Álgebra lineal Ejemplos

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

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

$2$ 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

$a_{12}$ is the determinant with row

$1$ and column

$2$ deleted.

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

Paso 1.1.4

Multiply element

$a_{12}$ by its cofactor.

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

Paso 1.1.5

The minor for

$a_{22}$ is the determinant with row

$2$ and column

$2$ deleted.

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

Paso 1.1.6

Multiply element

$a_{22}$ by its cofactor.

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

Paso 1.1.7

The minor for

$a_{32}$ is the determinant with row

$3$ and column

$2$ deleted.

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

Paso 1.1.8

Multiply element

$a_{32}$ by its cofactor.

$0 | \begin{array}{c} 4 & 4 \\ 1 & 2 \end{array} |$

Paso 1.1.9

Add the terms together.

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

Paso 1.2

Multiplica

$0$ por

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

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

Paso 1.3

Evalúa

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

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Paso 1.3.1

El determinante de una matriz

$2 \times 2$ puede obtenerse usando la fórmula

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

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

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

$2$ por

$1$ .

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

Paso 1.3.2.1.2

Multiplica

$- 3$ por

$2$ .

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

Paso 1.3.2.2

Resta

$6$ de

$2$ .

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

Paso 1.4

Evalúa

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

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Paso 1.4.1

El determinante de una matriz

$2 \times 2$ puede obtenerse usando la fórmula

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

$- 3 \cdot - 4 + 1 (4 \cdot 2 - 3 \cdot 4) + 0$

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

$2$ .

$- 3 \cdot - 4 + 1 (8 - 3 \cdot 4) + 0$

Paso 1.4.2.1.2

Multiplica

$- 3$ por

$4$ .

$- 3 \cdot - 4 + 1 (8 - 12) + 0$

Paso 1.4.2.2

Resta

$12$ de

$8$ .

$- 3 \cdot - 4 + 1 \cdot - 4 + 0$

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

$- 3$ por

$- 4$ .

$12 + 1 \cdot - 4 + 0$

Paso 1.5.1.2

Multiplica

$- 4$ por

$1$ .

$12 - 4 + 0$

Paso 1.5.2

Resta

$4$ de

$12$ .

$8 + 0$

Paso 1.5.3

Suma

$8$ y

$0$ .

$8$

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.

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

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

$1$ .

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

Paso 4.1.2

Simplifica

$R_{1}$ .

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

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

$0$ .

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

Paso 4.2.2

Simplifica

$R_{2}$ .

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

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

$0$ .

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

Paso 4.3.2

Simplifica

$R_{3}$ .

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

Paso 4.4

Multiply each element of

$R_{2}$ by

$4$ to make the entry at

$2, 2$ a

$1$ .

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Paso 4.4.1

Multiply each element of

$R_{2}$ by

$4$ to make the entry at

$2, 2$ a

$1$ .

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

Paso 4.4.2

Simplifica

$R_{2}$ .

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

Paso 4.5

Perform the row operation

$R_{3} = R_{3} + \frac{9}{4} R_{2}$ to make the entry at

$3, 2$ a

$0$ .

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Paso 4.5.1

Perform the row operation

$R_{3} = R_{3} + \frac{9}{4} R_{2}$ to make the entry at

$3, 2$ a

$0$ .

$[\begin{array}{c} 1 & \frac{3}{4} & 1 & \frac{1}{4} & 0 & 0 \\ 0 & 1 & 4 & - 1 & 4 & 0 \\ 0 + \frac{9}{4} \cdot 0 & - \frac{9}{4} + \frac{9}{4} \cdot 1 & - 1 + \frac{9}{4} \cdot 4 & - \frac{3}{4} + \frac{9}{4} \cdot - 1 & 0 + \frac{9}{4} \cdot 4 & 1 + \frac{9}{4} \cdot 0 \end{array}]$

Paso 4.5.2

Simplifica

$R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & 1 & \frac{1}{4} & 0 & 0 \\ 0 & 1 & 4 & - 1 & 4 & 0 \\ 0 & 0 & 8 & - 3 & 9 & 1 \end{array}]$

Paso 4.6

Multiply each element of

$R_{3}$ by

$\frac{1}{8}$ to make the entry at

$3, 3$ a

$1$ .

