Álgebra lineal Ejemplos

⎡ ⎢ ⎣ \begin{matrix} 4 & - 10 & 29 1 & - 2 & 5 - 3 & 7 & - 19 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 4 & - 10 & 29 \\ 1 & - 2 & 5 \\ - 3 & 7 & - 19 \end{array}]$

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} - 2 & 5 7 & - 19 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} - 2 & 5 \\ 7 & - 19 \end{array} |$

Paso 1.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

4 ∣ ∣ ∣ \begin{matrix} - 2 & 5 7 & - 19 \end{matrix} ∣ ∣ ∣

$4 | \begin{array}{c} - 2 & 5 \\ 7 & - 19 \end{array} |$

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} 1 & 5 - 3 & - 19 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 1 & 5 \\ - 3 & - 19 \end{array} |$

Paso 1.1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

10 ∣ ∣ ∣ \begin{matrix} 1 & 5 - 3 & - 19 \end{matrix} ∣ ∣ ∣

$10 | \begin{array}{c} 1 & 5 \\ - 3 & - 19 \end{array} |$

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} 1 & - 2 - 3 & 7 \end{matrix} ∣ ∣ ∣

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

Paso 1.1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

29 ∣ ∣ ∣ \begin{matrix} 1 & - 2 - 3 & 7 \end{matrix} ∣ ∣ ∣

$29 | \begin{array}{c} 1 & - 2 \\ - 3 & 7 \end{array} |$

Paso 1.1.9

Add the terms together.

4 ∣ ∣ ∣ \begin{matrix} - 2 & 5 7 & - 19 \end{matrix} ∣ ∣ ∣ + 10 ∣ ∣ ∣ \begin{matrix} 1 & 5 - 3 & - 19 \end{matrix} ∣ ∣ ∣ + 29 ∣ ∣ ∣ \begin{matrix} 1 & - 2 - 3 & 7 \end{matrix} ∣ ∣ ∣

$4 | \begin{array}{c} - 2 & 5 \\ 7 & - 19 \end{array} | + 10 | \begin{array}{c} 1 & 5 \\ - 3 & - 19 \end{array} | + 29 | \begin{array}{c} 1 & - 2 \\ - 3 & 7 \end{array} |$

4 ∣ ∣ ∣ \begin{matrix} - 2 & 5 7 & - 19 \end{matrix} ∣ ∣ ∣ + 10 ∣ ∣ ∣ \begin{matrix} 1 & 5 - 3 & - 19 \end{matrix} ∣ ∣ ∣ + 29 ∣ ∣ ∣ \begin{matrix} 1 & - 2 - 3 & 7 \end{matrix} ∣ ∣ ∣

$4 | \begin{array}{c} - 2 & 5 \\ 7 & - 19 \end{array} | + 10 | \begin{array}{c} 1 & 5 \\ - 3 & - 19 \end{array} | + 29 | \begin{array}{c} 1 & - 2 \\ - 3 & 7 \end{array} |$

Paso 1.2

Evalúa

∣ ∣ ∣ \begin{matrix} - 2 & 5 7 & - 19 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} - 2 & 5 \\ 7 & - 19 \end{array} |$ .

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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

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

4 (- 2 \cdot - 19 - 7 \cdot 5) + 10 ∣ ∣ ∣ \begin{matrix} 1 & 5 - 3 & - 19 \end{matrix} ∣ ∣ ∣ + 29 ∣ ∣ ∣ \begin{matrix} 1 & - 2 - 3 & 7 \end{matrix} ∣ ∣ ∣

$4 (- 2 \cdot - 19 - 7 \cdot 5) + 10 | \begin{array}{c} 1 & 5 \\ - 3 & - 19 \end{array} | + 29 | \begin{array}{c} 1 & - 2 \\ - 3 & 7 \end{array} |$

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

- 2

$- 2$ por

- 19

$- 19$ .

