Álgebra lineal Ejemplos

⎡ ⎢ ⎣ \begin{matrix} 1 & 7 & 1 2 & 9 & 0 1 & 8 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 4.5 & 0.5 & - 4.5 - 1 & 0 & 1 3.5 & - 0.5 & - 2.5 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 7 & 1 \\ 2 & 9 & 0 \\ 1 & 8 & 1 \end{array}] [\begin{array}{c} 4.5 & 0.5 & - 4.5 \\ - 1 & 0 & 1 \\ 3.5 & - 0.5 & - 2.5 \end{array}]$

Paso 1

Multiplica

⎡ ⎢ ⎣ \begin{matrix} 1 & 7 & 1 2 & 9 & 0 1 & 8 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 4.5 & 0.5 & - 4.5 - 1 & 0 & 1 3.5 & - 0.5 & - 2.5 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 7 & 1 \\ 2 & 9 & 0 \\ 1 & 8 & 1 \end{array}] [\begin{array}{c} 4.5 & 0.5 & - 4.5 \\ - 1 & 0 & 1 \\ 3.5 & - 0.5 & - 2.5 \end{array}]$ .

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Paso 1.1

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

$3 \times 3$ and the second matrix is

3 \times 3

$3 \times 3$ .

Paso 1.2

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

⎡ ⎢ ⎣ \begin{matrix} 1 \cdot 4.5 + 7 \cdot - 1 + 1 \cdot 3.5 & 1 \cdot 0.5 + 7 \cdot 0 + 1 \cdot - 0.5 & 1 \cdot - 4.5 + 7 \cdot 1 + 1 \cdot - 2.5 2 \cdot 4.5 + 9 \cdot - 1 + 0 \cdot 3.5 & 2 \cdot 0.5 + 9 \cdot 0 + 0 \cdot - 0.5 & 2 \cdot - 4.5 + 9 \cdot 1 + 0 \cdot - 2.5 1 \cdot 4.5 + 8 \cdot - 1 + 1 \cdot 3.5 & 1 \cdot 0.5 + 8 \cdot 0 + 1 \cdot - 0.5 & 1 \cdot - 4.5 + 8 \cdot 1 + 1 \cdot - 2.5 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 \cdot 4.5 + 7 \cdot - 1 + 1 \cdot 3.5 & 1 \cdot 0.5 + 7 \cdot 0 + 1 \cdot - 0.5 & 1 \cdot - 4.5 + 7 \cdot 1 + 1 \cdot - 2.5 \\ 2 \cdot 4.5 + 9 \cdot - 1 + 0 \cdot 3.5 & 2 \cdot 0.5 + 9 \cdot 0 + 0 \cdot - 0.5 & 2 \cdot - 4.5 + 9 \cdot 1 + 0 \cdot - 2.5 \\ 1 \cdot 4.5 + 8 \cdot - 1 + 1 \cdot 3.5 & 1 \cdot 0.5 + 8 \cdot 0 + 1 \cdot - 0.5 & 1 \cdot - 4.5 + 8 \cdot 1 + 1 \cdot - 2.5 \end{array}]$

Paso 1.3

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

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

Paso 2

Find the determinant.

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Paso 2.1

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 2.1.1

Consider the corresponding sign chart.

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

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

Paso 2.1.2

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

-

$-$ position on the sign chart.

Paso 2.1.3

The minor for

a_{11}

$a_{11}$ is the determinant with row

1

$1$ and column

1

$1$ deleted.

∣ ∣ ∣ \begin{matrix} 1 & 0 0 & 1 \end{matrix} ∣ ∣ ∣

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

Paso 2.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

1 ∣ ∣ ∣ \begin{matrix} 1 & 0 0 & 1 \end{matrix} ∣ ∣ ∣

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

Paso 2.1.5

The minor for

a_{12}

$a_{12}$ is the determinant with row

1

$1$ and column

2

$2$ deleted.