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Paso 4.6.1

Multiply each element of

$R_{3}$ by

$\frac{1}{8}$ to make the entry at

$3, 3$ a

$1$ .

$[\begin{array}{c} 1 & \frac{3}{4} & 1 & \frac{1}{4} & 0 & 0 \\ 0 & 1 & 4 & - 1 & 4 & 0 \\ \frac{0}{8} & \frac{0}{8} & \frac{8}{8} & \frac{- 3}{8} & \frac{9}{8} & \frac{1}{8} \end{array}]$

Paso 4.6.2

Simplifica

$R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & 1 & \frac{1}{4} & 0 & 0 \\ 0 & 1 & 4 & - 1 & 4 & 0 \\ 0 & 0 & 1 & - \frac{3}{8} & \frac{9}{8} & \frac{1}{8} \end{array}]$

Paso 4.7

Perform the row operation

$R_{2} = R_{2} - 4 R_{3}$ to make the entry at

$2, 3$ a

$0$ .

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Paso 4.7.1

Perform the row operation

$R_{2} = R_{2} - 4 R_{3}$ to make the entry at

$2, 3$ a

$0$ .

$[\begin{array}{c} 1 & \frac{3}{4} & 1 & \frac{1}{4} & 0 & 0 \\ 0 - 4 \cdot 0 & 1 - 4 \cdot 0 & 4 - 4 \cdot 1 & - 1 - 4 (- \frac{3}{8}) & 4 - 4 (\frac{9}{8}) & 0 - 4 (\frac{1}{8}) \\ 0 & 0 & 1 & - \frac{3}{8} & \frac{9}{8} & \frac{1}{8} \end{array}]$

Paso 4.7.2

Simplifica

$R_{2}$ .

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

Paso 4.8

Perform the row operation

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

$1, 3$ a

$0$ .

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Paso 4.8.1

Perform the row operation

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

$1, 3$ a

$0$ .

$[\begin{array}{c} 1 - 0 & \frac{3}{4} - 0 & 1 - 1 & \frac{1}{4} + \frac{3}{8} & 0 - \frac{9}{8} & 0 - \frac{1}{8} \\ 0 & 1 & 0 & \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} \\ 0 & 0 & 1 & - \frac{3}{8} & \frac{9}{8} & \frac{1}{8} \end{array}]$

Paso 4.8.2

Simplifica

$R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & 0 & \frac{5}{8} & - \frac{9}{8} & - \frac{1}{8} \\ 0 & 1 & 0 & \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} \\ 0 & 0 & 1 & - \frac{3}{8} & \frac{9}{8} & \frac{1}{8} \end{array}]$

Paso 4.9

Perform the row operation

$R_{1} = R_{1} - \frac{3}{4} R_{2}$ to make the entry at

$1, 2$ a

$0$ .

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Paso 4.9.1

Perform the row operation

$R_{1} = R_{1} - \frac{3}{4} R_{2}$ to make the entry at

$1, 2$ a

$0$ .

$[\begin{array}{c} 1 - \frac{3}{4} \cdot 0 & \frac{3}{4} - \frac{3}{4} \cdot 1 & 0 - \frac{3}{4} \cdot 0 & \frac{5}{8} - \frac{3}{4} \cdot \frac{1}{2} & - \frac{9}{8} - \frac{3}{4} (- \frac{1}{2}) & - \frac{1}{8} - \frac{3}{4} (- \frac{1}{2}) \\ 0 & 1 & 0 & \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} \\ 0 & 0 & 1 & - \frac{3}{8} & \frac{9}{8} & \frac{1}{8} \end{array}]$

Paso 4.9.2

Simplifica

$R_{1}$ .

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

Paso 5

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

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

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