4 (38 - 7 \cdot 5) + 10 ∣ ∣ ∣ \begin{matrix} 1 & 5 - 3 & - 19 \end{matrix} ∣ ∣ ∣ + 29 ∣ ∣ ∣ \begin{matrix} 1 & - 2 - 3 & 7 \end{matrix} ∣ ∣ ∣

$4 (38 - 7 \cdot 5) + 10 | \begin{array}{c} 1 & 5 \\ - 3 & - 19 \end{array} | + 29 | \begin{array}{c} 1 & - 2 \\ - 3 & 7 \end{array} |$

Paso 1.2.2.1.2

Multiplica

- 7

$- 7$ por

5

$5$ .

4 (38 - 35) + 10 ∣ ∣ ∣ \begin{matrix} 1 & 5 - 3 & - 19 \end{matrix} ∣ ∣ ∣ + 29 ∣ ∣ ∣ \begin{matrix} 1 & - 2 - 3 & 7 \end{matrix} ∣ ∣ ∣

$4 (38 - 35) + 10 | \begin{array}{c} 1 & 5 \\ - 3 & - 19 \end{array} | + 29 | \begin{array}{c} 1 & - 2 \\ - 3 & 7 \end{array} |$

4 (38 - 35) + 10 ∣ ∣ ∣ \begin{matrix} 1 & 5 - 3 & - 19 \end{matrix} ∣ ∣ ∣ + 29 ∣ ∣ ∣ \begin{matrix} 1 & - 2 - 3 & 7 \end{matrix} ∣ ∣ ∣

$4 (38 - 35) + 10 | \begin{array}{c} 1 & 5 \\ - 3 & - 19 \end{array} | + 29 | \begin{array}{c} 1 & - 2 \\ - 3 & 7 \end{array} |$

Paso 1.2.2.2

Resta

35

$35$ de

38

$38$ .

4 \cdot 3 + 10 ∣ ∣ ∣ \begin{matrix} 1 & 5 - 3 & - 19 \end{matrix} ∣ ∣ ∣ + 29 ∣ ∣ ∣ \begin{matrix} 1 & - 2 - 3 & 7 \end{matrix} ∣ ∣ ∣

$4 \cdot 3 + 10 | \begin{array}{c} 1 & 5 \\ - 3 & - 19 \end{array} | + 29 | \begin{array}{c} 1 & - 2 \\ - 3 & 7 \end{array} |$

4 \cdot 3 + 10 ∣ ∣ ∣ \begin{matrix} 1 & 5 - 3 & - 19 \end{matrix} ∣ ∣ ∣ + 29 ∣ ∣ ∣ \begin{matrix} 1 & - 2 - 3 & 7 \end{matrix} ∣ ∣ ∣

$4 \cdot 3 + 10 | \begin{array}{c} 1 & 5 \\ - 3 & - 19 \end{array} | + 29 | \begin{array}{c} 1 & - 2 \\ - 3 & 7 \end{array} |$

4 \cdot 3 + 10 ∣ ∣ ∣ \begin{matrix} 1 & 5 - 3 & - 19 \end{matrix} ∣ ∣ ∣ + 29 ∣ ∣ ∣ \begin{matrix} 1 & - 2 - 3 & 7 \end{matrix} ∣ ∣ ∣

$4 \cdot 3 + 10 | \begin{array}{c} 1 & 5 \\ - 3 & - 19 \end{array} | + 29 | \begin{array}{c} 1 & - 2 \\ - 3 & 7 \end{array} |$

Paso 1.3

Evalúa

∣ ∣ ∣ \begin{matrix} 1 & 5 - 3 & - 19 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 1 & 5 \\ - 3 & - 19 \end{array} |$ .

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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

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

4 \cdot 3 + 10 (1 \cdot - 19 - (- 3 \cdot 5)) + 29 ∣ ∣ ∣ \begin{matrix} 1 & - 2 - 3 & 7 \end{matrix} ∣ ∣ ∣

$4 \cdot 3 + 10 (1 \cdot - 19 - (- 3 \cdot 5)) + 29 | \begin{array}{c} 1 & - 2 \\ - 3 & 7 \end{array} |$

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

- 19

$- 19$ por

1

$1$ .

4 \cdot 3 + 10 (- 19 - (- 3 \cdot 5)) + 29 ∣ ∣ ∣ \begin{matrix} 1 & - 2 - 3 & 7 \end{matrix} ∣ ∣ ∣

$4 \cdot 3 + 10 (- 19 - (- 3 \cdot 5)) + 29 | \begin{array}{c} 1 & - 2 \\ - 3 & 7 \end{array} |$

Paso 1.3.2.1.2

Multiplica

- (- 3 \cdot 5)

$- (- 3 \cdot 5)$ .

