Matemática discreta Ejemplos

$[\begin{array}{c} 2 & 4 & 6 \\ - 1 & - 4 & - 3 \\ 0 & 1 & - 1 \end{array}]$

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 column $1$ 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.

$| \begin{array}{c} - 4 & - 3 \\ 1 & - 1 \end{array} |$

Paso 1.1.4

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

$2 | \begin{array}{c} - 4 & - 3 \\ 1 & - 1 \end{array} |$

Paso 1.1.5

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

$| \begin{array}{c} 4 & 6 \\ 1 & - 1 \end{array} |$

Paso 1.1.6

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

$1 | \begin{array}{c} 4 & 6 \\ 1 & - 1 \end{array} |$

Paso 1.1.7

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

$| \begin{array}{c} 4 & 6 \\ - 4 & - 3 \end{array} |$

Paso 1.1.8

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

$0 | \begin{array}{c} 4 & 6 \\ - 4 & - 3 \end{array} |$

Paso 1.1.9

Add the terms together.

$2 | \begin{array}{c} - 4 & - 3 \\ 1 & - 1 \end{array} | + 1 | \begin{array}{c} 4 & 6 \\ 1 & - 1 \end{array} | + 0 | \begin{array}{c} 4 & 6 \\ - 4 & - 3 \end{array} |$

Paso 1.2

Multiplica $0$ por $| \begin{array}{c} 4 & 6 \\ - 4 & - 3 \end{array} |$ .

$2 | \begin{array}{c} - 4 & - 3 \\ 1 & - 1 \end{array} | + 1 | \begin{array}{c} 4 & 6 \\ 1 & - 1 \end{array} | + 0$

Paso 1.3

Evalúa $| \begin{array}{c} - 4 & - 3 \\ 1 & - 1 \end{array} |$ .

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Paso 1.3.1

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

$2 (- 4 \cdot - 1 - 1 \cdot - 3) + 1 | \begin{array}{c} 4 & 6 \\ 1 & - 1 \end{array} | + 0$

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 $- 4$ por $- 1$ .

$2 (4 - 1 \cdot - 3) + 1 | \begin{array}{c} 4 & 6 \\ 1 & - 1 \end{array} | + 0$

Paso 1.3.2.1.2

Multiplica $- 1$ por $- 3$ .

$2 (4 + 3) + 1 | \begin{array}{c} 4 & 6 \\ 1 & - 1 \end{array} | + 0$

Paso 1.3.2.2

Suma $4$ y $3$ .

$2 \cdot 7 + 1 | \begin{array}{c} 4 & 6 \\ 1 & - 1 \end{array} | + 0$

Paso 1.4

Evalúa $| \begin{array}{c} 4 & 6 \\ 1 & - 1 \end{array} |$ .

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Paso 1.4.1

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

$2 \cdot 7 + 1 (4 \cdot - 1 - 1 \cdot 6) + 0$

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 $4$ por $- 1$ .

$2 \cdot 7 + 1 (- 4 - 1 \cdot 6) + 0$

Paso 1.4.2.1.2

Multiplica $- 1$ por $6$ .

$2 \cdot 7 + 1 (- 4 - 6) + 0$

Paso 1.4.2.2

Resta $6$ de $- 4$ .

$2 \cdot 7 + 1 \cdot - 10 + 0$

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 $2$ por $7$ .

$14 + 1 \cdot - 10 + 0$

Paso 1.5.1.2

Multiplica $- 10$ por $1$ .

$14 - 10 + 0$

Paso 1.5.2

Resta $10$ de $14$ .

$4 + 0$

Paso 1.5.3

Suma $4$ y $0$ .

$4$

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.

$[\begin{array}{c} 2 & 4 & 6 & 1 & 0 & 0 \\ - 1 & - 4 & - 3 & 0 & 1 & 0 \\ 0 & 1 & - 1 & 0 & 0 & 1 \end{array}]$

Paso 4

Obtén la forma escalonada reducida por filas.

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Paso 4.1

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

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Paso 4.1.1

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

$[\begin{array}{c} \frac{2}{2} & \frac{4}{2} & \frac{6}{2} & \frac{1}{2} & \frac{0}{2} & \frac{0}{2} \\ - 1 & - 4 & - 3 & 0 & 1 & 0 \\ 0 & 1 & - 1 & 0 & 0 & 1 \end{array}]$

Paso 4.1.2

Simplifica $R_{1}$ .

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

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 $0$ .

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

Paso 4.2.2

Simplifica $R_{2}$ .

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

Paso 4.3

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

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Paso 4.3.1

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

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

Paso 4.3.2

Simplifica $R_{2}$ .

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

Paso 4.4

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

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Paso 4.4.1

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

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

Paso 4.4.2

Simplifica $R_{3}$ .

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

Paso 4.5

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

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Paso 4.5.1

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

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

Paso 4.5.2

Simplifica $R_{3}$ .

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

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 $0$ .

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

Paso 4.6.2

Simplifica $R_{1}$ .

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

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 $0$ .

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

Paso 4.7.2

Simplifica $R_{1}$ .

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

Paso 5

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

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