Álgebra linear Exemplos

$- 4 x + 3 y + z = 17$ , $- x + 2 y + z = 7$ , $- 3 x + 3 y - z = 10$

Etapa 1

Encontre $A X = B$ do sistema de equações.

$[\begin{array}{c} - 4 & 3 & 1 \\ - 1 & 2 & 1 \\ - 3 & 3 & - 1 \end{array}] \cdot [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 17 \\ 7 \\ 10 \end{array}]$

Etapa 2

Encontre o inverso da matriz do coeficiente.

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Etapa 2.1

Find the determinant.

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Etapa 2.1.1

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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Etapa 2.1.1.1

Consider the corresponding sign chart.

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

Etapa 2.1.1.2

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

Etapa 2.1.1.3

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

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

Etapa 2.1.1.4

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

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

Etapa 2.1.1.5

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

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

Etapa 2.1.1.6

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

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

Etapa 2.1.1.7

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

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

Etapa 2.1.1.8

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

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

Etapa 2.1.1.9

Add the terms together.

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

Etapa 2.1.2

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

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Etapa 2.1.2.1

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

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

Etapa 2.1.2.2

Simplifique o determinante.

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Etapa 2.1.2.2.1

Simplifique cada termo.

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Etapa 2.1.2.2.1.1

Multiplique $2$ por $- 1$ .

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

Etapa 2.1.2.2.1.2

Multiplique $- 3$ por $1$ .

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

Etapa 2.1.2.2.2

Subtraia $3$ de $- 2$ .

$- 4 \cdot - 5 - 3 | \begin{array}{c} - 1 & 1 \\ - 3 & - 1 \end{array} | + 1 | \begin{array}{c} - 1 & 2 \\ - 3 & 3 \end{array} |$

Etapa 2.1.3

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

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Etapa 2.1.3.1

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

$- 4 \cdot - 5 - 3 (- - 1 - (- 3 \cdot 1)) + 1 | \begin{array}{c} - 1 & 2 \\ - 3 & 3 \end{array} |$

Etapa 2.1.3.2

Simplifique o determinante.

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Etapa 2.1.3.2.1

Simplifique cada termo.

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Etapa 2.1.3.2.1.1

Multiplique $- 1$ por $- 1$ .

$- 4 \cdot - 5 - 3 (1 - (- 3 \cdot 1)) + 1 | \begin{array}{c} - 1 & 2 \\ - 3 & 3 \end{array} |$

Etapa 2.1.3.2.1.2

Multiplique $- (- 3 \cdot 1)$ .

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Etapa 2.1.3.2.1.2.1

Multiplique $- 3$ por $1$ .

$- 4 \cdot - 5 - 3 (1 - - 3) + 1 | \begin{array}{c} - 1 & 2 \\ - 3 & 3 \end{array} |$

Etapa 2.1.3.2.1.2.2

Multiplique $- 1$ por $- 3$ .

$- 4 \cdot - 5 - 3 (1 + 3) + 1 | \begin{array}{c} - 1 & 2 \\ - 3 & 3 \end{array} |$

Etapa 2.1.3.2.2

Some $1$ e $3$ .

$- 4 \cdot - 5 - 3 \cdot 4 + 1 | \begin{array}{c} - 1 & 2 \\ - 3 & 3 \end{array} |$

Etapa 2.1.4

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

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Etapa 2.1.4.1

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

$- 4 \cdot - 5 - 3 \cdot 4 + 1 (- 1 \cdot 3 - (- 3 \cdot 2))$

Etapa 2.1.4.2

Simplifique o determinante.

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Etapa 2.1.4.2.1

Simplifique cada termo.

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Etapa 2.1.4.2.1.1

Multiplique $- 1$ por $3$ .

$- 4 \cdot - 5 - 3 \cdot 4 + 1 (- 3 - (- 3 \cdot 2))$

Etapa 2.1.4.2.1.2

Multiplique $- (- 3 \cdot 2)$ .

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Etapa 2.1.4.2.1.2.1

Multiplique $- 3$ por $2$ .

$- 4 \cdot - 5 - 3 \cdot 4 + 1 (- 3 - - 6)$

Etapa 2.1.4.2.1.2.2

Multiplique $- 1$ por $- 6$ .

$- 4 \cdot - 5 - 3 \cdot 4 + 1 (- 3 + 6)$

Etapa 2.1.4.2.2

Some $- 3$ e $6$ .

$- 4 \cdot - 5 - 3 \cdot 4 + 1 \cdot 3$

Etapa 2.1.5

Simplifique o determinante.

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Etapa 2.1.5.1

Simplifique cada termo.

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Etapa 2.1.5.1.1

Multiplique $- 4$ por $- 5$ .

$20 - 3 \cdot 4 + 1 \cdot 3$

Etapa 2.1.5.1.2

Multiplique $- 3$ por $4$ .

