유한 수학 예제

$x + y - z = 7$ , $- x - 5 y - 11 z = - 43$ , $2 x - y + 3 z = 1$

단계 1

Write the system as a matrix.

$[\begin{array}{c} 1 & 1 & - 1 & 7 \\ - 1 & - 5 & - 11 & - 43 \\ 2 & - 1 & 3 & 1 \end{array}]$

단계 2

기약 행 사다리꼴을 구합니다.

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단계 2.1

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

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단계 2.1.1

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

$[\begin{array}{c} 1 & 1 & - 1 & 7 \\ - 1 + 1 \cdot 1 & - 5 + 1 \cdot 1 & - 11 - 1 & - 43 + 1 \cdot 7 \\ 2 & - 1 & 3 & 1 \end{array}]$

단계 2.1.2

$R_{2}$ 을 간단히 합니다.

$[\begin{array}{c} 1 & 1 & - 1 & 7 \\ 0 & - 4 & - 12 & - 36 \\ 2 & - 1 & 3 & 1 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & 1 & - 1 & 7 \\ 0 & - 4 & - 12 & - 36 \\ 2 - 2 \cdot 1 & - 1 - 2 \cdot 1 & 3 - 2 \cdot - 1 & 1 - 2 \cdot 7 \end{array}]$

단계 2.2.2

$R_{3}$ 을 간단히 합니다.

$[\begin{array}{c} 1 & 1 & - 1 & 7 \\ 0 & - 4 & - 12 & - 36 \\ 0 & - 3 & 5 & - 13 \end{array}]$

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 $1$ .

$[\begin{array}{c} 1 & 1 & - 1 & 7 \\ - \frac{1}{4} \cdot 0 & - \frac{1}{4} \cdot - 4 & - \frac{1}{4} \cdot - 12 & - \frac{1}{4} \cdot - 36 \\ 0 & - 3 & 5 & - 13 \end{array}]$

단계 2.3.2

$R_{2}$ 을 간단히 합니다.

$[\begin{array}{c} 1 & 1 & - 1 & 7 \\ 0 & 1 & 3 & 9 \\ 0 & - 3 & 5 & - 13 \end{array}]$

단계 2.4

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

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단계 2.4.1

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

$[\begin{array}{c} 1 & 1 & - 1 & 7 \\ 0 & 1 & 3 & 9 \\ 0 + 3 \cdot 0 & - 3 + 3 \cdot 1 & 5 + 3 \cdot 3 & - 13 + 3 \cdot 9 \end{array}]$

단계 2.4.2

$R_{3}$ 을 간단히 합니다.

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

단계 2.5

Multiply each element of $R_{3}$ by $\frac{1}{14}$ to make the entry at $3, 3$ a $1$ .

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단계 2.5.1

Multiply each element of $R_{3}$ by $\frac{1}{14}$ to make the entry at $3, 3$ a $1$ .

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

단계 2.5.2

$R_{3}$ 을 간단히 합니다.

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

단계 2.6

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

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단계 2.6.1

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

$[\begin{array}{c} 1 & 1 & - 1 & 7 \\ 0 - 3 \cdot 0 & 1 - 3 \cdot 0 & 3 - 3 \cdot 1 & 9 - 3 \cdot 1 \\ 0 & 0 & 1 & 1 \end{array}]$

단계 2.6.2

$R_{2}$ 을 간단히 합니다.

$[\begin{array}{c} 1 & 1 & - 1 & 7 \\ 0 & 1 & 0 & 6 \\ 0 & 0 & 1 & 1 \end{array}]$

단계 2.7

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

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단계 2.7.1

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

$[\begin{array}{c} 1 + 0 & 1 + 0 & - 1 + 1 \cdot 1 & 7 + 1 \cdot 1 \\ 0 & 1 & 0 & 6 \\ 0 & 0 & 1 & 1 \end{array}]$

단계 2.7.2

$R_{1}$ 을 간단히 합니다.

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

단계 2.8

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

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단계 2.8.1

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

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

단계 2.8.2

$R_{1}$ 을 간단히 합니다.

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

단계 3

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

$x = 2$

$y = 6$

$z = 1$

단계 4

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

$(2, 6, 1)$