Lineare Algebra Beispiele

$[\begin{array}{c} \frac{d}{a d - b c} & - \frac{b}{a d - b c} \\ - \frac{c}{a d - b c} & \frac{a}{a d - b d} \end{array}] [\begin{array}{c} - 6 & k \\ 0 & - 6 \end{array}] [\begin{array}{c} a & b \\ c & d \end{array}] = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Schritt 1

Faktorisiere $d$ aus $a d - b d$ heraus.

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Schritt 1.1

Faktorisiere $d$ aus $a d$ heraus.

$[\begin{array}{c} \frac{d}{a d - b c} & - \frac{b}{a d - b c} \\ - \frac{c}{a d - b c} & \frac{a}{d a - b d} \end{array}] [\begin{array}{c} - 6 & k \\ 0 & - 6 \end{array}] [\begin{array}{c} a & b \\ c & d \end{array}] = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Schritt 1.2

Faktorisiere $d$ aus $- b d$ heraus.

$[\begin{array}{c} \frac{d}{a d - b c} & - \frac{b}{a d - b c} \\ - \frac{c}{a d - b c} & \frac{a}{d a + d (- b)} \end{array}] [\begin{array}{c} - 6 & k \\ 0 & - 6 \end{array}] [\begin{array}{c} a & b \\ c & d \end{array}] = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Schritt 1.3

Faktorisiere $d$ aus $d a + d (- b)$ heraus.

$[\begin{array}{c} \frac{d}{a d - b c} & - \frac{b}{a d - b c} \\ - \frac{c}{a d - b c} & \frac{a}{d (a - b)} \end{array}] [\begin{array}{c} - 6 & k \\ 0 & - 6 \end{array}] [\begin{array}{c} a & b \\ c & d \end{array}] = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Schritt 2

Multipliziere $[\begin{array}{c} \frac{d}{a d - b c} & - \frac{b}{a d - b c} \\ - \frac{c}{a d - b c} & \frac{a}{d (a - b)} \end{array}] [\begin{array}{c} - 6 & k \\ 0 & - 6 \end{array}]$ .

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Schritt 2.1

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 $2 \times 2$ and the second matrix is $2 \times 2$ .

Schritt 2.2

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

$[\begin{array}{c} \frac{d}{a d - b c} \cdot - 6 - \frac{b}{a d - b c} \cdot 0 & \frac{d}{a d - b c} k - \frac{b}{a d - b c} \cdot - 6 \\ - \frac{c}{a d - b c} \cdot - 6 + \frac{a}{d (a - b)} \cdot 0 & - \frac{c}{a d - b c} k + \frac{a}{d (a - b)} \cdot - 6 \end{array}] [\begin{array}{c} a & b \\ c & d \end{array}] = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Schritt 2.3

Vereinfache jedes Element der Matrix durch Ausmultiplizieren aller Ausdrücke.

$[\begin{array}{c} - \frac{6 d}{a d - b c} & \frac{d k + 6 b}{a d - b c} \\ \frac{6 c}{a d - b c} & - \frac{k c d a - k c d b + 6 a^{2} d - 6 a b c}{d (a - b) (a d - b c)} \end{array}] [\begin{array}{c} a & b \\ c & d \end{array}] = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Schritt 3

Multipliziere $[\begin{array}{c} - \frac{6 d}{a d - b c} & \frac{d k + 6 b}{a d - b c} \\ \frac{6 c}{a d - b c} & - \frac{k c d a - k c d b + 6 a^{2} d - 6 a b c}{d (a - b) (a d - b c)} \end{array}] [\begin{array}{c} a & b \\ c & d \end{array}]$ .

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Schritt 3.1

Schritt 3.2

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

$[\begin{array}{c} - \frac{6 d}{a d - b c} a + \frac{d k + 6 b}{a d - b c} c & - \frac{6 d}{a d - b c} b + \frac{d k + 6 b}{a d - b c} d \\ \frac{6 c}{a d - b c} a - \frac{k c d a - k c d b + 6 a^{2} d - 6 a b c}{d (a - b) (a d - b c)} c & \frac{6 c}{a d - b c} b - \frac{k c d a - k c d b + 6 a^{2} d - 6 a b c}{d (a - b) (a d - b c)} d \end{array}] = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Schritt 3.3

Vereinfache jedes Element der Matrix durch Ausmultiplizieren aller Ausdrücke.