∣ ∣ ∣ \begin{matrix} 0 & 0 0 & 1 \end{matrix} ∣ ∣ ∣

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

Paso 2.1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

0 ∣ ∣ ∣ \begin{matrix} 0 & 0 0 & 1 \end{matrix} ∣ ∣ ∣

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

Paso 2.1.7

The minor for

a_{13}

$a_{13}$ is the determinant with row

1

$1$ and column

3

$3$ deleted.

∣ ∣ ∣ \begin{matrix} 0 & 1 0 & 0 \end{matrix} ∣ ∣ ∣

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

Paso 2.1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

0 ∣ ∣ ∣ \begin{matrix} 0 & 1 0 & 0 \end{matrix} ∣ ∣ ∣

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

Paso 2.1.9

Add the terms together.

1 ∣ ∣ ∣ \begin{matrix} 1 & 0 0 & 1 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 0 & 0 0 & 1 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 0 & 1 0 & 0 \end{matrix} ∣ ∣ ∣

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

1 ∣ ∣ ∣ \begin{matrix} 1 & 0 0 & 1 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 0 & 0 0 & 1 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 0 & 1 0 & 0 \end{matrix} ∣ ∣ ∣

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

Paso 2.2

Multiplica

0

$0$ por

∣ ∣ ∣ \begin{matrix} 0 & 0 0 & 1 \end{matrix} ∣ ∣ ∣

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

1 ∣ ∣ ∣ \begin{matrix} 1 & 0 0 & 1 \end{matrix} ∣ ∣ ∣ + 0 + 0 ∣ ∣ ∣ \begin{matrix} 0 & 1 0 & 0 \end{matrix} ∣ ∣ ∣

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

Paso 2.3

Multiplica

0

$0$ por

∣ ∣ ∣ \begin{matrix} 0 & 1 0 & 0 \end{matrix} ∣ ∣ ∣

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

1 ∣ ∣ ∣ \begin{matrix} 1 & 0 0 & 1 \end{matrix} ∣ ∣ ∣ + 0 + 0

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

Paso 2.4

Evalúa

∣ ∣ ∣ \begin{matrix} 1 & 0 0 & 1 \end{matrix} ∣ ∣ ∣

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

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Paso 2.4.1

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

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

1 (1 \cdot 1 + 0 \cdot 0) + 0 + 0

$1 (1 \cdot 1 + 0 \cdot 0) + 0 + 0$

Paso 2.4.2

Simplifica el determinante.

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Paso 2.4.2.1

Simplifica cada término.

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Paso 2.4.2.1.1

Multiplica

1

$1$ por

1

$1$ .

1 (1 + 0 \cdot 0) + 0 + 0

$1 (1 + 0 \cdot 0) + 0 + 0$

Paso 2.4.2.1.2

Multiplica

0

$0$ por

0

$0$ .

1 (1 + 0) + 0 + 0

$1 (1 + 0) + 0 + 0$

1 (1 + 0) + 0 + 0

$1 (1 + 0) + 0 + 0$

Paso 2.4.2.2

Suma

1

$1$ y

0

$0$ .

1 \cdot 1 + 0 + 0

$1 \cdot 1 + 0 + 0$

1 \cdot 1 + 0 + 0

$1 \cdot 1 + 0 + 0$

1 \cdot 1 + 0 + 0

$1 \cdot 1 + 0 + 0$

Paso 2.5

Simplifica el determinante.

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Paso 2.5.1

Multiplica

1

$1$ por

1

$1$ .

1 + 0 + 0

$1 + 0 + 0$

Paso 2.5.2

Suma

1

$1$ y

0

$0$ .

1 + 0

$1 + 0$

Paso 2.5.3

Suma

1

$1$ y

0

$0$ .

1

$1$

1

$1$

1

$1$

Paso 3

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

Paso 4

Set up a

3 \times 6

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

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 0 & 1 & 0 & 0 & 1 & 0 0 & 0 & 1 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

Paso 5

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

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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