Precálculo Ejemplos

$x - 2 y + 3 z = 6$ , $2 x + y + z = 2$ , $- 2 x + 4 y - 6 z = - 12$

Paso 1

Write the system as a matrix.

$[\begin{array}{c} 1 & - 2 & 3 & 6 \\ 2 & 1 & 1 & 2 \\ - 2 & 4 & - 6 & - 12 \end{array}]$

Paso 2

Obtén la forma escalonada reducida por filas.

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Paso 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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Paso 2.1.1

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

$[\begin{array}{c} 1 & - 2 & 3 & 6 \\ 2 - 2 \cdot 1 & 1 - 2 \cdot - 2 & 1 - 2 \cdot 3 & 2 - 2 \cdot 6 \\ - 2 & 4 & - 6 & - 12 \end{array}]$

Paso 2.1.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & 3 & 6 \\ 0 & 5 & - 5 & - 10 \\ - 2 & 4 & - 6 & - 12 \end{array}]$

Paso 2.2

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

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Paso 2.2.1

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

$[\begin{array}{c} 1 & - 2 & 3 & 6 \\ 0 & 5 & - 5 & - 10 \\ - 2 + 2 \cdot 1 & 4 + 2 \cdot - 2 & - 6 + 2 \cdot 3 & - 12 + 2 \cdot 6 \end{array}]$

Paso 2.2.2

Simplifica $R_{3}$ .

$[\begin{array}{c} 1 & - 2 & 3 & 6 \\ 0 & 5 & - 5 & - 10 \\ 0 & 0 & 0 & 0 \end{array}]$

Paso 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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Paso 2.3.1

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

$[\begin{array}{c} 1 & - 2 & 3 & 6 \\ \frac{0}{5} & \frac{5}{5} & \frac{- 5}{5} & \frac{- 10}{5} \\ 0 & 0 & 0 & 0 \end{array}]$

Paso 2.3.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & 3 & 6 \\ 0 & 1 & - 1 & - 2 \\ 0 & 0 & 0 & 0 \end{array}]$

Paso 2.4

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

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Paso 2.4.1

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

$[\begin{array}{c} 1 + 2 \cdot 0 & - 2 + 2 \cdot 1 & 3 + 2 \cdot - 1 & 6 + 2 \cdot - 2 \\ 0 & 1 & - 1 & - 2 \\ 0 & 0 & 0 & 0 \end{array}]$

Paso 2.4.2

Simplifica $R_{1}$ .

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

Paso 3

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

$x + z = 2$

$y - z = - 2$

$0 = 0$

Paso 4

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

$(2 - z, - 2 + z, z)$