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Paso 1.3.2.1.2.1

Multiplica

- 3

$- 3$ por

5

$5$ .

4 \cdot 3 + 10 (- 19 - - 15) + 29 ∣ ∣ ∣ \begin{matrix} 1 & - 2 - 3 & 7 \end{matrix} ∣ ∣ ∣

$4 \cdot 3 + 10 (- 19 - - 15) + 29 | \begin{array}{c} 1 & - 2 \\ - 3 & 7 \end{array} |$

Paso 1.3.2.1.2.2

Multiplica

- 1

$- 1$ por

- 15

$- 15$ .

4 \cdot 3 + 10 (- 19 + 15) + 29 ∣ ∣ ∣ \begin{matrix} 1 & - 2 - 3 & 7 \end{matrix} ∣ ∣ ∣

$4 \cdot 3 + 10 (- 19 + 15) + 29 | \begin{array}{c} 1 & - 2 \\ - 3 & 7 \end{array} |$

4 \cdot 3 + 10 (- 19 + 15) + 29 ∣ ∣ ∣ \begin{matrix} 1 & - 2 - 3 & 7 \end{matrix} ∣ ∣ ∣

$4 \cdot 3 + 10 (- 19 + 15) + 29 | \begin{array}{c} 1 & - 2 \\ - 3 & 7 \end{array} |$

4 \cdot 3 + 10 (- 19 + 15) + 29 ∣ ∣ ∣ \begin{matrix} 1 & - 2 - 3 & 7 \end{matrix} ∣ ∣ ∣

$4 \cdot 3 + 10 (- 19 + 15) + 29 | \begin{array}{c} 1 & - 2 \\ - 3 & 7 \end{array} |$

Paso 1.3.2.2

Suma

- 19

$- 19$ y

15

$15$ .

4 \cdot 3 + 10 \cdot - 4 + 29 ∣ ∣ ∣ \begin{matrix} 1 & - 2 - 3 & 7 \end{matrix} ∣ ∣ ∣

$4 \cdot 3 + 10 \cdot - 4 + 29 | \begin{array}{c} 1 & - 2 \\ - 3 & 7 \end{array} |$

4 \cdot 3 + 10 \cdot - 4 + 29 ∣ ∣ ∣ \begin{matrix} 1 & - 2 - 3 & 7 \end{matrix} ∣ ∣ ∣

$4 \cdot 3 + 10 \cdot - 4 + 29 | \begin{array}{c} 1 & - 2 \\ - 3 & 7 \end{array} |$

4 \cdot 3 + 10 \cdot - 4 + 29 ∣ ∣ ∣ \begin{matrix} 1 & - 2 - 3 & 7 \end{matrix} ∣ ∣ ∣

$4 \cdot 3 + 10 \cdot - 4 + 29 | \begin{array}{c} 1 & - 2 \\ - 3 & 7 \end{array} |$

Paso 1.4

Evalúa

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

$| \begin{array}{c} 1 & - 2 \\ - 3 & 7 \end{array} |$ .

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Paso 1.4.1

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

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

4 \cdot 3 + 10 \cdot - 4 + 29 (1 \cdot 7 - (- 3 \cdot - 2))

$4 \cdot 3 + 10 \cdot - 4 + 29 (1 \cdot 7 - (- 3 \cdot - 2))$

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

7

$7$ por

1

$1$ .

4 \cdot 3 + 10 \cdot - 4 + 29 (7 - (- 3 \cdot - 2))

$4 \cdot 3 + 10 \cdot - 4 + 29 (7 - (- 3 \cdot - 2))$

Paso 1.4.2.1.2

Multiplica

- (- 3 \cdot - 2)

$- (- 3 \cdot - 2)$ .

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Paso 1.4.2.1.2.1

Multiplica

- 3

$- 3$ por

- 2

$- 2$ .

