有限数学示例

$2 x - y + 5 z = - 7$ , $x + y - 2 x = - 2$ , $x - y + 4 z = - 4$

解题步骤 1

从 $x$ 中减去 $2 x$ 。

$2 x - y + 5 z = - 7$

$- x + y = - 2$

$x - y + 4 z = - 4$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} 2 & - 1 & 5 & - 7 \\ - 1 & 1 & 0 & - 2 \\ 1 & - 1 & 4 & - 4 \end{array}]$

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 $\frac{1}{2}$ 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 $\frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{2}{2} & \frac{- 1}{2} & \frac{5}{2} & \frac{- 7}{2} \\ - 1 & 1 & 0 & - 2 \\ 1 & - 1 & 4 & - 4 \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{5}{2} & - \frac{7}{2} \\ - 1 & 1 & 0 & - 2 \\ 1 & - 1 & 4 & - 4 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{5}{2} & - \frac{7}{2} \\ - 1 + 1 \cdot 1 & 1 - \frac{1}{2} & 0 + \frac{5}{2} & - 2 - \frac{7}{2} \\ 1 & - 1 & 4 & - 4 \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{5}{2} & - \frac{7}{2} \\ 0 & \frac{1}{2} & \frac{5}{2} & - \frac{11}{2} \\ 1 & - 1 & 4 & - 4 \end{array}]$

解题步骤 3.3

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

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解题步骤 3.3.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{5}{2} & - \frac{7}{2} \\ 0 & \frac{1}{2} & \frac{5}{2} & - \frac{11}{2} \\ 1 - 1 & - 1 + \frac{1}{2} & 4 - \frac{5}{2} & - 4 + \frac{7}{2} \end{array}]$

解题步骤 3.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{5}{2} & - \frac{7}{2} \\ 0 & \frac{1}{2} & \frac{5}{2} & - \frac{11}{2} \\ 0 & - \frac{1}{2} & \frac{3}{2} & - \frac{1}{2} \end{array}]$

解题步骤 3.4

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

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解题步骤 3.4.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{5}{2} & - \frac{7}{2} \\ 2 \cdot 0 & 2 (\frac{1}{2}) & 2 (\frac{5}{2}) & 2 (- \frac{11}{2}) \\ 0 & - \frac{1}{2} & \frac{3}{2} & - \frac{1}{2} \end{array}]$

解题步骤 3.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{5}{2} & - \frac{7}{2} \\ 0 & 1 & 5 & - 11 \\ 0 & - \frac{1}{2} & \frac{3}{2} & - \frac{1}{2} \end{array}]$

解题步骤 3.5

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

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解题步骤 3.5.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{5}{2} & - \frac{7}{2} \\ 0 & 1 & 5 & - 11 \\ 0 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & \frac{3}{2} + \frac{1}{2} \cdot 5 & - \frac{1}{2} + \frac{1}{2} \cdot - 11 \end{array}]$

解题步骤 3.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{5}{2} & - \frac{7}{2} \\ 0 & 1 & 5 & - 11 \\ 0 & 0 & 4 & - 6 \end{array}]$

解题步骤 3.6

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

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解题步骤 3.6.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{5}{2} & - \frac{7}{2} \\ 0 & 1 & 5 & - 11 \\ \frac{0}{4} & \frac{0}{4} & \frac{4}{4} & \frac{- 6}{4} \end{array}]$

解题步骤 3.6.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{5}{2} & - \frac{7}{2} \\ 0 & 1 & 5 & - 11 \\ 0 & 0 & 1 & - \frac{3}{2} \end{array}]$

解题步骤 3.7

Perform the row operation $R_{2} = R_{2} - 5 R_{3}$ to make the entry at $2, 3$ a $0$ .

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解题步骤 3.7.1

Perform the row operation $R_{2} = R_{2} - 5 R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{5}{2} & - \frac{7}{2} \\ 0 - 5 \cdot 0 & 1 - 5 \cdot 0 & 5 - 5 \cdot 1 & - 11 - 5 (- \frac{3}{2}) \\ 0 & 0 & 1 & - \frac{3}{2} \end{array}]$

解题步骤 3.7.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{5}{2} & - \frac{7}{2} \\ 0 & 1 & 0 & - \frac{7}{2} \\ 0 & 0 & 1 & - \frac{3}{2} \end{array}]$

解题步骤 3.8

Perform the row operation $R_{1} = R_{1} - \frac{5}{2} R_{3}$ to make the entry at $1, 3$ a $0$ .

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解题步骤 3.8.1

Perform the row operation $R_{1} = R_{1} - \frac{5}{2} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{5}{2} \cdot 0 & - \frac{1}{2} - \frac{5}{2} \cdot 0 & \frac{5}{2} - \frac{5}{2} \cdot 1 & - \frac{7}{2} - \frac{5}{2} (- \frac{3}{2}) \\ 0 & 1 & 0 & - \frac{7}{2} \\ 0 & 0 & 1 & - \frac{3}{2} \end{array}]$

解题步骤 3.8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{1}{4} \\ 0 & 1 & 0 & - \frac{7}{2} \\ 0 & 0 & 1 & - \frac{3}{2} \end{array}]$

解题步骤 3.9

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

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解题步骤 3.9.1

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

$[\begin{array}{c} 1 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & 0 + \frac{1}{2} \cdot 0 & \frac{1}{4} + \frac{1}{2} (- \frac{7}{2}) \\ 0 & 1 & 0 & - \frac{7}{2} \\ 0 & 0 & 1 & - \frac{3}{2} \end{array}]$

解题步骤 3.9.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & - \frac{3}{2} \\ 0 & 1 & 0 & - \frac{7}{2} \\ 0 & 0 & 1 & - \frac{3}{2} \end{array}]$

解题步骤 4

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

$x = - \frac{3}{2}$

$y = - \frac{7}{2}$

$z = - \frac{3}{2}$

解题步骤 5

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

$(- \frac{3}{2}, - \frac{7}{2}, - \frac{3}{2})$

有限数学 示例

有限数学示例