有限数学示例

$- 4 x - y + 7 z = 9$ , $8 y + 6 z = - 6$ , $4 x + 2 y + 6 z = 1 - 22$

解题步骤 1

从 $1$ 中减去 $22$ 。

$- 4 x - y + 7 z = 9$

$8 y + 6 z = - 6$

$4 x + 2 y + 6 z = - 21$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} - 4 & - 1 & 7 & 9 \\ 0 & 8 & 6 & - 6 \\ 4 & 2 & 6 & - 21 \end{array}]$

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 $- \frac{1}{4}$ 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 $- \frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{4} \cdot - 4 & - \frac{1}{4} \cdot - 1 & - \frac{1}{4} \cdot 7 & - \frac{1}{4} \cdot 9 \\ 0 & 8 & 6 & - 6 \\ 4 & 2 & 6 & - 21 \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{1}{4} & - \frac{7}{4} & - \frac{9}{4} \\ 0 & 8 & 6 & - 6 \\ 4 & 2 & 6 & - 21 \end{array}]$

解题步骤 3.2

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 3.2.1

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

$[\begin{array}{c} 1 & \frac{1}{4} & - \frac{7}{4} & - \frac{9}{4} \\ 0 & 8 & 6 & - 6 \\ 4 - 4 \cdot 1 & 2 - 4 (\frac{1}{4}) & 6 - 4 (- \frac{7}{4}) & - 21 - 4 (- \frac{9}{4}) \end{array}]$

解题步骤 3.2.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{1}{4} & - \frac{7}{4} & - \frac{9}{4} \\ 0 & 8 & 6 & - 6 \\ 0 & 1 & 13 & - 12 \end{array}]$

解题步骤 3.3

Multiply each element of $R_{2}$ by $\frac{1}{8}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 3.3.1

Multiply each element of $R_{2}$ by $\frac{1}{8}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{4} & - \frac{7}{4} & - \frac{9}{4} \\ \frac{0}{8} & \frac{8}{8} & \frac{6}{8} & \frac{- 6}{8} \\ 0 & 1 & 13 & - 12 \end{array}]$

解题步骤 3.3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{1}{4} & - \frac{7}{4} & - \frac{9}{4} \\ 0 & 1 & \frac{3}{4} & - \frac{3}{4} \\ 0 & 1 & 13 & - 12 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & \frac{1}{4} & - \frac{7}{4} & - \frac{9}{4} \\ 0 & 1 & \frac{3}{4} & - \frac{3}{4} \\ 0 - 0 & 1 - 1 & 13 - \frac{3}{4} & - 12 + \frac{3}{4} \end{array}]$

解题步骤 3.4.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{1}{4} & - \frac{7}{4} & - \frac{9}{4} \\ 0 & 1 & \frac{3}{4} & - \frac{3}{4} \\ 0 & 0 & \frac{49}{4} & - \frac{45}{4} \end{array}]$

解题步骤 3.5

Multiply each element of $R_{3}$ by $\frac{4}{49}$ to make the entry at $3, 3$ a $1$ .

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解题步骤 3.5.1

Multiply each element of $R_{3}$ by $\frac{4}{49}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{4} & - \frac{7}{4} & - \frac{9}{4} \\ 0 & 1 & \frac{3}{4} & - \frac{3}{4} \\ \frac{4}{49} \cdot 0 & \frac{4}{49} \cdot 0 & \frac{4}{49} \cdot \frac{49}{4} & \frac{4}{49} (- \frac{45}{4}) \end{array}]$

解题步骤 3.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{1}{4} & - \frac{7}{4} & - \frac{9}{4} \\ 0 & 1 & \frac{3}{4} & - \frac{3}{4} \\ 0 & 0 & 1 & - \frac{45}{49} \end{array}]$

解题步骤 3.6

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

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解题步骤 3.6.1

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

$[\begin{array}{c} 1 & \frac{1}{4} & - \frac{7}{4} & - \frac{9}{4} \\ 0 - \frac{3}{4} \cdot 0 & 1 - \frac{3}{4} \cdot 0 & \frac{3}{4} - \frac{3}{4} \cdot 1 & - \frac{3}{4} - \frac{3}{4} (- \frac{45}{49}) \\ 0 & 0 & 1 & - \frac{45}{49} \end{array}]$

解题步骤 3.6.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{1}{4} & - \frac{7}{4} & - \frac{9}{4} \\ 0 & 1 & 0 & - \frac{3}{49} \\ 0 & 0 & 1 & - \frac{45}{49} \end{array}]$

解题步骤 3.7

Perform the row operation $R_{1} = R_{1} + \frac{7}{4} R_{3}$ to make the entry at $1, 3$ a $0$ .

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解题步骤 3.7.1

Perform the row operation $R_{1} = R_{1} + \frac{7}{4} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 + \frac{7}{4} \cdot 0 & \frac{1}{4} + \frac{7}{4} \cdot 0 & - \frac{7}{4} + \frac{7}{4} \cdot 1 & - \frac{9}{4} + \frac{7}{4} (- \frac{45}{49}) \\ 0 & 1 & 0 & - \frac{3}{49} \\ 0 & 0 & 1 & - \frac{45}{49} \end{array}]$

解题步骤 3.7.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{1}{4} & 0 & - \frac{27}{7} \\ 0 & 1 & 0 & - \frac{3}{49} \\ 0 & 0 & 1 & - \frac{45}{49} \end{array}]$

解题步骤 3.8

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

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解题步骤 3.8.1

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

$[\begin{array}{c} 1 - \frac{1}{4} \cdot 0 & \frac{1}{4} - \frac{1}{4} \cdot 1 & 0 - \frac{1}{4} \cdot 0 & - \frac{27}{7} - \frac{1}{4} (- \frac{3}{49}) \\ 0 & 1 & 0 & - \frac{3}{49} \\ 0 & 0 & 1 & - \frac{45}{49} \end{array}]$

解题步骤 3.8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & - \frac{753}{196} \\ 0 & 1 & 0 & - \frac{3}{49} \\ 0 & 0 & 1 & - \frac{45}{49} \end{array}]$

解题步骤 4

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

$x = - \frac{753}{196}$

$y = - \frac{3}{49}$

$z = - \frac{45}{49}$

解题步骤 5

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

$(- \frac{753}{196}, - \frac{3}{49}, - \frac{45}{49})$

有限数学 示例

有限数学示例