4 \cdot 3 + 10 \cdot - 4 + 29 (7 - 1 \cdot 6)

$4 \cdot 3 + 10 \cdot - 4 + 29 (7 - 1 \cdot 6)$

Paso 1.4.2.1.2.2

Multiplica

- 1

$- 1$ por

6

$6$ .

4 \cdot 3 + 10 \cdot - 4 + 29 (7 - 6)

$4 \cdot 3 + 10 \cdot - 4 + 29 (7 - 6)$

4 \cdot 3 + 10 \cdot - 4 + 29 (7 - 6)

$4 \cdot 3 + 10 \cdot - 4 + 29 (7 - 6)$

4 \cdot 3 + 10 \cdot - 4 + 29 (7 - 6)

$4 \cdot 3 + 10 \cdot - 4 + 29 (7 - 6)$

Paso 1.4.2.2

Resta

6

$6$ de

7

$7$ .

4 \cdot 3 + 10 \cdot - 4 + 29 \cdot 1

$4 \cdot 3 + 10 \cdot - 4 + 29 \cdot 1$

4 \cdot 3 + 10 \cdot - 4 + 29 \cdot 1

$4 \cdot 3 + 10 \cdot - 4 + 29 \cdot 1$

4 \cdot 3 + 10 \cdot - 4 + 29 \cdot 1

$4 \cdot 3 + 10 \cdot - 4 + 29 \cdot 1$

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

$4$ por

3

$3$ .

12 + 10 \cdot - 4 + 29 \cdot 1

$12 + 10 \cdot - 4 + 29 \cdot 1$

Paso 1.5.1.2

Multiplica

10

$10$ por

- 4

$- 4$ .

12 - 40 + 29 \cdot 1

$12 - 40 + 29 \cdot 1$

Paso 1.5.1.3

Multiplica

$29$ por

$1$ .

$12 - 40 + 29$

Paso 1.5.2

Resta

$40$ de

$12$ .

$- 28 + 29$

Paso 1.5.3

Suma

$- 28$ y

$29$ .

$1$

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.

$[\begin{array}{c} 4 & - 10 & 29 & 1 & 0 & 0 \\ 1 & - 2 & 5 & 0 & 1 & 0 \\ - 3 & 7 & - 19 & 0 & 0 & 1 \end{array}]$

Paso 4

Obtén la forma escalonada reducida por filas.

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Paso 4.1

Multiply each element of

$R_{1}$ by

$\frac{1}{4}$ to make the entry at

$1, 1$ a

$1$ .

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Paso 4.1.1

Multiply each element of

$R_{1}$ by

$\frac{1}{4}$ to make the entry at

$1, 1$ a

$1$ .

$[\begin{array}{c} \frac{4}{4} & \frac{- 10}{4} & \frac{29}{4} & \frac{1}{4} & \frac{0}{4} & \frac{0}{4} \\ 1 & - 2 & 5 & 0 & 1 & 0 \\ - 3 & 7 & - 19 & 0 & 0 & 1 \end{array}]$

Paso 4.1.2

Simplifica

$R_{1}$ .

$[\begin{array}{c} 1 & - \frac{5}{2} & \frac{29}{4} & \frac{1}{4} & 0 & 0 \\ 1 & - 2 & 5 & 0 & 1 & 0 \\ - 3 & 7 & - 19 & 0 & 0 & 1 \end{array}]$

Paso 4.2

Perform the row operation

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

$2, 1$ a

$0$ .

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Paso 4.2.1

Perform the row operation

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

$2, 1$ a

$0$ .

$[\begin{array}{c} 1 & - \frac{5}{2} & \frac{29}{4} & \frac{1}{4} & 0 & 0 \\ 1 - 1 & - 2 + \frac{5}{2} & 5 - \frac{29}{4} & 0 - \frac{1}{4} & 1 - 0 & 0 - 0 \\ - 3 & 7 & - 19 & 0 & 0 & 1 \end{array}]$

Paso 4.2.2

Simplifica

$R_{2}$ .

$[\begin{array}{c} 1 & - \frac{5}{2} & \frac{29}{4} & \frac{1}{4} & 0 & 0 \\ 0 & \frac{1}{2} & - \frac{9}{4} & - \frac{1}{4} & 1 & 0 \\ - 3 & 7 & - 19 & 0 & 0 & 1 \end{array}]$

Paso 4.3

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 4.3.1

Perform the row operation

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

$3, 1$ a

$0$ .

