유한 수학 예제

$30 x + 20 y + 20 z = 390$ , $10 x + 20 y + 60 z = 210$

단계 1

Write the system as a matrix.

$[\begin{array}{c} 30 & 20 & 20 & 390 \\ 10 & 20 & 60 & 210 \end{array}]$

단계 2

기약 행 사다리꼴을 구합니다.

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단계 2.1

Multiply each element of $R_{1}$ by $\frac{1}{30}$ to make the entry at $1, 1$ a $1$ .

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단계 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{30}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{30}{30} & \frac{20}{30} & \frac{20}{30} & \frac{390}{30} \\ 10 & 20 & 60 & 210 \end{array}]$

단계 2.1.2

$R_{1}$ 을 간단히 합니다.

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{2}{3} & 13 \\ 10 & 20 & 60 & 210 \end{array}]$

단계 2.2

Perform the row operation $R_{2} = R_{2} - 10 R_{1}$ to make the entry at $2, 1$ a $0$ .

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단계 2.2.1

Perform the row operation $R_{2} = R_{2} - 10 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{2}{3} & 13 \\ 10 - 10 \cdot 1 & 20 - 10 (\frac{2}{3}) & 60 - 10 (\frac{2}{3}) & 210 - 10 \cdot 13 \end{array}]$

단계 2.2.2

$R_{2}$ 을 간단히 합니다.

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{2}{3} & 13 \\ 0 & \frac{40}{3} & \frac{160}{3} & 80 \end{array}]$

단계 2.3

Multiply each element of $R_{2}$ by $\frac{3}{40}$ to make the entry at $2, 2$ a $1$ .

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단계 2.3.1

Multiply each element of $R_{2}$ by $\frac{3}{40}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{2}{3} & 13 \\ \frac{3}{40} \cdot 0 & \frac{3}{40} \cdot \frac{40}{3} & \frac{3}{40} \cdot \frac{160}{3} & \frac{3}{40} \cdot 80 \end{array}]$

단계 2.3.2

$R_{2}$ 을 간단히 합니다.

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{2}{3} & 13 \\ 0 & 1 & 4 & 6 \end{array}]$

단계 2.4

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

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단계 2.4.1

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

$[\begin{array}{c} 1 - \frac{2}{3} \cdot 0 & \frac{2}{3} - \frac{2}{3} \cdot 1 & \frac{2}{3} - \frac{2}{3} \cdot 4 & 13 - \frac{2}{3} \cdot 6 \\ 0 & 1 & 4 & 6 \end{array}]$

단계 2.4.2

$R_{1}$ 을 간단히 합니다.

$[\begin{array}{c} 1 & 0 & - 2 & 9 \\ 0 & 1 & 4 & 6 \end{array}]$

단계 3

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

$x - 2 z = 9$

$y + 4 z = 6$

단계 4

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

$(9 + 2 z, 6 - 4 z, z)$