Álgebra lineal Ejemplos

⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 2 & 5 & 7 3 & 7 & 9 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 2 & 3 \\ 2 & 5 & 7 \\ 3 & 7 & 9 \end{array}]$

Paso 1

Find the determinant.

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Paso 1.1

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 row

1

$1$ by its cofactor and add.

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Paso 1.1.1

Consider the corresponding sign chart.

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

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

Paso 1.1.2

The cofactor is the minor with the sign changed if the indices match a

-

$-$ position on the sign chart.

Paso 1.1.3

The minor for

a_{11}

$a_{11}$ is the determinant with row

1

$1$ and column

1

$1$ deleted.

∣ ∣ ∣ \begin{matrix} 5 & 7 7 & 9 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 5 & 7 \\ 7 & 9 \end{array} |$

Paso 1.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

1 ∣ ∣ ∣ \begin{matrix} 5 & 7 7 & 9 \end{matrix} ∣ ∣ ∣

$1 | \begin{array}{c} 5 & 7 \\ 7 & 9 \end{array} |$

Paso 1.1.5

The minor for

a_{12}

$a_{12}$ is the determinant with row

1

$1$ and column

2

$2$ deleted.

∣ ∣ ∣ \begin{matrix} 2 & 7 3 & 9 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 2 & 7 \\ 3 & 9 \end{array} |$

Paso 1.1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

- 2 ∣ ∣ ∣ \begin{matrix} 2 & 7 3 & 9 \end{matrix} ∣ ∣ ∣

$- 2 | \begin{array}{c} 2 & 7 \\ 3 & 9 \end{array} |$

Paso 1.1.7

The minor for

a_{13}

$a_{13}$ is the determinant with row

1

$1$ and column

3

$3$ deleted.

∣ ∣ ∣ \begin{matrix} 2 & 5 3 & 7 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

Paso 1.1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

3 ∣ ∣ ∣ \begin{matrix} 2 & 5 3 & 7 \end{matrix} ∣ ∣ ∣

$3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

Paso 1.1.9

Add the terms together.

1 ∣ ∣ ∣ \begin{matrix} 5 & 7 7 & 9 \end{matrix} ∣ ∣ ∣ - 2 ∣ ∣ ∣ \begin{matrix} 2 & 7 3 & 9 \end{matrix} ∣ ∣ ∣ + 3 ∣ ∣ ∣ \begin{matrix} 2 & 5 3 & 7 \end{matrix} ∣ ∣ ∣

$1 | \begin{array}{c} 5 & 7 \\ 7 & 9 \end{array} | - 2 | \begin{array}{c} 2 & 7 \\ 3 & 9 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

1 ∣ ∣ ∣ \begin{matrix} 5 & 7 7 & 9 \end{matrix} ∣ ∣ ∣ - 2 ∣ ∣ ∣ \begin{matrix} 2 & 7 3 & 9 \end{matrix} ∣ ∣ ∣ + 3 ∣ ∣ ∣ \begin{matrix} 2 & 5 3 & 7 \end{matrix} ∣ ∣ ∣

$1 | \begin{array}{c} 5 & 7 \\ 7 & 9 \end{array} | - 2 | \begin{array}{c} 2 & 7 \\ 3 & 9 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

Paso 1.2

Evalúa

∣ ∣ ∣ \begin{matrix} 5 & 7 7 & 9 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 5 & 7 \\ 7 & 9 \end{array} |$ .

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Paso 1.2.1

El determinante de una matriz

2 \times 2

$2 \times 2$ puede obtenerse usando la fórmula

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

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

1 (5 \cdot 9 - 7 \cdot 7) - 2 ∣ ∣ ∣ \begin{matrix} 2 & 7 3 & 9 \end{matrix} ∣ ∣ ∣ + 3 ∣ ∣ ∣ \begin{matrix} 2 & 5 3 & 7 \end{matrix} ∣ ∣ ∣

$1 (5 \cdot 9 - 7 \cdot 7) - 2 | \begin{array}{c} 2 & 7 \\ 3 & 9 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

Paso 1.2.2

Simplifica el determinante.

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Paso 1.2.2.1

Simplifica cada término.

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Paso 1.2.2.1.1

Multiplica

5

$5$ por

9

$9$ .

