Matemática discreta Ejemplos

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

Paso 1

Find the determinant.

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Paso 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 row $1$ by its cofactor and add.

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Paso 1.1.1

Consider the corresponding sign chart.

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

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

$| \begin{array}{c} 1 & 2 \\ 0 & 1 \end{array} |$

Paso 1.1.4

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

$1 | \begin{array}{c} 1 & 2 \\ 0 & 1 \end{array} |$

Paso 1.1.5

The minor for $a_{12}$ is the determinant with row $1$ and column $2$ deleted.

$| \begin{array}{c} 1 & 2 \\ 0 & 1 \end{array} |$

Paso 1.1.6

Multiply element $a_{12}$ by its cofactor.

$0 | \begin{array}{c} 1 & 2 \\ 0 & 1 \end{array} |$

Paso 1.1.7

The minor for $a_{13}$ is the determinant with row $1$ and column $3$ deleted.

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

Paso 1.1.8

Multiply element $a_{13}$ by its cofactor.

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

Paso 1.1.9

Add the terms together.

$1 | \begin{array}{c} 1 & 2 \\ 0 & 1 \end{array} | + 0 | \begin{array}{c} 1 & 2 \\ 0 & 1 \end{array} | + d | \begin{array}{c} 1 & 1 \\ 0 & 0 \end{array} |$

Paso 1.2

Multiplica $0$ por $| \begin{array}{c} 1 & 2 \\ 0 & 1 \end{array} |$ .

$1 | \begin{array}{c} 1 & 2 \\ 0 & 1 \end{array} | + 0 + d | \begin{array}{c} 1 & 1 \\ 0 & 0 \end{array} |$

Paso 1.3

Evalúa $| \begin{array}{c} 1 & 2 \\ 0 & 1 \end{array} |$ .

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Paso 1.3.1

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

$1 (1 \cdot 1 + 0 \cdot 2) + 0 + d | \begin{array}{c} 1 & 1 \\ 0 & 0 \end{array} |$

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 $1$ por $1$ .

$1 (1 + 0 \cdot 2) + 0 + d | \begin{array}{c} 1 & 1 \\ 0 & 0 \end{array} |$

Paso 1.3.2.1.2

Multiplica $0$ por $2$ .

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

Paso 1.3.2.2

Suma $1$ y $0$ .

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

Paso 1.4

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

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Paso 1.4.1

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

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

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 $0$ por $1$ .

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

Paso 1.4.2.1.2

Multiplica $0$ por $1$ .

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

Paso 1.4.2.2

Suma $0$ y $0$ .

$1 \cdot 1 + 0 + d \cdot 0$

Paso 1.5

Simplifica el determinante.

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Paso 1.5.1

Suma $1 \cdot 1$ y $0$ .

$1 \cdot 1 + d \cdot 0$

Paso 1.5.2

Simplifica cada término.

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Paso 1.5.2.1

Multiplica $1$ por $1$ .

$1 + d \cdot 0$

Paso 1.5.2.2

Multiplica $d$ por $0$ .

$1 + 0$

Paso 1.5.3

Suma $1$ y $0$ .

$1$

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.

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

Paso 4

Obtén la forma escalonada reducida por filas.

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Paso 4.1

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

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Paso 4.1.1

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

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

Paso 4.1.2

Simplifica $R_{2}$ .

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

Paso 4.2

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

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Paso 4.2.1

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

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

Paso 4.2.2

Simplifica $R_{2}$ .

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

Paso 4.3

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

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Paso 4.3.1

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

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

Paso 4.3.2

Simplifica $R_{1}$ .

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

Paso 5

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

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