有限数学示例

$3 x + 6 y - 5 z = 35$ , $x + 2 y + 4 z = - 11$ , $4 x - 4 y + 6 z = - 28$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 3 & 6 & - 5 & 35 \\ 1 & 2 & 4 & - 11 \\ 4 & - 4 & 6 & - 28 \end{array}]$

解题步骤 2

求行简化阶梯形矩阵。

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解题步骤 2.1

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 2.1.1

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

$[\begin{array}{c} \frac{3}{3} & \frac{6}{3} & \frac{- 5}{3} & \frac{35}{3} \\ 1 & 2 & 4 & - 11 \\ 4 & - 4 & 6 & - 28 \end{array}]$

解题步骤 2.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 2 & - \frac{5}{3} & \frac{35}{3} \\ 1 & 2 & 4 & - 11 \\ 4 & - 4 & 6 & - 28 \end{array}]$

解题步骤 2.2

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 2.2.1

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

$[\begin{array}{c} 1 & 2 & - \frac{5}{3} & \frac{35}{3} \\ 1 - 1 & 2 - 2 & 4 + \frac{5}{3} & - 11 - \frac{35}{3} \\ 4 & - 4 & 6 & - 28 \end{array}]$

解题步骤 2.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 2 & - \frac{5}{3} & \frac{35}{3} \\ 0 & 0 & \frac{17}{3} & - \frac{68}{3} \\ 4 & - 4 & 6 & - 28 \end{array}]$

解题步骤 2.3

Perform the row operation $R_{3} = R_{3} - 4 R_{1}$ to make the entry at $3, 1$ a $0$ .

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解题步骤 2.3.1

Perform the row operation $R_{3} = R_{3} - 4 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 2 & - \frac{5}{3} & \frac{35}{3} \\ 0 & 0 & \frac{17}{3} & - \frac{68}{3} \\ 4 - 4 \cdot 1 & - 4 - 4 \cdot 2 & 6 - 4 (- \frac{5}{3}) & - 28 - 4 (\frac{35}{3}) \end{array}]$

解题步骤 2.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & 2 & - \frac{5}{3} & \frac{35}{3} \\ 0 & 0 & \frac{17}{3} & - \frac{68}{3} \\ 0 & - 12 & \frac{38}{3} & - \frac{224}{3} \end{array}]$

解题步骤 2.4

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 2$ .

$[\begin{array}{c} 1 & 2 & - \frac{5}{3} & \frac{35}{3} \\ 0 & - 12 & \frac{38}{3} & - \frac{224}{3} \\ 0 & 0 & \frac{17}{3} & - \frac{68}{3} \end{array}]$

解题步骤 2.5

Multiply each element of $R_{2}$ by $- \frac{1}{12}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 2.5.1

Multiply each element of $R_{2}$ by $- \frac{1}{12}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 2 & - \frac{5}{3} & \frac{35}{3} \\ - \frac{1}{12} \cdot 0 & - \frac{1}{12} \cdot - 12 & - \frac{1}{12} \cdot \frac{38}{3} & - \frac{1}{12} (- \frac{224}{3}) \\ 0 & 0 & \frac{17}{3} & - \frac{68}{3} \end{array}]$

解题步骤 2.5.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 2 & - \frac{5}{3} & \frac{35}{3} \\ 0 & 1 & - \frac{19}{18} & \frac{56}{9} \\ 0 & 0 & \frac{17}{3} & - \frac{68}{3} \end{array}]$

解题步骤 2.6

Multiply each element of $R_{3}$ by $\frac{3}{17}$ to make the entry at $3, 3$ a $1$ .

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解题步骤 2.6.1

Multiply each element of $R_{3}$ by $\frac{3}{17}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 2 & - \frac{5}{3} & \frac{35}{3} \\ 0 & 1 & - \frac{19}{18} & \frac{56}{9} \\ \frac{3}{17} \cdot 0 & \frac{3}{17} \cdot 0 & \frac{3}{17} \cdot \frac{17}{3} & \frac{3}{17} (- \frac{68}{3}) \end{array}]$

解题步骤 2.6.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & 2 & - \frac{5}{3} & \frac{35}{3} \\ 0 & 1 & - \frac{19}{18} & \frac{56}{9} \\ 0 & 0 & 1 & - 4 \end{array}]$

解题步骤 2.7

Perform the row operation $R_{2} = R_{2} + \frac{19}{18} R_{3}$ to make the entry at $2, 3$ a $0$ .

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解题步骤 2.7.1

Perform the row operation $R_{2} = R_{2} + \frac{19}{18} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 2 & - \frac{5}{3} & \frac{35}{3} \\ 0 + \frac{19}{18} \cdot 0 & 1 + \frac{19}{18} \cdot 0 & - \frac{19}{18} + \frac{19}{18} \cdot 1 & \frac{56}{9} + \frac{19}{18} \cdot - 4 \\ 0 & 0 & 1 & - 4 \end{array}]$

解题步骤 2.7.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 2 & - \frac{5}{3} & \frac{35}{3} \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & - 4 \end{array}]$

解题步骤 2.8

Perform the row operation $R_{1} = R_{1} + \frac{5}{3} R_{3}$ to make the entry at $1, 3$ a $0$ .

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解题步骤 2.8.1

Perform the row operation $R_{1} = R_{1} + \frac{5}{3} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 + \frac{5}{3} \cdot 0 & 2 + \frac{5}{3} \cdot 0 & - \frac{5}{3} + \frac{5}{3} \cdot 1 & \frac{35}{3} + \frac{5}{3} \cdot - 4 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & - 4 \end{array}]$

解题步骤 2.8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 2 & 0 & 5 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & - 4 \end{array}]$

解题步骤 2.9

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

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解题步骤 2.9.1

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

$[\begin{array}{c} 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & 0 - 2 \cdot 0 & 5 - 2 \cdot 2 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & - 4 \end{array}]$

解题步骤 2.9.2

化简 $R_{1}$ 。

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

解题步骤 3

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

$x = 1$

$y = 2$

$z = - 4$

解题步骤 4

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

$(1, 2, - 4)$

有限数学 示例

有限数学示例