$20 - 12 + 1 \cdot 3$

Etapa 2.1.5.1.3

Multiplique $3$ por $1$ .

$20 - 12 + 3$

Etapa 2.1.5.2

Subtraia $12$ de $20$ .

$8 + 3$

Etapa 2.1.5.3

Some $8$ e $3$ .

$11$

Etapa 2.2

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

Etapa 2.3

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

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

Etapa 2.4

Encontre a forma escalonada reduzida por linhas.

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Etapa 2.4.1

Multiply each element of $R_{1}$ by $- \frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

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Etapa 2.4.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

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

Etapa 2.4.1.2

Simplifique $R_{1}$ .

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

Etapa 2.4.2

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

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Etapa 2.4.2.1

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

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

Etapa 2.4.2.2

Simplifique $R_{2}$ .

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

Etapa 2.4.3

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

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Etapa 2.4.3.1

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

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

Etapa 2.4.3.2

Simplifique $R_{3}$ .

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

Etapa 2.4.4

Multiply each element of $R_{2}$ by $\frac{4}{5}$ to make the entry at $2, 2$ a $1$ .

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Etapa 2.4.4.1

Multiply each element of $R_{2}$ by $\frac{4}{5}$ to make the entry at $2, 2$ a $1$ .

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

Etapa 2.4.4.2

Simplifique $R_{2}$ .

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

Etapa 2.4.5

Perform the row operation $R_{3} = R_{3} - \frac{3}{4} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Etapa 2.4.5.1

Perform the row operation $R_{3} = R_{3} - \frac{3}{4} R_{2}$ to make the entry at $3, 2$ a $0$ .

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

Etapa 2.4.5.2

Simplifique $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{3}{4} & - \frac{1}{4} & - \frac{1}{4} & 0 & 0 \\ 0 & 1 & \frac{3}{5} & - \frac{1}{5} & \frac{4}{5} & 0 \\ 0 & 0 & - \frac{11}{5} & - \frac{3}{5} & - \frac{3}{5} & 1 \end{array}]$

Etapa 2.4.6

Multiply each element of $R_{3}$ by $- \frac{5}{11}$ to make the entry at $3, 3$ a $1$ .

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Etapa 2.4.6.1

Multiply each element of $R_{3}$ by $- \frac{5}{11}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{3}{4} & - \frac{1}{4} & - \frac{1}{4} & 0 & 0 \\ 0 & 1 & \frac{3}{5} & - \frac{1}{5} & \frac{4}{5} & 0 \\ - \frac{5}{11} \cdot 0 & - \frac{5}{11} \cdot 0 & - \frac{5}{11} (- \frac{11}{5}) & - \frac{5}{11} (- \frac{3}{5}) & - \frac{5}{11} (- \frac{3}{5}) & - \frac{5}{11} \cdot 1 \end{array}]$

Etapa 2.4.6.2

Simplifique $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{3}{4} & - \frac{1}{4} & - \frac{1}{4} & 0 & 0 \\ 0 & 1 & \frac{3}{5} & - \frac{1}{5} & \frac{4}{5} & 0 \\ 0 & 0 & 1 & \frac{3}{11} & \frac{3}{11} & - \frac{5}{11} \end{array}]$

Etapa 2.4.7

Perform the row operation $R_{2} = R_{2} - \frac{3}{5} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Etapa 2.4.7.1

Perform the row operation $R_{2} = R_{2} - \frac{3}{5} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - \frac{3}{4} & - \frac{1}{4} & - \frac{1}{4} & 0 & 0 \\ 0 - \frac{3}{5} \cdot 0 & 1 - \frac{3}{5} \cdot 0 & \frac{3}{5} - \frac{3}{5} \cdot 1 & - \frac{1}{5} - \frac{3}{5} \cdot \frac{3}{11} & \frac{4}{5} - \frac{3}{5} \cdot \frac{3}{11} & 0 - \frac{3}{5} (- \frac{5}{11}) \\ 0 & 0 & 1 & \frac{3}{11} & \frac{3}{11} & - \frac{5}{11} \end{array}]$

Etapa 2.4.7.2

Simplifique $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{3}{4} & - \frac{1}{4} & - \frac{1}{4} & 0 & 0 \\ 0 & 1 & 0 & - \frac{4}{11} & \frac{7}{11} & \frac{3}{11} \\ 0 & 0 & 1 & \frac{3}{11} & \frac{3}{11} & - \frac{5}{11} \end{array}]$

Etapa 2.4.8

Perform the row operation $R_{1} = R_{1} + \frac{1}{4} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Etapa 2.4.8.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{4} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 + \frac{1}{4} \cdot 0 & - \frac{3}{4} + \frac{1}{4} \cdot 0 & - \frac{1}{4} + \frac{1}{4} \cdot 1 & - \frac{1}{4} + \frac{1}{4} \cdot \frac{3}{11} & 0 + \frac{1}{4} \cdot \frac{3}{11} & 0 + \frac{1}{4} (- \frac{5}{11}) \\ 0 & 1 & 0 & - \frac{4}{11} & \frac{7}{11} & \frac{3}{11} \\ 0 & 0 & 1 & \frac{3}{11} & \frac{3}{11} & - \frac{5}{11} \end{array}]$

