有限数学示例

$x + 2 y + z = 72$ , $2 x + y + z = 66$ , $3 x + y + 2 z = 102$ , $p - 60 x - 48 y - 42 z = 0$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 0 & 1 & 2 & 1 & 72 \\ 0 & 2 & 1 & 1 & 66 \\ 0 & 3 & 1 & 2 & 102 \\ 1 & - 60 & - 48 & - 42 & 0 \end{array}]$

解题步骤 2

求行简化阶梯形矩阵。

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解题步骤 2.1

Swap $R_{4}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} 1 & - 60 & - 48 & - 42 & 0 \\ 0 & 2 & 1 & 1 & 66 \\ 0 & 3 & 1 & 2 & 102 \\ 0 & 1 & 2 & 1 & 72 \end{array}]$

解题步骤 2.2

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

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解题步骤 2.2.1

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

$[\begin{array}{c} 1 & - 60 & - 48 & - 42 & 0 \\ \frac{0}{2} & \frac{2}{2} & \frac{1}{2} & \frac{1}{2} & \frac{66}{2} \\ 0 & 3 & 1 & 2 & 102 \\ 0 & 1 & 2 & 1 & 72 \end{array}]$

解题步骤 2.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 60 & - 48 & - 42 & 0 \\ 0 & 1 & \frac{1}{2} & \frac{1}{2} & 33 \\ 0 & 3 & 1 & 2 & 102 \\ 0 & 1 & 2 & 1 & 72 \end{array}]$

解题步骤 2.3

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

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解题步骤 2.3.1

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

$[\begin{array}{c} 1 & - 60 & - 48 & - 42 & 0 \\ 0 & 1 & \frac{1}{2} & \frac{1}{2} & 33 \\ 0 - 3 \cdot 0 & 3 - 3 \cdot 1 & 1 - 3 (\frac{1}{2}) & 2 - 3 (\frac{1}{2}) & 102 - 3 \cdot 33 \\ 0 & 1 & 2 & 1 & 72 \end{array}]$

解题步骤 2.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 60 & - 48 & - 42 & 0 \\ 0 & 1 & \frac{1}{2} & \frac{1}{2} & 33 \\ 0 & 0 & - \frac{1}{2} & \frac{1}{2} & 3 \\ 0 & 1 & 2 & 1 & 72 \end{array}]$

解题步骤 2.4

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

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解题步骤 2.4.1

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

$[\begin{array}{c} 1 & - 60 & - 48 & - 42 & 0 \\ 0 & 1 & \frac{1}{2} & \frac{1}{2} & 33 \\ 0 & 0 & - \frac{1}{2} & \frac{1}{2} & 3 \\ 0 - 0 & 1 - 1 & 2 - \frac{1}{2} & 1 - \frac{1}{2} & 72 - 33 \end{array}]$

解题步骤 2.4.2

化简 $R_{4}$ 。

$[\begin{array}{c} 1 & - 60 & - 48 & - 42 & 0 \\ 0 & 1 & \frac{1}{2} & \frac{1}{2} & 33 \\ 0 & 0 & - \frac{1}{2} & \frac{1}{2} & 3 \\ 0 & 0 & \frac{3}{2} & \frac{1}{2} & 39 \end{array}]$

解题步骤 2.5

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

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解题步骤 2.5.1

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

$[\begin{array}{c} 1 & - 60 & - 48 & - 42 & 0 \\ 0 & 1 & \frac{1}{2} & \frac{1}{2} & 33 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 (- \frac{1}{2}) & - 2 (\frac{1}{2}) & - 2 \cdot 3 \\ 0 & 0 & \frac{3}{2} & \frac{1}{2} & 39 \end{array}]$

解题步骤 2.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 60 & - 48 & - 42 & 0 \\ 0 & 1 & \frac{1}{2} & \frac{1}{2} & 33 \\ 0 & 0 & 1 & - 1 & - 6 \\ 0 & 0 & \frac{3}{2} & \frac{1}{2} & 39 \end{array}]$

解题步骤 2.6

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

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解题步骤 2.6.1

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

$[\begin{array}{c} 1 & - 60 & - 48 & - 42 & 0 \\ 0 & 1 & \frac{1}{2} & \frac{1}{2} & 33 \\ 0 & 0 & 1 & - 1 & - 6 \\ 0 - \frac{3}{2} \cdot 0 & 0 - \frac{3}{2} \cdot 0 & \frac{3}{2} - \frac{3}{2} \cdot 1 & \frac{1}{2} - \frac{3}{2} \cdot - 1 & 39 - \frac{3}{2} \cdot - 6 \end{array}]$

解题步骤 2.6.2

化简 $R_{4}$ 。

$[\begin{array}{c} 1 & - 60 & - 48 & - 42 & 0 \\ 0 & 1 & \frac{1}{2} & \frac{1}{2} & 33 \\ 0 & 0 & 1 & - 1 & - 6 \\ 0 & 0 & 0 & 2 & 48 \end{array}]$

解题步骤 2.7

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

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解题步骤 2.7.1

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

$[\begin{array}{c} 1 & - 60 & - 48 & - 42 & 0 \\ 0 & 1 & \frac{1}{2} & \frac{1}{2} & 33 \\ 0 & 0 & 1 & - 1 & - 6 \\ \frac{0}{2} & \frac{0}{2} & \frac{0}{2} & \frac{2}{2} & \frac{48}{2} \end{array}]$

解题步骤 2.7.2

化简 $R_{4}$ 。

$[\begin{array}{c} 1 & - 60 & - 48 & - 42 & 0 \\ 0 & 1 & \frac{1}{2} & \frac{1}{2} & 33 \\ 0 & 0 & 1 & - 1 & - 6 \\ 0 & 0 & 0 & 1 & 24 \end{array}]$

