Álgebra linear Exemplos

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

Etapa 1

Find the inverse of $[\begin{array}{c} 1 & - 1 & 2 \\ 2 & 0 & 1 \\ 1 & 1 & 1 \end{array}]$ .

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Etapa 1.1

Reescreva.

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

Etapa 1.2

Find the determinant.

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Etapa 1.2.1

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 $2$ by its cofactor and add.

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Etapa 1.2.1.1

Consider the corresponding sign chart.

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

Etapa 1.2.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Etapa 1.2.1.3

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

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

Etapa 1.2.1.4

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

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

Etapa 1.2.1.5

The minor for $a_{22}$ is the determinant with row $2$ and column $2$ deleted.

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

Etapa 1.2.1.6

Multiply element $a_{22}$ by its cofactor.

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

Etapa 1.2.1.7

The minor for $a_{23}$ is the determinant with row $2$ and column $3$ deleted.

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

Etapa 1.2.1.8

Multiply element $a_{23}$ by its cofactor.

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

Etapa 1.2.1.9

Add the terms together.

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

Etapa 1.2.2

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

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

Etapa 1.2.3

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

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Etapa 1.2.3.1

O determinante de uma matriz $2 \times 2$ pode ser encontrado ao usar a fórmula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

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

Etapa 1.2.3.2

Simplifique o determinante.

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Etapa 1.2.3.2.1

Simplifique cada termo.

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Etapa 1.2.3.2.1.1

Multiplique $- 1$ por $1$ .

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

Etapa 1.2.3.2.1.2

Multiplique $- 1$ por $2$ .

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

Etapa 1.2.3.2.2

Subtraia $2$ de $- 1$ .

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

Etapa 1.2.4

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

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Etapa 1.2.4.1

O determinante de uma matriz $2 \times 2$ pode ser encontrado ao usar a fórmula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

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

Etapa 1.2.4.2

Simplifique o determinante.

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Etapa 1.2.4.2.1

Simplifique cada termo.

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Etapa 1.2.4.2.1.1

Multiplique $1$ por $1$ .

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

Etapa 1.2.4.2.1.2

Multiplique $- 1$ por $- 1$ .

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

Etapa 1.2.4.2.2

Some $1$ e $1$ .

$- 2 \cdot - 3 + 0 - 1 \cdot 2$

Etapa 1.2.5

Simplifique o determinante.

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Etapa 1.2.5.1

Simplifique cada termo.

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Etapa 1.2.5.1.1

Multiplique $- 2$ por $- 3$ .

$6 + 0 - 1 \cdot 2$

Etapa 1.2.5.1.2

Multiplique $- 1$ por $2$ .

$6 + 0 - 2$

Etapa 1.2.5.2

Some $6$ e $0$ .

$6 - 2$

Etapa 1.2.5.3

Subtraia $2$ de $6$ .

$4$

Etapa 1.3

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

Etapa 1.4

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

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

Etapa 1.5

Encontre a forma escalonada reduzida por linhas.

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Etapa 1.5.1

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

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Etapa 1.5.1.1

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

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

Etapa 1.5.1.2

Simplifique $R_{2}$ .

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

Etapa 1.5.2

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

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Etapa 1.5.2.1

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

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

Etapa 1.5.2.2

Simplifique $R_{3}$ .

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

Etapa 1.5.3

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

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Etapa 1.5.3.1

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

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

Etapa 1.5.3.2

Simplifique $R_{2}$ .

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

Etapa 1.5.4

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

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Etapa 1.5.4.1

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

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

Etapa 1.5.4.2

Simplifique $R_{3}$ .

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

Etapa 1.5.5

Multiply each element of $R_{3}$ by $\frac{1}{2}$ to make the entry at $3, 3$ a $1$ .

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Etapa 1.5.5.1

Multiply each element of $R_{3}$ by $\frac{1}{2}$ to make the entry at $3, 3$ a $1$ .

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

Etapa 1.5.5.2

Simplifique $R_{3}$ .

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

Etapa 1.5.6

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

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Etapa 1.5.6.1

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

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

Etapa 1.5.6.2

Simplifique $R_{2}$ .

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

Etapa 1.5.7

Perform the row operation $R_{1} = R_{1} - 2 R_{3}$ to make the entry at $1, 3$ a $0$ .

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Etapa 1.5.7.1

Perform the row operation $R_{1} = R_{1} - 2 R_{3}$ to make the entry at $1, 3$ a $0$ .

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

Etapa 1.5.7.2

Simplifique $R_{1}$ .

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

Etapa 1.5.8

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

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Etapa 1.5.8.1

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

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

Etapa 1.5.8.2

Simplifique $R_{1}$ .

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

Etapa 1.6

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

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

Etapa 2

Multiply both sides by the inverse of $[\begin{array}{c} 1 & - 1 & 2 \\ 2 & 0 & 1 \\ 1 & 1 & 1 \end{array}]$ .

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

Etapa 3

Simplifique a equação.

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Etapa 3.1

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

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Etapa 3.1.1

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 $3 \times 3$ and the second matrix is $3 \times 3$ .

Etapa 3.1.2

Multiplique cada linha na primeira matriz por cada coluna na segunda matriz.

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

Etapa 3.1.3

Simplifique cada elemento da matriz multiplicando todas as expressões.

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

Etapa 3.2

Multiplying the identity matrix by any matrix $A$ is the matrix $A$ itself.

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

Etapa 3.3

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

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Etapa 3.3.1

Etapa 3.3.2

Multiplique cada linha na primeira matriz por cada coluna na segunda matriz.

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

Etapa 3.3.3

Simplifique cada elemento da matriz multiplicando todas as expressões.

$x = [\begin{array}{c} - \frac{1}{2} & \frac{1}{4} & - 2 \\ \frac{1}{2} & - \frac{7}{4} & 1 \\ 2 & - \frac{1}{2} & 2 \end{array}]$