$[\begin{array}{c} 1 & - \frac{5}{2} & \frac{29}{4} & \frac{1}{4} & 0 & 0 \\ 0 & \frac{1}{2} & - \frac{9}{4} & - \frac{1}{4} & 1 & 0 \\ - 3 + 3 \cdot 1 & 7 + 3 (- \frac{5}{2}) & - 19 + 3 (\frac{29}{4}) & 0 + 3 (\frac{1}{4}) & 0 + 3 \cdot 0 & 1 + 3 \cdot 0 \end{array}]$

Paso 4.3.2

Simplifica

$R_{3}$ .

$[\begin{array}{c} 1 & - \frac{5}{2} & \frac{29}{4} & \frac{1}{4} & 0 & 0 \\ 0 & \frac{1}{2} & - \frac{9}{4} & - \frac{1}{4} & 1 & 0 \\ 0 & - \frac{1}{2} & \frac{11}{4} & \frac{3}{4} & 0 & 1 \end{array}]$

Paso 4.4

Multiply each element of

$R_{2}$ by

$2$ to make the entry at

$2, 2$ a

$1$ .

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Paso 4.4.1

Multiply each element of

$R_{2}$ by

$2$ to make the entry at

$2, 2$ a

$1$ .

$[\begin{array}{c} 1 & - \frac{5}{2} & \frac{29}{4} & \frac{1}{4} & 0 & 0 \\ 2 \cdot 0 & 2 (\frac{1}{2}) & 2 (- \frac{9}{4}) & 2 (- \frac{1}{4}) & 2 \cdot 1 & 2 \cdot 0 \\ 0 & - \frac{1}{2} & \frac{11}{4} & \frac{3}{4} & 0 & 1 \end{array}]$

Paso 4.4.2

Simplifica

$R_{2}$ .

$[\begin{array}{c} 1 & - \frac{5}{2} & \frac{29}{4} & \frac{1}{4} & 0 & 0 \\ 0 & 1 & - \frac{9}{2} & - \frac{1}{2} & 2 & 0 \\ 0 & - \frac{1}{2} & \frac{11}{4} & \frac{3}{4} & 0 & 1 \end{array}]$

Paso 4.5

Perform the row operation

$R_{3} = R_{3} + \frac{1}{2} R_{2}$ to make the entry at

$3, 2$ a

$0$ .

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Paso 4.5.1

Perform the row operation

$R_{3} = R_{3} + \frac{1}{2} R_{2}$ to make the entry at

$3, 2$ a

$0$ .

$[\begin{array}{c} 1 & - \frac{5}{2} & \frac{29}{4} & \frac{1}{4} & 0 & 0 \\ 0 & 1 & - \frac{9}{2} & - \frac{1}{2} & 2 & 0 \\ 0 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & \frac{11}{4} + \frac{1}{2} (- \frac{9}{2}) & \frac{3}{4} + \frac{1}{2} (- \frac{1}{2}) & 0 + \frac{1}{2} \cdot 2 & 1 + \frac{1}{2} \cdot 0 \end{array}]$

Paso 4.5.2

Simplifica

$R_{3}$ .

$[\begin{array}{c} 1 & - \frac{5}{2} & \frac{29}{4} & \frac{1}{4} & 0 & 0 \\ 0 & 1 & - \frac{9}{2} & - \frac{1}{2} & 2 & 0 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2} & 1 & 1 \end{array}]$

Paso 4.6

Multiply each element of

$R_{3}$ by

$2$ to make the entry at

$3, 3$ a

$1$ .

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Paso 4.6.1

Multiply each element of

$R_{3}$ by

$2$ to make the entry at

$3, 3$ a

$1$ .

$[\begin{array}{c} 1 & - \frac{5}{2} & \frac{29}{4} & \frac{1}{4} & 0 & 0 \\ 0 & 1 & - \frac{9}{2} & - \frac{1}{2} & 2 & 0 \\ 2 \cdot 0 & 2 \cdot 0 & 2 (\frac{1}{2}) & 2 (\frac{1}{2}) & 2 \cdot 1 & 2 \cdot 1 \end{array}]$

Paso 4.6.2

Simplifica

$R_{3}$ .

