有限数学示例

$3 x + 3 y + 2 z = 3405$ , $1 x + 2 y + 1 z = 1715$ , $2 x + 1 y + 2 z = 1930$

解题步骤 1

Move variables to the left and constant terms to the right.

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解题步骤 1.1

化简每一项。

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解题步骤 1.1.1

将 $x$ 乘以 $1$ 。

$3 x + 3 y + 2 z = 3405$

$x + 2 y + 1 z = 1715$

$2 x + 1 y + 2 z = 1930$

解题步骤 1.1.2

将 $z$ 乘以 $1$ 。

$3 x + 3 y + 2 z = 3405$

$x + 2 y + z = 1715$

$2 x + 1 y + 2 z = 1930$

$3 x + 3 y + 2 z = 3405$

$x + 2 y + z = 1715$

$2 x + 1 y + 2 z = 1930$

解题步骤 1.2

将 $y$ 乘以 $1$ 。

$3 x + 3 y + 2 z = 3405$

$x + 2 y + z = 1715$

$2 x + y + 2 z = 1930$

$3 x + 3 y + 2 z = 3405$

$x + 2 y + z = 1715$

$2 x + y + 2 z = 1930$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} 3 & 3 & 2 & 3405 \\ 1 & 2 & 1 & 1715 \\ 2 & 1 & 2 & 1930 \end{array}]$

解题步骤 3

求行简化阶梯形矩阵。

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解题步骤 3.1

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

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解题步骤 3.1.1

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

$[\begin{array}{c} \frac{3}{3} & \frac{3}{3} & \frac{2}{3} & \frac{3405}{3} \\ 1 & 2 & 1 & 1715 \\ 2 & 1 & 2 & 1930 \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 1 & \frac{2}{3} & 1135 \\ 1 & 2 & 1 & 1715 \\ 2 & 1 & 2 & 1930 \end{array}]$

解题步骤 3.2

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

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解题步骤 3.2.1

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

$[\begin{array}{c} 1 & 1 & \frac{2}{3} & 1135 \\ 1 - 1 & 2 - 1 & 1 - \frac{2}{3} & 1715 - 1135 \\ 2 & 1 & 2 & 1930 \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 1 & \frac{2}{3} & 1135 \\ 0 & 1 & \frac{1}{3} & 580 \\ 2 & 1 & 2 & 1930 \end{array}]$

解题步骤 3.3

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

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解题步骤 3.3.1

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

$[\begin{array}{c} 1 & 1 & \frac{2}{3} & 1135 \\ 0 & 1 & \frac{1}{3} & 580 \\ 2 - 2 \cdot 1 & 1 - 2 \cdot 1 & 2 - 2 (\frac{2}{3}) & 1930 - 2 \cdot 1135 \end{array}]$

解题步骤 3.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & 1 & \frac{2}{3} & 1135 \\ 0 & 1 & \frac{1}{3} & 580 \\ 0 & - 1 & \frac{2}{3} & - 340 \end{array}]$

解题步骤 3.4

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

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解题步骤 3.4.1

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

$[\begin{array}{c} 1 & 1 & \frac{2}{3} & 1135 \\ 0 & 1 & \frac{1}{3} & 580 \\ 0 + 0 & - 1 + 1 \cdot 1 & \frac{2}{3} + \frac{1}{3} & - 340 + 1 \cdot 580 \end{array}]$

解题步骤 3.4.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & 1 & \frac{2}{3} & 1135 \\ 0 & 1 & \frac{1}{3} & 580 \\ 0 & 0 & 1 & 240 \end{array}]$

解题步骤 3.5

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

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解题步骤 3.5.1

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

$[\begin{array}{c} 1 & 1 & \frac{2}{3} & 1135 \\ 0 - \frac{1}{3} \cdot 0 & 1 - \frac{1}{3} \cdot 0 & \frac{1}{3} - \frac{1}{3} \cdot 1 & 580 - \frac{1}{3} \cdot 240 \\ 0 & 0 & 1 & 240 \end{array}]$

解题步骤 3.5.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 1 & \frac{2}{3} & 1135 \\ 0 & 1 & 0 & 500 \\ 0 & 0 & 1 & 240 \end{array}]$

解题步骤 3.6

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

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解题步骤 3.6.1

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

$[\begin{array}{c} 1 - \frac{2}{3} \cdot 0 & 1 - \frac{2}{3} \cdot 0 & \frac{2}{3} - \frac{2}{3} \cdot 1 & 1135 - \frac{2}{3} \cdot 240 \\ 0 & 1 & 0 & 500 \\ 0 & 0 & 1 & 240 \end{array}]$

解题步骤 3.6.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 1 & 0 & 975 \\ 0 & 1 & 0 & 500 \\ 0 & 0 & 1 & 240 \end{array}]$

解题步骤 3.7

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

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解题步骤 3.7.1

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

$[\begin{array}{c} 1 - 0 & 1 - 1 & 0 - 0 & 975 - 500 \\ 0 & 1 & 0 & 500 \\ 0 & 0 & 1 & 240 \end{array}]$

解题步骤 3.7.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & 475 \\ 0 & 1 & 0 & 500 \\ 0 & 0 & 1 & 240 \end{array}]$

解题步骤 4

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

$x = 475$

$y = 500$

$z = 240$

解题步骤 5

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

$(475, 500, 240)$

有限数学 示例

有限数学示例