Precálculo Ejemplos

$2 x + 3 y = 11$ , $3 x + 3 y = 9$ , $x + y + 3 z = 1$

Paso 1

Write the system as a matrix.

$[\begin{array}{c} 2 & 3 & 0 & 11 \\ 3 & 3 & 0 & 9 \\ 1 & 1 & 3 & 1 \end{array}]$

Paso 2

Obtén la forma escalonada reducida por filas.

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Paso 2.1

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

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Paso 2.1.1

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

$[\begin{array}{c} \frac{2}{2} & \frac{3}{2} & \frac{0}{2} & \frac{11}{2} \\ 3 & 3 & 0 & 9 \\ 1 & 1 & 3 & 1 \end{array}]$

Paso 2.1.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 0 & \frac{11}{2} \\ 3 & 3 & 0 & 9 \\ 1 & 1 & 3 & 1 \end{array}]$

Paso 2.2

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

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Paso 2.2.1

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

$[\begin{array}{c} 1 & \frac{3}{2} & 0 & \frac{11}{2} \\ 3 - 3 \cdot 1 & 3 - 3 (\frac{3}{2}) & 0 - 3 \cdot 0 & 9 - 3 (\frac{11}{2}) \\ 1 & 1 & 3 & 1 \end{array}]$

Paso 2.2.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 0 & \frac{11}{2} \\ 0 & - \frac{3}{2} & 0 & - \frac{15}{2} \\ 1 & 1 & 3 & 1 \end{array}]$

Paso 2.3

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

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Paso 2.3.1

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

$[\begin{array}{c} 1 & \frac{3}{2} & 0 & \frac{11}{2} \\ 0 & - \frac{3}{2} & 0 & - \frac{15}{2} \\ 1 - 1 & 1 - \frac{3}{2} & 3 - 0 & 1 - \frac{11}{2} \end{array}]$

Paso 2.3.2

Simplifica $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 0 & \frac{11}{2} \\ 0 & - \frac{3}{2} & 0 & - \frac{15}{2} \\ 0 & - \frac{1}{2} & 3 & - \frac{9}{2} \end{array}]$

Paso 2.4

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

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Paso 2.4.1

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

$[\begin{array}{c} 1 & \frac{3}{2} & 0 & \frac{11}{2} \\ - \frac{2}{3} \cdot 0 & - \frac{2}{3} (- \frac{3}{2}) & - \frac{2}{3} \cdot 0 & - \frac{2}{3} (- \frac{15}{2}) \\ 0 & - \frac{1}{2} & 3 & - \frac{9}{2} \end{array}]$

Paso 2.4.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 0 & \frac{11}{2} \\ 0 & 1 & 0 & 5 \\ 0 & - \frac{1}{2} & 3 & - \frac{9}{2} \end{array}]$

Paso 2.5

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

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Paso 2.5.1

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

$[\begin{array}{c} 1 & \frac{3}{2} & 0 & \frac{11}{2} \\ 0 & 1 & 0 & 5 \\ 0 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & 3 + \frac{1}{2} \cdot 0 & - \frac{9}{2} + \frac{1}{2} \cdot 5 \end{array}]$

Paso 2.5.2

Simplifica $R_{3}$ .

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

Paso 2.6

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

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Paso 2.6.1

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

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

Paso 2.6.2

Simplifica $R_{3}$ .

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

Paso 2.7

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

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Paso 2.7.1

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

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

Paso 2.7.2

Simplifica $R_{1}$ .

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

Paso 3

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

$x = - 2$

$y = 5$

$z = - \frac{2}{3}$

Paso 4

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

$(- 2, 5, - \frac{2}{3})$