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Schritt 3.3.1

Wende das Distributivgesetz an.

$[\begin{array}{c} - \frac{6 a d - c d k - 6 c b}{a d - b c} & \frac{d^{2} k}{a d - b c} \\ - \frac{c (6 a d b + k c d a - k c d b - 6 a b c)}{d (a - b) (a d - b c)} & \frac{6 c b a + 6 c b (- b) - (k c d a - k c d b + 6 a^{2} d - 6 a b c)}{(a - b) (a d - b c)} \end{array}] = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Schritt 3.3.2

Multipliziere $b$ mit $b$ durch Addieren der Exponenten.

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Schritt 3.3.2.1

Bewege $b$ .

$[\begin{array}{c} - \frac{6 a d - c d k - 6 c b}{a d - b c} & \frac{d^{2} k}{a d - b c} \\ - \frac{c (6 a d b + k c d a - k c d b - 6 a b c)}{d (a - b) (a d - b c)} & \frac{6 c b a + 6 c (b \cdot b) \cdot - 1 - (k c d a - k c d b + 6 a^{2} d - 6 a b c)}{(a - b) (a d - b c)} \end{array}] = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Schritt 3.3.2.2

Mutltipliziere $b$ mit $b$ .

$[\begin{array}{c} - \frac{6 a d - c d k - 6 c b}{a d - b c} & \frac{d^{2} k}{a d - b c} \\ - \frac{c (6 a d b + k c d a - k c d b - 6 a b c)}{d (a - b) (a d - b c)} & \frac{6 c b a + 6 c b^{2} \cdot - 1 - (k c d a - k c d b + 6 a^{2} d - 6 a b c)}{(a - b) (a d - b c)} \end{array}] = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Schritt 3.3.3

Mutltipliziere $- 1$ mit $6$ .

$[\begin{array}{c} - \frac{6 a d - c d k - 6 c b}{a d - b c} & \frac{d^{2} k}{a d - b c} \\ - \frac{c (6 a d b + k c d a - k c d b - 6 a b c)}{d (a - b) (a d - b c)} & \frac{6 c b a - 6 c b^{2} - (k c d a - k c d b + 6 a^{2} d - 6 a b c)}{(a - b) (a d - b c)} \end{array}] = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Schritt 3.3.4

Wende das Distributivgesetz an.

$[\begin{array}{c} - \frac{6 a d - c d k - 6 c b}{a d - b c} & \frac{d^{2} k}{a d - b c} \\ - \frac{c (6 a d b + k c d a - k c d b - 6 a b c)}{d (a - b) (a d - b c)} & \frac{6 c b a - 6 c b^{2} - (k c d a) - (- k c d b) - (6 a^{2} d) - (- 6 a b c)}{(a - b) (a d - b c)} \end{array}] = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Schritt 3.3.5

Vereinfache.

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Schritt 3.3.5.1

Multipliziere $- (- k c d b)$ .

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Schritt 3.3.5.1.1

Mutltipliziere $- 1$ mit $- 1$ .

$[\begin{array}{c} - \frac{6 a d - c d k - 6 c b}{a d - b c} & \frac{d^{2} k}{a d - b c} \\ - \frac{c (6 a d b + k c d a - k c d b - 6 a b c)}{d (a - b) (a d - b c)} & \frac{6 c b a - 6 c b^{2} - k c d a + 1 (k c d b) - (6 a^{2} d) - (- 6 a b c)}{(a - b) (a d - b c)} \end{array}] = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Schritt 3.3.5.1.2

Mutltipliziere $k$ mit $1$ .

