Pré-cálculo Exemplos

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

Etapa 1

Write the system as a matrix.

$[\begin{array}{c} 1 & - 2 & 3 & 7 \\ 2 & 1 & 1 & 4 \\ - 3 & 2 & - 2 & - 10 \end{array}]$

Etapa 2

Encontre a forma escalonada reduzida por linhas.

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Etapa 2.1

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 2.1.1

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

$[\begin{array}{c} 1 & - 2 & 3 & 7 \\ 2 - 2 \cdot 1 & 1 - 2 \cdot - 2 & 1 - 2 \cdot 3 & 4 - 2 \cdot 7 \\ - 3 & 2 & - 2 & - 10 \end{array}]$

Etapa 2.1.2

Simplifique $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & 3 & 7 \\ 0 & 5 & - 5 & - 10 \\ - 3 & 2 & - 2 & - 10 \end{array}]$

Etapa 2.2

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 2.2.1

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

$[\begin{array}{c} 1 & - 2 & 3 & 7 \\ 0 & 5 & - 5 & - 10 \\ - 3 + 3 \cdot 1 & 2 + 3 \cdot - 2 & - 2 + 3 \cdot 3 & - 10 + 3 \cdot 7 \end{array}]$

Etapa 2.2.2

Simplifique $R_{3}$ .

$[\begin{array}{c} 1 & - 2 & 3 & 7 \\ 0 & 5 & - 5 & - 10 \\ 0 & - 4 & 7 & 11 \end{array}]$

Etapa 2.3

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

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Etapa 2.3.1

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

$[\begin{array}{c} 1 & - 2 & 3 & 7 \\ \frac{0}{5} & \frac{5}{5} & \frac{- 5}{5} & \frac{- 10}{5} \\ 0 & - 4 & 7 & 11 \end{array}]$

Etapa 2.3.2

Simplifique $R_{2}$ .

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

Etapa 2.4

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

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Etapa 2.4.1

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

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

Etapa 2.4.2

Simplifique $R_{3}$ .

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

Etapa 2.5

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

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Etapa 2.5.1

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

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

Etapa 2.5.2

Simplifique $R_{3}$ .

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

Etapa 2.6

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

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Etapa 2.6.1

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

$[\begin{array}{c} 1 & - 2 & 3 & 7 \\ 0 + 0 & 1 + 0 & - 1 + 1 \cdot 1 & - 2 + 1 \cdot 1 \\ 0 & 0 & 1 & 1 \end{array}]$

Etapa 2.6.2

Simplifique $R_{2}$ .

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

Etapa 2.7

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

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Etapa 2.7.1

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

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

Etapa 2.7.2

Simplifique $R_{1}$ .

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

Etapa 2.8

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

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Etapa 2.8.1

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

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

Etapa 2.8.2

Simplifique $R_{1}$ .

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

Etapa 3

Use the result matrix to declare the final solution to the system of equations.

$x = 2$

$y = - 1$

$z = 1$

Etapa 4

The solution is the set of ordered pairs that make the system true.

$(2, - 1, 1)$