Álgebra lineal Ejemplos

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

Paso 1

Factoriza $d$ de $a d - b d$ .

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Paso 1.1

Factoriza $d$ de $a d$ .

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

Paso 1.2

Factoriza $d$ de $- b d$ .

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

Paso 1.3

Factoriza $d$ de $d a + d (- b)$ .

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

Paso 2

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

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Paso 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 $2 \times 2$ .

Paso 2.2

Multiplica cada fila en la primera matriz por cada columna en la segunda matriz.

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

Paso 2.3

Simplifica cada elemento de la matriz mediante la multiplicación de todas las expresiones.

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

Paso 3

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

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Paso 3.1

Paso 3.2

Multiplica cada fila en la primera matriz por cada columna en la segunda matriz.

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

Paso 3.3

Simplifica cada elemento de la matriz mediante la multiplicación de todas las expresiones.

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Paso 3.3.1

Aplica la propiedad distributiva.

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

Paso 3.3.2

Multiplica $b$ por $b$ sumando los exponentes.

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Paso 3.3.2.1

Mueve $b$ .

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

Paso 3.3.2.2

Multiplica $b$ por $b$ .

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

Paso 3.3.3

Multiplica $- 1$ por $6$ .

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

Paso 3.3.4

Aplica la propiedad distributiva.

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

Paso 3.3.5

Simplifica.

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Paso 3.3.5.1

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

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Paso 3.3.5.1.1

Multiplica $- 1$ por $- 1$ .

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

Paso 3.3.5.1.2

Multiplica $k$ por $1$ .

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

Paso 3.3.5.2

Multiplica $6$ por $- 1$ .

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

Paso 3.3.5.3

Multiplica $- 6$ por $- 1$ .

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

Paso 3.3.6

Elimina los paréntesis.

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

Paso 3.3.7

Suma $6 c b a$ y $6 a b c$ .

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Paso 3.3.7.1

Mueve $b$ .

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

Paso 3.3.7.2

Suma $6 a b c$ y $6 a b c$ .

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

Paso 4

Write as a linear system of equations.

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

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

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

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

Paso 5

Reduce el sistema.

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