & 6 & 1 \end{array}]$

Schritt 3.2

Multiplying the identity matrix by any matrix

$A$ is the matrix

$A$ itself.

$X = [\begin{array}{c} 3 & - 1 & 3 \\ - 2 & 1 & - 2 \\ - 4 & 2 & - 5 \end{array}] [\begin{array}{c} 6 & - 1 & 9 & 11 & 12 \\ 8 & 0 & - 7 & 0 & 1 \\ 7 & 9 & - 11 & 6 & 1 \end{array}]$

Schritt 3.3

Multipliziere

$[\begin{array}{c} 3 & - 1 & 3 \\ - 2 & 1 & - 2 \\ - 4 & 2 & - 5 \end{array}] [\begin{array}{c} 6 & - 1 & 9 & 11 & 12 \\ 8 & 0 & - 7 & 0 & 1 \\ 7 & 9 & - 11 & 6 & 1 \end{array}]$ .

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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

$3 \times 3$ and the second matrix is

$3 \times 5$ .

Schritt 3.3.2

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

$X = [\begin{array}{c} 3 \cdot 6 - 1 \cdot 8 + 3 \cdot 7 & 3 \cdot - 1 - 0 + 3 \cdot 9 & 3 \cdot 9 - - 7 + 3 \cdot - 11 & 3 \cdot 11 - 0 + 3 \cdot 6 & 3 \cdot 12 - 1 \cdot 1 + 3 \cdot 1 \\ - 2 \cdot 6 + 1 \cdot 8 - 2 \cdot 7 & - 2 \cdot - 1 + 1 \cdot 0 - 2 \cdot 9 & - 2 \cdot 9 + 1 \cdot - 7 - 2 \cdot - 11 & - 2 \cdot 11 + 1 \cdot 0 - 2 \cdot 6 & - 2 \cdot 12 + 1 \cdot 1 - 2 \cdot 1 \\ - 4 \cdot 6 + 2 \cdot 8 - 5 \cdot 7 & - 4 \cdot - 1 + 2 \cdot 0 - 5 \cdot 9 & - 4 \cdot 9 + 2 \cdot - 7 - 5 \cdot - 11 & - 4 \cdot 11 + 2 \cdot 0 - 5 \cdot 6 & - 4 \cdot 12 + 2 \cdot 1 - 5 \cdot 1 \end{array}]$

Schritt 3.3.3

Vereinfache jedes Element der Matrix durch Ausmultiplizieren aller Ausdrücke.

$X = [\begin{array}{c} 31 & 24 & 1 & 51 & 38 \\ - 18 & - 16 & - 3 & - 34 & - 25 \\ - 43 & - 41 & 5 & - 74 & - 51 \end{array}]$