1 (45 - 7 \cdot 7) - 2 ∣ ∣ ∣ \begin{matrix} 2 & 7 3 & 9 \end{matrix} ∣ ∣ ∣ + 3 ∣ ∣ ∣ \begin{matrix} 2 & 5 3 & 7 \end{matrix} ∣ ∣ ∣

$1 (45 - 7 \cdot 7) - 2 | \begin{array}{c} 2 & 7 \\ 3 & 9 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

Paso 1.2.2.1.2

Multiplica

- 7

$- 7$ por

7

$7$ .

1 (45 - 49) - 2 ∣ ∣ ∣ \begin{matrix} 2 & 7 3 & 9 \end{matrix} ∣ ∣ ∣ + 3 ∣ ∣ ∣ \begin{matrix} 2 & 5 3 & 7 \end{matrix} ∣ ∣ ∣

$1 (45 - 49) - 2 | \begin{array}{c} 2 & 7 \\ 3 & 9 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

1 (45 - 49) - 2 ∣ ∣ ∣ \begin{matrix} 2 & 7 3 & 9 \end{matrix} ∣ ∣ ∣ + 3 ∣ ∣ ∣ \begin{matrix} 2 & 5 3 & 7 \end{matrix} ∣ ∣ ∣

$1 (45 - 49) - 2 | \begin{array}{c} 2 & 7 \\ 3 & 9 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

Paso 1.2.2.2

Resta

49

$49$ de

45

$45$ .

1 \cdot - 4 - 2 ∣ ∣ ∣ \begin{matrix} 2 & 7 3 & 9 \end{matrix} ∣ ∣ ∣ + 3 ∣ ∣ ∣ \begin{matrix} 2 & 5 3 & 7 \end{matrix} ∣ ∣ ∣

$1 \cdot - 4 - 2 | \begin{array}{c} 2 & 7 \\ 3 & 9 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

1 \cdot - 4 - 2 ∣ ∣ ∣ \begin{matrix} 2 & 7 3 & 9 \end{matrix} ∣ ∣ ∣ + 3 ∣ ∣ ∣ \begin{matrix} 2 & 5 3 & 7 \end{matrix} ∣ ∣ ∣

$1 \cdot - 4 - 2 | \begin{array}{c} 2 & 7 \\ 3 & 9 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

1 \cdot - 4 - 2 ∣ ∣ ∣ \begin{matrix} 2 & 7 3 & 9 \end{matrix} ∣ ∣ ∣ + 3 ∣ ∣ ∣ \begin{matrix} 2 & 5 3 & 7 \end{matrix} ∣ ∣ ∣

$1 \cdot - 4 - 2 | \begin{array}{c} 2 & 7 \\ 3 & 9 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

Paso 1.3

Evalúa

∣ ∣ ∣ \begin{matrix} 2 & 7 3 & 9 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 2 & 7 \\ 3 & 9 \end{array} |$ .

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Paso 1.3.1

El determinante de una matriz

2 \times 2

$2 \times 2$ puede obtenerse usando la fórmula

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

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

1 \cdot - 4 - 2 (2 \cdot 9 - 3 \cdot 7) + 3 ∣ ∣ ∣ \begin{matrix} 2 & 5 3 & 7 \end{matrix} ∣ ∣ ∣

$1 \cdot - 4 - 2 (2 \cdot 9 - 3 \cdot 7) + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

Paso 1.3.2

Simplifica el determinante.

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Paso 1.3.2.1

Simplifica cada término.

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Paso 1.3.2.1.1

Multiplica

$2$ por

$9$ .

$1 \cdot - 4 - 2 (18 - 3 \cdot 7) + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

Paso 1.3.2.1.2

Multiplica

$- 3$ por

$7$ .

$1 \cdot - 4 - 2 (18 - 21) + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

Paso 1.3.2.2

Resta

$21$ de

$18$ .

$1 \cdot - 4 - 2 \cdot - 3 + 3 | \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$

Paso 1.4

Evalúa

$| \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$ .

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Paso 1.4.1

El determinante de una matriz

$2 \times 2$ puede obtenerse usando la fórmula

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

$1 \cdot - 4 - 2 \cdot - 3 + 3 (2 \cdot 7 - 3 \cdot 5)$

Paso 1.4.2

Simplifica el determinante.

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Paso 1.4.2.1

Simplifica cada término.