$[\begin{array}{c} 1 & - \frac{5}{2} & \frac{29}{4} & \frac{1}{4} & 0 & 0 \\ 0 & 1 & - \frac{9}{2} & - \frac{1}{2} & 2 & 0 \\ 0 & 0 & 1 & 1 & 2 & 2 \end{array}]$

Paso 4.7

Perform the row operation

$R_{2} = R_{2} + \frac{9}{2} R_{3}$ to make the entry at

$2, 3$ a

$0$ .

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Paso 4.7.1

Perform the row operation

$R_{2} = R_{2} + \frac{9}{2} R_{3}$ to make the entry at

$2, 3$ a

$0$ .

$[\begin{array}{c} 1 & - \frac{5}{2} & \frac{29}{4} & \frac{1}{4} & 0 & 0 \\ 0 + \frac{9}{2} \cdot 0 & 1 + \frac{9}{2} \cdot 0 & - \frac{9}{2} + \frac{9}{2} \cdot 1 & - \frac{1}{2} + \frac{9}{2} \cdot 1 & 2 + \frac{9}{2} \cdot 2 & 0 + \frac{9}{2} \cdot 2 \\ 0 & 0 & 1 & 1 & 2 & 2 \end{array}]$

Paso 4.7.2

Simplifica

$R_{2}$ .

$[\begin{array}{c} 1 & - \frac{5}{2} & \frac{29}{4} & \frac{1}{4} & 0 & 0 \\ 0 & 1 & 0 & 4 & 11 & 9 \\ 0 & 0 & 1 & 1 & 2 & 2 \end{array}]$

Paso 4.8

Perform the row operation

$R_{1} = R_{1} - \frac{29}{4} R_{3}$ to make the entry at

$1, 3$ a

$0$ .

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Paso 4.8.1

Perform the row operation

$R_{1} = R_{1} - \frac{29}{4} R_{3}$ to make the entry at

$1, 3$ a

$0$ .

$[\begin{array}{c} 1 - \frac{29}{4} \cdot 0 & - \frac{5}{2} - \frac{29}{4} \cdot 0 & \frac{29}{4} - \frac{29}{4} \cdot 1 & \frac{1}{4} - \frac{29}{4} \cdot 1 & 0 - \frac{29}{4} \cdot 2 & 0 - \frac{29}{4} \cdot 2 \\ 0 & 1 & 0 & 4 & 11 & 9 \\ 0 & 0 & 1 & 1 & 2 & 2 \end{array}]$

Paso 4.8.2

Simplifica

$R_{1}$ .

$[\begin{array}{c} 1 & - \frac{5}{2} & 0 & - 7 & - \frac{29}{2} & - \frac{29}{2} \\ 0 & 1 & 0 & 4 & 11 & 9 \\ 0 & 0 & 1 & 1 & 2 & 2 \end{array}]$

Paso 4.9

Perform the row operation

$R_{1} = R_{1} + \frac{5}{2} R_{2}$ to make the entry at

$1, 2$ a

$0$ .

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Paso 4.9.1

Perform the row operation

$R_{1} = R_{1} + \frac{5}{2} R_{2}$ to make the entry at

$1, 2$ a

$0$ .

$[\begin{array}{c} 1 + \frac{5}{2} \cdot 0 & - \frac{5}{2} + \frac{5}{2} \cdot 1 & 0 + \frac{5}{2} \cdot 0 & - 7 + \frac{5}{2} \cdot 4 & - \frac{29}{2} + \frac{5}{2} \cdot 11 & - \frac{29}{2} + \frac{5}{2} \cdot 9 \\ 0 & 1 & 0 & 4 & 11 & 9 \\ 0 & 0 & 1 & 1 & 2 & 2 \end{array}]$

Paso 4.9.2

Simplifica

$R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 3 & 13 & 8 \\ 0 & 1 & 0 & 4 & 11 & 9 \\ 0 & 0 & 1 & 1 & 2 & 2 \end{array}]$

Paso 5

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

$[\begin{array}{c} 3 & 13 & 8 \\ 4 & 11 & 9 \\ 1 & 2 & 2 \end{array}]$