解题步骤 2.8

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

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解题步骤 2.8.1

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

$[\begin{array}{c} 1 & - 60 & - 48 & - 42 & 0 \\ 0 & 1 & \frac{1}{2} & \frac{1}{2} & 33 \\ 0 + 0 & 0 + 0 & 1 + 0 & - 1 + 1 \cdot 1 & - 6 + 1 \cdot 24 \\ 0 & 0 & 0 & 1 & 24 \end{array}]$

解题步骤 2.8.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 60 & - 48 & - 42 & 0 \\ 0 & 1 & \frac{1}{2} & \frac{1}{2} & 33 \\ 0 & 0 & 1 & 0 & 18 \\ 0 & 0 & 0 & 1 & 24 \end{array}]$

解题步骤 2.9

Perform the row operation $R_{2} = R_{2} - \frac{1}{2} R_{4}$ to make the entry at $2, 4$ a $0$ .

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解题步骤 2.9.1

Perform the row operation $R_{2} = R_{2} - \frac{1}{2} R_{4}$ to make the entry at $2, 4$ a $0$ .

$[\begin{array}{c} 1 & - 60 & - 48 & - 42 & 0 \\ 0 - \frac{1}{2} \cdot 0 & 1 - \frac{1}{2} \cdot 0 & \frac{1}{2} - \frac{1}{2} \cdot 0 & \frac{1}{2} - \frac{1}{2} \cdot 1 & 33 - \frac{1}{2} \cdot 24 \\ 0 & 0 & 1 & 0 & 18 \\ 0 & 0 & 0 & 1 & 24 \end{array}]$

解题步骤 2.9.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 60 & - 48 & - 42 & 0 \\ 0 & 1 & \frac{1}{2} & 0 & 21 \\ 0 & 0 & 1 & 0 & 18 \\ 0 & 0 & 0 & 1 & 24 \end{array}]$

解题步骤 2.10

Perform the row operation $R_{1} = R_{1} + 42 R_{4}$ to make the entry at $1, 4$ a $0$ .

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解题步骤 2.10.1

Perform the row operation $R_{1} = R_{1} + 42 R_{4}$ to make the entry at $1, 4$ a $0$ .

$[\begin{array}{c} 1 + 42 \cdot 0 & - 60 + 42 \cdot 0 & - 48 + 42 \cdot 0 & - 42 + 42 \cdot 1 & 0 + 42 \cdot 24 \\ 0 & 1 & \frac{1}{2} & 0 & 21 \\ 0 & 0 & 1 & 0 & 18 \\ 0 & 0 & 0 & 1 & 24 \end{array}]$

解题步骤 2.10.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - 60 & - 48 & 0 & 1008 \\ 0 & 1 & \frac{1}{2} & 0 & 21 \\ 0 & 0 & 1 & 0 & 18 \\ 0 & 0 & 0 & 1 & 24 \end{array}]$

解题步骤 2.11

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

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解题步骤 2.11.1

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

$[\begin{array}{c} 1 & - 60 & - 48 & 0 & 1008 \\ 0 - \frac{1}{2} \cdot 0 & 1 - \frac{1}{2} \cdot 0 & \frac{1}{2} - \frac{1}{2} \cdot 1 & 0 - \frac{1}{2} \cdot 0 & 21 - \frac{1}{2} \cdot 18 \\ 0 & 0 & 1 & 0 & 18 \\ 0 & 0 & 0 & 1 & 24 \end{array}]$

解题步骤 2.11.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 60 & - 48 & 0 & 1008 \\ 0 & 1 & 0 & 0 & 12 \\ 0 & 0 & 1 & 0 & 18 \\ 0 & 0 & 0 & 1 & 24 \end{array}]$

解题步骤 2.12

Perform the row operation $R_{1} = R_{1} + 48 R_{3}$ to make the entry at $1, 3$ a $0$ .

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解题步骤 2.12.1

Perform the row operation $R_{1} = R_{1} + 48 R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 + 48 \cdot 0 & - 60 + 48 \cdot 0 & - 48 + 48 \cdot 1 & 0 + 48 \cdot 0 & 1008 + 48 \cdot 18 \\ 0 & 1 & 0 & 0 & 12 \\ 0 & 0 & 1 & 0 & 18 \\ 0 & 0 & 0 & 1 & 24 \end{array}]$

解题步骤 2.12.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - 60 & 0 & 0 & 1872 \\ 0 & 1 & 0 & 0 & 12 \\ 0 & 0 & 1 & 0 & 18 \\ 0 & 0 & 0 & 1 & 24 \end{array}]$

解题步骤 2.13

Perform the row operation $R_{1} = R_{1} + 60 R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 2.13.1

Perform the row operation $R_{1} = R_{1} + 60 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + 60 \cdot 0 & - 60 + 60 \cdot 1 & 0 + 60 \cdot 0 & 0 + 60 \cdot 0 & 1872 + 60 \cdot 12 \\ 0 & 1 & 0 & 0 & 12 \\ 0 & 0 & 1 & 0 & 18 \\ 0 & 0 & 0 & 1 & 24 \end{array}]$

解题步骤 2.13.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & 0 & 2592 \\ 0 & 1 & 0 & 0 & 12 \\ 0 & 0 & 1 & 0 & 18 \\ 0 & 0 & 0 & 1 & 24 \end{array}]$

解题步骤 3

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

$p = 2592$

$x = 12$

$y = 18$

$z = 24$

解题步骤 4

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

$(2592, 12, 18, 24)$

有限数学 示例

有限数学示例