Álgebra lineal Ejemplos

$x + y + z = 8$ , $x - 3 y = 0$ , $5 y - z = 10$

Paso 1

Obtén la forma $A X = B$ del sistema de ecuaciones.

$[\begin{array}{c} 1 & 1 & 1 \\ 1 & - 3 & 0 \\ 0 & 5 & - 1 \end{array}] \cdot [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 8 \\ 0 \\ 10 \end{array}]$

Paso 2

Obtén la inversa de la matriz de coeficientes.

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Paso 2.1

Find the determinant.

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Paso 2.1.1

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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Paso 2.1.1.1

Consider the corresponding sign chart.

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

Paso 2.1.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Paso 2.1.1.3

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

$| \begin{array}{c} - 3 & 0 \\ 5 & - 1 \end{array} |$

Paso 2.1.1.4

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

$1 | \begin{array}{c} - 3 & 0 \\ 5 & - 1 \end{array} |$

Paso 2.1.1.5

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

$| \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} |$

Paso 2.1.1.6

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

$- 1 | \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} |$

Paso 2.1.1.7

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

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

Paso 2.1.1.8

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

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

Paso 2.1.1.9

Add the terms together.

$1 | \begin{array}{c} - 3 & 0 \\ 5 & - 1 \end{array} | - 1 | \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} | + 0 | \begin{array}{c} 1 & 1 \\ - 3 & 0 \end{array} |$

Paso 2.1.2

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

$1 | \begin{array}{c} - 3 & 0 \\ 5 & - 1 \end{array} | - 1 | \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} | + 0$

Paso 2.1.3

Evalúa $| \begin{array}{c} - 3 & 0 \\ 5 & - 1 \end{array} |$ .

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Paso 2.1.3.1

El determinante de una matriz $2 \times 2$ puede obtenerse usando la fórmula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$1 (- 3 \cdot - 1 - 5 \cdot 0) - 1 | \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} | + 0$

Paso 2.1.3.2

Simplifica el determinante.

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Paso 2.1.3.2.1

Simplifica cada término.

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Paso 2.1.3.2.1.1

Multiplica $- 3$ por $- 1$ .

$1 (3 - 5 \cdot 0) - 1 | \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} | + 0$

Paso 2.1.3.2.1.2

Multiplica $- 5$ por $0$ .

$1 (3 + 0) - 1 | \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} | + 0$

Paso 2.1.3.2.2

Suma $3$ y $0$ .

$1 \cdot 3 - 1 | \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} | + 0$

Paso 2.1.4

Evalúa $| \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} |$ .

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Paso 2.1.4.1

El determinante de una matriz $2 \times 2$ puede obtenerse usando la fórmula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$1 \cdot 3 - 1 (1 \cdot - 1 - 5 \cdot 1) + 0$

Paso 2.1.4.2

Simplifica el determinante.

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Paso 2.1.4.2.1

Simplifica cada término.

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Paso 2.1.4.2.1.1

Multiplica $- 1$ por $1$ .

$1 \cdot 3 - 1 (- 1 - 5 \cdot 1) + 0$

Paso 2.1.4.2.1.2

Multiplica $- 5$ por $1$ .

$1 \cdot 3 - 1 (- 1 - 5) + 0$

Paso 2.1.4.2.2

Resta $5$ de $- 1$ .

$1 \cdot 3 - 1 \cdot - 6 + 0$

Paso 2.1.5

Simplifica el determinante.

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Paso 2.1.5.1

Simplifica cada término.

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Paso 2.1.5.1.1

Multiplica $3$ por $1$ .

$3 - 1 \cdot - 6 + 0$

Paso 2.1.5.1.2

Multiplica $- 1$ por $- 6$ .

$3 + 6 + 0$

Paso 2.1.5.2

Suma $3$ y $6$ .

$9 + 0$

Paso 2.1.5.3

Suma $9$ y $0$ .

$9$

Paso 2.2

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

Paso 2.3

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

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

Paso 2.4

Obtén la forma escalonada reducida por filas.

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Paso 2.4.1

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

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Paso 2.4.1.1

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

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

Paso 2.4.1.2

Simplifica $R_{2}$ .

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

Paso 2.4.2

Multiply each element of $R_{2}$ by $- \frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

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Paso 2.4.2.1

Multiply each element of $R_{2}$ by $- \frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

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

Paso 2.4.2.2

Simplifica $R_{2}$ .

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

Paso 2.4.3

Perform the row operation $R_{3} = R_{3} - 5 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Paso 2.4.3.1

Perform the row operation $R_{3} = R_{3} - 5 R_{2}$ to make the entry at $3, 2$ a $0$ .

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

Paso 2.4.3.2

Simplifica $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & \frac{1}{4} & \frac{1}{4} & - \frac{1}{4} & 0 \\ 0 & 0 & - \frac{9}{4} & - \frac{5}{4} & \frac{5}{4} & 1 \end{array}]$

Paso 2.4.4

Multiply each element of $R_{3}$ by $- \frac{4}{9}$ to make the entry at $3, 3$ a $1$ .

