유한 수학 예제

$7 x - y + 2 z = - 4$ , $4 y + 3 z = 0$ , $2 x + 3 y + 3 z = - 1$

단계 1

Write the system as a matrix.

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

단계 2

기약 행 사다리꼴을 구합니다.

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단계 2.1

Multiply each element of $R_{1}$ by $\frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

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단계 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

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

단계 2.1.2

$R_{1}$ 을 간단히 합니다.

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

단계 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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단계 2.2.1

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

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

단계 2.2.2

$R_{3}$ 을 간단히 합니다.

$[\begin{array}{c} 1 & - \frac{1}{7} & \frac{2}{7} & - \frac{4}{7} \\ 0 & 4 & 3 & 0 \\ 0 & \frac{23}{7} & \frac{17}{7} & \frac{1}{7} \end{array}]$

단계 2.3

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

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단계 2.3.1

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{7} & \frac{2}{7} & - \frac{4}{7} \\ \frac{0}{4} & \frac{4}{4} & \frac{3}{4} & \frac{0}{4} \\ 0 & \frac{23}{7} & \frac{17}{7} & \frac{1}{7} \end{array}]$

단계 2.3.2

$R_{2}$ 을 간단히 합니다.

$[\begin{array}{c} 1 & - \frac{1}{7} & \frac{2}{7} & - \frac{4}{7} \\ 0 & 1 & \frac{3}{4} & 0 \\ 0 & \frac{23}{7} & \frac{17}{7} & \frac{1}{7} \end{array}]$

단계 2.4

Perform the row operation $R_{3} = R_{3} - \frac{23}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

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단계 2.4.1

Perform the row operation $R_{3} = R_{3} - \frac{23}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{7} & \frac{2}{7} & - \frac{4}{7} \\ 0 & 1 & \frac{3}{4} & 0 \\ 0 - \frac{23}{7} \cdot 0 & \frac{23}{7} - \frac{23}{7} \cdot 1 & \frac{17}{7} - \frac{23}{7} \cdot \frac{3}{4} & \frac{1}{7} - \frac{23}{7} \cdot 0 \end{array}]$

단계 2.4.2

$R_{3}$ 을 간단히 합니다.

$[\begin{array}{c} 1 & - \frac{1}{7} & \frac{2}{7} & - \frac{4}{7} \\ 0 & 1 & \frac{3}{4} & 0 \\ 0 & 0 & - \frac{1}{28} & \frac{1}{7} \end{array}]$

단계 2.5

Multiply each element of $R_{3}$ by $- 28$ to make the entry at $3, 3$ a $1$ .

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단계 2.5.1

Multiply each element of $R_{3}$ by $- 28$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{7} & \frac{2}{7} & - \frac{4}{7} \\ 0 & 1 & \frac{3}{4} & 0 \\ - 28 \cdot 0 & - 28 \cdot 0 & - 28 (- \frac{1}{28}) & - 28 (\frac{1}{7}) \end{array}]$

단계 2.5.2

$R_{3}$ 을 간단히 합니다.

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

단계 2.6

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

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단계 2.6.1

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

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

단계 2.6.2

$R_{2}$ 을 간단히 합니다.

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

단계 2.7

Perform the row operation $R_{1} = R_{1} - \frac{2}{7} R_{3}$ to make the entry at $1, 3$ a $0$ .

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단계 2.7.1

Perform the row operation $R_{1} = R_{1} - \frac{2}{7} R_{3}$ to make the entry at $1, 3$ a $0$ .

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

단계 2.7.2

$R_{1}$ 을 간단히 합니다.

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

단계 2.8

Perform the row operation $R_{1} = R_{1} + \frac{1}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

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단계 2.8.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

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

단계 2.8.2

$R_{1}$ 을 간단히 합니다.

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

단계 3

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

$x = 1$

$y = 3$

$z = - 4$

단계 4

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

$(1, 3, - 4)$