Álgebra lineal Ejemplos

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

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.

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

Paso 1.1.4

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

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

Paso 1.1.5

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

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

Paso 1.1.6

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

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

Paso 1.1.7

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

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

Paso 1.1.8

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

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

Paso 1.1.9

Add the terms together.

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

Paso 1.2

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

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

Paso 1.3

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

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

Paso 1.4

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

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Paso 1.4.1

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

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

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 $1$ por $1$ .

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

Paso 1.4.2.1.2

Multiplica $0$ por $0$ .

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

Paso 1.4.2.2

Suma $1$ y $0$ .

$1 \cdot 1 + 0 + 0$

Paso 1.5

Simplifica el determinante.

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Paso 1.5.1

Multiplica $1$ por $1$ .

$1 + 0 + 0$

Paso 1.5.2

Suma $1$ y $0$ .

$1 + 0$

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.

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

Paso 4

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

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