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Paso 1.4.2.1.1

Multiplica

$2$ por

$7$ .

$1 \cdot - 4 - 2 \cdot - 3 + 3 (14 - 3 \cdot 5)$

Paso 1.4.2.1.2

Multiplica

$- 3$ por

$5$ .

$1 \cdot - 4 - 2 \cdot - 3 + 3 (14 - 15)$

Paso 1.4.2.2

Resta

$15$ de

$14$ .

$1 \cdot - 4 - 2 \cdot - 3 + 3 \cdot - 1$

Paso 1.5

Simplifica el determinante.

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Paso 1.5.1

Simplifica cada término.

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Paso 1.5.1.1

Multiplica

$- 4$ por

$1$ .

$- 4 - 2 \cdot - 3 + 3 \cdot - 1$

Paso 1.5.1.2

Multiplica

$- 2$ por

$- 3$ .

$- 4 + 6 + 3 \cdot - 1$

Paso 1.5.1.3

Multiplica

$3$ por

$- 1$ .

$- 4 + 6 - 3$

Paso 1.5.2

Suma

$- 4$ y

$6$ .

$2 - 3$

Paso 1.5.3

Resta

$3$ de

$2$ .

$- 1$

Paso 2

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

Paso 3

Set up a

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

$[\begin{array}{c} 1 & 2 & 3 & 1 & 0 & 0 \\ 2 & 5 & 7 & 0 & 1 & 0 \\ 3 & 7 & 9 & 0 & 0 & 1 \end{array}]$

Paso 4

Obtén la forma escalonada reducida por filas.

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Paso 4.1

Perform the row operation

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

$2, 1$ a

$0$ .

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Paso 4.1.1

Perform the row operation

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

$2, 1$ a

$0$ .

$[\begin{array}{c} 1 & 2 & 3 & 1 & 0 & 0 \\ 2 - 2 \cdot 1 & 5 - 2 \cdot 2 & 7 - 2 \cdot 3 & 0 - 2 \cdot 1 & 1 - 2 \cdot 0 & 0 - 2 \cdot 0 \\ 3 & 7 & 9 & 0 & 0 & 1 \end{array}]$

Paso 4.1.2

Simplifica

$R_{2}$ .

$[\begin{array}{c} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 1 & - 2 & 1 & 0 \\ 3 & 7 & 9 & 0 & 0 & 1 \end{array}]$

Paso 4.2

Perform the row operation

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

$3, 1$ a

$0$ .

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Paso 4.2.1

Perform the row operation

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

$3, 1$ a

$0$ .

$[\begin{array}{c} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 1 & - 2 & 1 & 0 \\ 3 - 3 \cdot 1 & 7 - 3 \cdot 2 & 9 - 3 \cdot 3 & 0 - 3 \cdot 1 & 0 - 3 \cdot 0 & 1 - 3 \cdot 0 \end{array}]$

Paso 4.2.2

Simplifica

$R_{3}$ .

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

Paso 4.3

Perform the row operation

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

$3, 2$ a

$0$ .

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Paso 4.3.1

Perform the row operation

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

$3, 2$ a

$0$ .

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

Paso 4.3.2

Simplifica

$R_{3}$ .

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

Paso 4.4

Multiply each element of

$R_{3}$ by

$- 1$ to make the entry at

$3, 3$ a

$1$ .

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Paso 4.4.1

Multiply each element of

$R_{3}$ by

$- 1$ to make the entry at

$3, 3$ a

$1$ .

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

Paso 4.4.2

Simplifica

$R_{3}$ .

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

Paso 4.5

Perform the row operation

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

$2, 3$ a

$0$ .

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Paso 4.5.1

Perform the row operation

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

$2, 3$ a

$0$ .

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

Paso 4.5.2

Simplifica

$R_{2}$ .

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

Paso 4.6

Perform the row operation

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

$1, 3$ a

$0$ .

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Paso 4.6.1

Perform the row operation

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

$1, 3$ a

$0$ .

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

Paso 4.6.2

Simplifica

$R_{1}$ .

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

Paso 4.7

Perform the row operation

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

$1, 2$ a

$0$ .

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Paso 4.7.1

Perform the row operation

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

$1, 2$ a

$0$ .

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

Paso 4.7.2

Simplifica

$R_{1}$ .

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

Paso 5

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

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