$[\begin{array}{c} - \frac{6 a d - c d k - 6 c b}{a d - b c} & \frac{d^{2} k}{a d - b c} \\ - \frac{c (6 a d b + k c d a - k c d b - 6 a b c)}{d (a - b) (a d - b c)} & \frac{6 c b a - 6 c b^{2} - k c d a + k (c d b) - (6 a^{2} d) - (- 6 a b c)}{(a - b) (a d - b c)} \end{array}] = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Schritt 3.3.5.2

Mutltipliziere $6$ mit $- 1$ .

$[\begin{array}{c} - \frac{6 a d - c d k - 6 c b}{a d - b c} & \frac{d^{2} k}{a d - b c} \\ - \frac{c (6 a d b + k c d a - k c d b - 6 a b c)}{d (a - b) (a d - b c)} & \frac{6 c b a - 6 c b^{2} - k c d a + k (c d b) - 6 (a^{2} d) - (- 6 a b c)}{(a - b) (a d - b c)} \end{array}] = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Schritt 3.3.5.3

Mutltipliziere $- 6$ mit $- 1$ .

$[\begin{array}{c} - \frac{6 a d - c d k - 6 c b}{a d - b c} & \frac{d^{2} k}{a d - b c} \\ - \frac{c (6 a d b + k c d a - k c d b - 6 a b c)}{d (a - b) (a d - b c)} & \frac{6 c b a - 6 c b^{2} - k c d a + k (c d b) - 6 (a^{2} d) + 6 (a b c)}{(a - b) (a d - b c)} \end{array}] = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Schritt 3.3.6

Entferne die Klammern.

$[\begin{array}{c} - \frac{6 a d - c d k - 6 c b}{a d - b c} & \frac{d^{2} k}{a d - b c} \\ - \frac{c (6 a d b + k c d a - k c d b - 6 a b c)}{d (a - b) (a d - b c)} & \frac{6 c b a - 6 c b^{2} - k c d a + k c d b - 6 a^{2} d + 6 a b c}{(a - b) (a d - b c)} \end{array}] = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Schritt 3.3.7

Addiere $6 c b a$ und $6 a b c$ .

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Schritt 3.3.7.1

Bewege $b$ .

$[\begin{array}{c} - \frac{6 a d - c d k - 6 c b}{a d - b c} & \frac{d^{2} k}{a d - b c} \\ - \frac{c (6 a d b + k c d a - k c d b - 6 a b c)}{d (a - b) (a d - b c)} & \frac{6 a b c + 6 a b c - 6 c b^{2} - k c d a + k c d b - 6 a^{2} d}{(a - b) (a d - b c)} \end{array}] = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Schritt 3.3.7.2

Addiere $6 a b c$ und $6 a b c$ .

$[\begin{array}{c} - \frac{6 a d - c d k - 6 c b}{a d - b c} & \frac{d^{2} k}{a d - b c} \\ - \frac{c (6 a d b + k c d a - k c d b - 6 a b c)}{d (a - b) (a d - b c)} & \frac{12 a b c - 6 c b^{2} - k c d a + k c d b - 6 a^{2} d}{(a - b) (a d - b c)} \end{array}] = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Schritt 4

Write as a linear system of equations.

$- \frac{6 a d - c d k - 6 c b}{a d - b c} = 1$

$\frac{d^{2} k}{a d - b c} = 0$

$- \frac{c (6 a d b + k c d a - k c d b - 6 a b c)}{d (a - b) (a d - b c)} = 0$

$\frac{12 a b c - 6 c b^{2} - k c d a + k c d b - 6 a^{2} d}{(a - b) (a d - b c)} = 1$

Schritt 5

Vereinfache das System.

$- \frac{6 a d - c d k - 6 c b}{a d - b c} = 1, \frac{d^{2} k}{a d - b c} = 0, - \frac{c (6 a d b + k c d a - k c d b - 6 a b c)}{d (a - b) (a d - b c)} = 0, \frac{12 a b c - 6 c b^{2} - k c d a + k c d b - 6 a^{2} d}{(a - b) (a d - b c)} = 1$