Matemática discreta Ejemplos

$3 x - 4 y + 7 z = 24$ , $y + 9 z = - 7$ , $- 2 x + 3 y = 12$

Paso 1

Write the system as a matrix.

$[\begin{array}{c} 3 & - 4 & 7 & 24 \\ 0 & 1 & 9 & - 7 \\ - 2 & 3 & 0 & 12 \end{array}]$

Paso 2

Obtén la forma escalonada reducida por filas.

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Paso 2.1

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

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Paso 2.1.1

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

$[\begin{array}{c} \frac{3}{3} & \frac{- 4}{3} & \frac{7}{3} & \frac{24}{3} \\ 0 & 1 & 9 & - 7 \\ - 2 & 3 & 0 & 12 \end{array}]$

Paso 2.1.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{7}{3} & 8 \\ 0 & 1 & 9 & - 7 \\ - 2 & 3 & 0 & 12 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{7}{3} & 8 \\ 0 & 1 & 9 & - 7 \\ - 2 + 2 \cdot 1 & 3 + 2 (- \frac{4}{3}) & 0 + 2 (\frac{7}{3}) & 12 + 2 \cdot 8 \end{array}]$

Paso 2.2.2

Simplifica $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{7}{3} & 8 \\ 0 & 1 & 9 & - 7 \\ 0 & \frac{1}{3} & \frac{14}{3} & 28 \end{array}]$

Paso 2.3

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

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Paso 2.3.1

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

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{7}{3} & 8 \\ 0 & 1 & 9 & - 7 \\ 0 - \frac{1}{3} \cdot 0 & \frac{1}{3} - \frac{1}{3} \cdot 1 & \frac{14}{3} - \frac{1}{3} \cdot 9 & 28 - \frac{1}{3} \cdot - 7 \end{array}]$

Paso 2.3.2

Simplifica $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{7}{3} & 8 \\ 0 & 1 & 9 & - 7 \\ 0 & 0 & \frac{5}{3} & \frac{91}{3} \end{array}]$

Paso 2.4

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

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Paso 2.4.1

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

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{7}{3} & 8 \\ 0 & 1 & 9 & - 7 \\ \frac{3}{5} \cdot 0 & \frac{3}{5} \cdot 0 & \frac{3}{5} \cdot \frac{5}{3} & \frac{3}{5} \cdot \frac{91}{3} \end{array}]$

Paso 2.4.2

Simplifica $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{7}{3} & 8 \\ 0 & 1 & 9 & - 7 \\ 0 & 0 & 1 & \frac{91}{5} \end{array}]$

Paso 2.5

Perform the row operation $R_{2} = R_{2} - 9 R_{3}$ to make the entry at $2, 3$ a $0$ .

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Paso 2.5.1

Perform the row operation $R_{2} = R_{2} - 9 R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{7}{3} & 8 \\ 0 - 9 \cdot 0 & 1 - 9 \cdot 0 & 9 - 9 \cdot 1 & - 7 - 9 (\frac{91}{5}) \\ 0 & 0 & 1 & \frac{91}{5} \end{array}]$

Paso 2.5.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{7}{3} & 8 \\ 0 & 1 & 0 & - \frac{854}{5} \\ 0 & 0 & 1 & \frac{91}{5} \end{array}]$

Paso 2.6

Perform the row operation $R_{1} = R_{1} - \frac{7}{3} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Paso 2.6.1

Perform the row operation $R_{1} = R_{1} - \frac{7}{3} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{7}{3} \cdot 0 & - \frac{4}{3} - \frac{7}{3} \cdot 0 & \frac{7}{3} - \frac{7}{3} \cdot 1 & 8 - \frac{7}{3} \cdot \frac{91}{5} \\ 0 & 1 & 0 & - \frac{854}{5} \\ 0 & 0 & 1 & \frac{91}{5} \end{array}]$

Paso 2.6.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & 0 & - \frac{517}{15} \\ 0 & 1 & 0 & - \frac{854}{5} \\ 0 & 0 & 1 & \frac{91}{5} \end{array}]$

Paso 2.7

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

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Paso 2.7.1

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

$[\begin{array}{c} 1 + \frac{4}{3} \cdot 0 & - \frac{4}{3} + \frac{4}{3} \cdot 1 & 0 + \frac{4}{3} \cdot 0 & - \frac{517}{15} + \frac{4}{3} (- \frac{854}{5}) \\ 0 & 1 & 0 & - \frac{854}{5} \\ 0 & 0 & 1 & \frac{91}{5} \end{array}]$

Paso 2.7.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & - \frac{1311}{5} \\ 0 & 1 & 0 & - \frac{854}{5} \\ 0 & 0 & 1 & \frac{91}{5} \end{array}]$

Paso 3

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

$x = - \frac{1311}{5}$

$y = - \frac{854}{5}$

$z = \frac{91}{5}$

Paso 4

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

$(- \frac{1311}{5}, - \frac{854}{5}, \frac{91}{5})$