Etapa 2.4.8.2

Simplifique $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{3}{4} & 0 & - \frac{2}{11} & \frac{3}{44} & - \frac{5}{44} \\ 0 & 1 & 0 & - \frac{4}{11} & \frac{7}{11} & \frac{3}{11} \\ 0 & 0 & 1 & \frac{3}{11} & \frac{3}{11} & - \frac{5}{11} \end{array}]$

Etapa 2.4.9

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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Etapa 2.4.9.1

Perform the row operation $R_{1} = R_{1} + \frac{3}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{3}{4} \cdot 0 & - \frac{3}{4} + \frac{3}{4} \cdot 1 & 0 + \frac{3}{4} \cdot 0 & - \frac{2}{11} + \frac{3}{4} (- \frac{4}{11}) & \frac{3}{44} + \frac{3}{4} \cdot \frac{7}{11} & - \frac{5}{44} + \frac{3}{4} \cdot \frac{3}{11} \\ 0 & 1 & 0 & - \frac{4}{11} & \frac{7}{11} & \frac{3}{11} \\ 0 & 0 & 1 & \frac{3}{11} & \frac{3}{11} & - \frac{5}{11} \end{array}]$

Etapa 2.4.9.2

Simplifique $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & - \frac{5}{11} & \frac{6}{11} & \frac{1}{11} \\ 0 & 1 & 0 & - \frac{4}{11} & \frac{7}{11} & \frac{3}{11} \\ 0 & 0 & 1 & \frac{3}{11} & \frac{3}{11} & - \frac{5}{11} \end{array}]$

Etapa 2.5

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

$[\begin{array}{c} - \frac{5}{11} & \frac{6}{11} & \frac{1}{11} \\ - \frac{4}{11} & \frac{7}{11} & \frac{3}{11} \\ \frac{3}{11} & \frac{3}{11} & - \frac{5}{11} \end{array}]$

Etapa 3

Multiplique pela esquerda os dois lados da equação da matriz pela matriz inversa.

$([\begin{array}{c} - \frac{5}{11} & \frac{6}{11} & \frac{1}{11} \\ - \frac{4}{11} & \frac{7}{11} & \frac{3}{11} \\ \frac{3}{11} & \frac{3}{11} & - \frac{5}{11} \end{array}] \cdot [\begin{array}{c} - 4 & 3 & 1 \\ - 1 & 2 & 1 \\ - 3 & 3 & - 1 \end{array}]) \cdot [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - \frac{5}{11} & \frac{6}{11} & \frac{1}{11} \\ - \frac{4}{11} & \frac{7}{11} & \frac{3}{11} \\ \frac{3}{11} & \frac{3}{11} & - \frac{5}{11} \end{array}] \cdot [\begin{array}{c} 17 \\ 7 \\ 10 \end{array}]$

Etapa 4

Qualquer matriz multiplicada por seu inverso é sempre igual a $1$ . $A \cdot A^{- 1} = 1$ .

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - \frac{5}{11} & \frac{6}{11} & \frac{1}{11} \\ - \frac{4}{11} & \frac{7}{11} & \frac{3}{11} \\ \frac{3}{11} & \frac{3}{11} & - \frac{5}{11} \end{array}] \cdot [\begin{array}{c} 17 \\ 7 \\ 10 \end{array}]$

Etapa 5

Multiplique $[\begin{array}{c} - \frac{5}{11} & \frac{6}{11} & \frac{1}{11} \\ - \frac{4}{11} & \frac{7}{11} & \frac{3}{11} \\ \frac{3}{11} & \frac{3}{11} & - \frac{5}{11} \end{array}] [\begin{array}{c} 17 \\ 7 \\ 10 \end{array}]$ .

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Etapa 5.1

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 $3 \times 1$ .

Etapa 5.2

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

$[\begin{array}{c} - \frac{5}{11} \cdot 17 + \frac{6}{11} \cdot 7 + \frac{1}{11} \cdot 10 \\ - \frac{4}{11} \cdot 17 + \frac{7}{11} \cdot 7 + \frac{3}{11} \cdot 10 \\ \frac{3}{11} \cdot 17 + \frac{3}{11} \cdot 7 - \frac{5}{11} \cdot 10 \end{array}]$

Etapa 5.3

Simplifique cada elemento da matriz multiplicando todas as expressões.

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

Etapa 6

Simplifique os lados esquerdo e direito.

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

Etapa 7

Encontre a solução.

$x = - 3$

$y = 1$

$z = 2$