유한 수학 예제

$x + 3 y + 2 z = 8$ , $3 x + 1 y + 3 z = - 10$ , $- 2 x - 2 y - z = 10$

단계 1

$y$ 에 $1$ 을 곱합니다.

$x + 3 y + 2 z = 8$

$3 x + y + 3 z = - 10$

$- 2 x - 2 y - z = 10$

단계 2

Write the system as a matrix.

$[\begin{array}{c} 1 & 3 & 2 & 8 \\ 3 & 1 & 3 & - 10 \\ - 2 & - 2 & - 1 & 10 \end{array}]$

단계 3

기약 행 사다리꼴을 구합니다.

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단계 3.1

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

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단계 3.1.1

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

$[\begin{array}{c} 1 & 3 & 2 & 8 \\ 3 - 3 \cdot 1 & 1 - 3 \cdot 3 & 3 - 3 \cdot 2 & - 10 - 3 \cdot 8 \\ - 2 & - 2 & - 1 & 10 \end{array}]$

단계 3.1.2

$R_{2}$ 을 간단히 합니다.

$[\begin{array}{c} 1 & 3 & 2 & 8 \\ 0 & - 8 & - 3 & - 34 \\ - 2 & - 2 & - 1 & 10 \end{array}]$

단계 3.2

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

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단계 3.2.1

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

$[\begin{array}{c} 1 & 3 & 2 & 8 \\ 0 & - 8 & - 3 & - 34 \\ - 2 + 2 \cdot 1 & - 2 + 2 \cdot 3 & - 1 + 2 \cdot 2 & 10 + 2 \cdot 8 \end{array}]$

단계 3.2.2

$R_{3}$ 을 간단히 합니다.

$[\begin{array}{c} 1 & 3 & 2 & 8 \\ 0 & - 8 & - 3 & - 34 \\ 0 & 4 & 3 & 26 \end{array}]$

단계 3.3

Multiply each element of $R_{2}$ by $- \frac{1}{8}$ to make the entry at $2, 2$ a $1$ .

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단계 3.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{8}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 3 & 2 & 8 \\ - \frac{1}{8} \cdot 0 & - \frac{1}{8} \cdot - 8 & - \frac{1}{8} \cdot - 3 & - \frac{1}{8} \cdot - 34 \\ 0 & 4 & 3 & 26 \end{array}]$

단계 3.3.2

$R_{2}$ 을 간단히 합니다.

$[\begin{array}{c} 1 & 3 & 2 & 8 \\ 0 & 1 & \frac{3}{8} & \frac{17}{4} \\ 0 & 4 & 3 & 26 \end{array}]$

단계 3.4

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

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단계 3.4.1

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

$[\begin{array}{c} 1 & 3 & 2 & 8 \\ 0 & 1 & \frac{3}{8} & \frac{17}{4} \\ 0 - 4 \cdot 0 & 4 - 4 \cdot 1 & 3 - 4 (\frac{3}{8}) & 26 - 4 (\frac{17}{4}) \end{array}]$

단계 3.4.2

$R_{3}$ 을 간단히 합니다.

$[\begin{array}{c} 1 & 3 & 2 & 8 \\ 0 & 1 & \frac{3}{8} & \frac{17}{4} \\ 0 & 0 & \frac{3}{2} & 9 \end{array}]$

단계 3.5

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

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단계 3.5.1

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

$[\begin{array}{c} 1 & 3 & 2 & 8 \\ 0 & 1 & \frac{3}{8} & \frac{17}{4} \\ \frac{2}{3} \cdot 0 & \frac{2}{3} \cdot 0 & \frac{2}{3} \cdot \frac{3}{2} & \frac{2}{3} \cdot 9 \end{array}]$

단계 3.5.2

$R_{3}$ 을 간단히 합니다.

$[\begin{array}{c} 1 & 3 & 2 & 8 \\ 0 & 1 & \frac{3}{8} & \frac{17}{4} \\ 0 & 0 & 1 & 6 \end{array}]$

단계 3.6

Perform the row operation $R_{2} = R_{2} - \frac{3}{8} R_{3}$ to make the entry at $2, 3$ a $0$ .

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단계 3.6.1

Perform the row operation $R_{2} = R_{2} - \frac{3}{8} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 3 & 2 & 8 \\ 0 - \frac{3}{8} \cdot 0 & 1 - \frac{3}{8} \cdot 0 & \frac{3}{8} - \frac{3}{8} \cdot 1 & \frac{17}{4} - \frac{3}{8} \cdot 6 \\ 0 & 0 & 1 & 6 \end{array}]$

단계 3.6.2

$R_{2}$ 을 간단히 합니다.

$[\begin{array}{c} 1 & 3 & 2 & 8 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 6 \end{array}]$

단계 3.7

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

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단계 3.7.1

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

$[\begin{array}{c} 1 - 2 \cdot 0 & 3 - 2 \cdot 0 & 2 - 2 \cdot 1 & 8 - 2 \cdot 6 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 6 \end{array}]$

단계 3.7.2

$R_{1}$ 을 간단히 합니다.

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

단계 3.8

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

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단계 3.8.1

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

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

단계 3.8.2

$R_{1}$ 을 간단히 합니다.

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

단계 4

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

$x = - 10$

$y = 2$

$z = 6$

단계 5

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

$(- 10, 2, 6)$