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Paso 2.4.4.1

Multiply each element of $R_{3}$ by $- \frac{4}{9}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & \frac{1}{4} & \frac{1}{4} & - \frac{1}{4} & 0 \\ - \frac{4}{9} \cdot 0 & - \frac{4}{9} \cdot 0 & - \frac{4}{9} (- \frac{9}{4}) & - \frac{4}{9} (- \frac{5}{4}) & - \frac{4}{9} \cdot \frac{5}{4} & - \frac{4}{9} \cdot 1 \end{array}]$

Paso 2.4.4.2

Simplifica $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & \frac{1}{4} & \frac{1}{4} & - \frac{1}{4} & 0 \\ 0 & 0 & 1 & \frac{5}{9} & - \frac{5}{9} & - \frac{4}{9} \end{array}]$

Paso 2.4.5

Perform the row operation $R_{2} = R_{2} - \frac{1}{4} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Paso 2.4.5.1

Perform the row operation $R_{2} = R_{2} - \frac{1}{4} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 - \frac{1}{4} \cdot 0 & 1 - \frac{1}{4} \cdot 0 & \frac{1}{4} - \frac{1}{4} \cdot 1 & \frac{1}{4} - \frac{1}{4} \cdot \frac{5}{9} & - \frac{1}{4} - \frac{1}{4} (- \frac{5}{9}) & 0 - \frac{1}{4} (- \frac{4}{9}) \\ 0 & 0 & 1 & \frac{5}{9} & - \frac{5}{9} & - \frac{4}{9} \end{array}]$

Paso 2.4.5.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & \frac{1}{9} & - \frac{1}{9} & \frac{1}{9} \\ 0 & 0 & 1 & \frac{5}{9} & - \frac{5}{9} & - \frac{4}{9} \end{array}]$

Paso 2.4.6

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

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Paso 2.4.6.1

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

$[\begin{array}{c} 1 - 0 & 1 - 0 & 1 - 1 & 1 - \frac{5}{9} & 0 + \frac{5}{9} & 0 + \frac{4}{9} \\ 0 & 1 & 0 & \frac{1}{9} & - \frac{1}{9} & \frac{1}{9} \\ 0 & 0 & 1 & \frac{5}{9} & - \frac{5}{9} & - \frac{4}{9} \end{array}]$

Paso 2.4.6.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & 1 & 0 & \frac{4}{9} & \frac{5}{9} & \frac{4}{9} \\ 0 & 1 & 0 & \frac{1}{9} & - \frac{1}{9} & \frac{1}{9} \\ 0 & 0 & 1 & \frac{5}{9} & - \frac{5}{9} & - \frac{4}{9} \end{array}]$

Paso 2.4.7

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

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Paso 2.4.7.1

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

$[\begin{array}{c} 1 - 0 & 1 - 1 & 0 - 0 & \frac{4}{9} - \frac{1}{9} & \frac{5}{9} + \frac{1}{9} & \frac{4}{9} - \frac{1}{9} \\ 0 & 1 & 0 & \frac{1}{9} & - \frac{1}{9} & \frac{1}{9} \\ 0 & 0 & 1 & \frac{5}{9} & - \frac{5}{9} & - \frac{4}{9} \end{array}]$

Paso 2.4.7.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{1}{3} & \frac{2}{3} & \frac{1}{3} \\ 0 & 1 & 0 & \frac{1}{9} & - \frac{1}{9} & \frac{1}{9} \\ 0 & 0 & 1 & \frac{5}{9} & - \frac{5}{9} & - \frac{4}{9} \end{array}]$

Paso 2.5

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

$[\begin{array}{c} \frac{1}{3} & \frac{2}{3} & \frac{1}{3} \\ \frac{1}{9} & - \frac{1}{9} & \frac{1}{9} \\ \frac{5}{9} & - \frac{5}{9} & - \frac{4}{9} \end{array}]$

Paso 3

Multiplica por la izquierda ambos lados de la ecuación de matriz por la matriz inversa.

$([\begin{array}{c} \frac{1}{3} & \frac{2}{3} & \frac{1}{3} \\ \frac{1}{9} & - \frac{1}{9} & \frac{1}{9} \\ \frac{5}{9} & - \frac{5}{9} & - \frac{4}{9} \end{array}] \cdot [\begin{array}{c} 1 & 1 & 1 \\ 1 & - 3 & 0 \\ 0 & 5 & - 1 \end{array}]) \cdot [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{1}{3} & \frac{2}{3} & \frac{1}{3} \\ \frac{1}{9} & - \frac{1}{9} & \frac{1}{9} \\ \frac{5}{9} & - \frac{5}{9} & - \frac{4}{9} \end{array}] \cdot [\begin{array}{c} 8 \\ 0 \\ 10 \end{array}]$

Paso 4

Cualquier matriz multiplicada por su inversa es igual a $1$ todo el tiempo. $A \cdot A^{- 1} = 1$ .

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{1}{3} & \frac{2}{3} & \frac{1}{3} \\ \frac{1}{9} & - \frac{1}{9} & \frac{1}{9} \\ \frac{5}{9} & - \frac{5}{9} & - \frac{4}{9} \end{array}] \cdot [\begin{array}{c} 8 \\ 0 \\ 10 \end{array}]$

Paso 5

Multiplica $[\begin{array}{c} \frac{1}{3} & \frac{2}{3} & \frac{1}{3} \\ \frac{1}{9} & - \frac{1}{9} & \frac{1}{9} \\ \frac{5}{9} & - \frac{5}{9} & - \frac{4}{9} \end{array}] [\begin{array}{c} 8 \\ 0 \\ 10 \end{array}]$ .

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Paso 5.1

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 $3 \times 1$ .

Paso 5.2

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

$[\begin{array}{c} \frac{1}{3} \cdot 8 + \frac{2}{3} \cdot 0 + \frac{1}{3} \cdot 10 \\ \frac{1}{9} \cdot 8 - \frac{1}{9} \cdot 0 + \frac{1}{9} \cdot 10 \\ \frac{5}{9} \cdot 8 - \frac{5}{9} \cdot 0 - \frac{4}{9} \cdot 10 \end{array}]$

Paso 5.3

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

$[\begin{array}{c} 6 \\ 2 \\ 0 \end{array}]$

Paso 6

Simplifica los lados izquierdo y derecho.

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 6 \\ 2 \\ 0 \end{array}]$

Paso 7

Obtén la solución.

$x = 6$

$y = 2$

$z = 0$