线性代数示例

$4 a - b - 3 c = 27$ , $- 4 a - 4 b - 2 c = 32$ , $4 a + 9 b + 7 c = 37$

解题步骤 1

以矩阵形式书写方程组。

$[\begin{array}{c} 4 & - 1 & - 3 & 27 \\ - 4 & - 4 & - 2 & 32 \\ 4 & 9 & 7 & 37 \end{array}]$

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

$[\begin{array}{c} \frac{4}{4} & \frac{- 1}{4} & \frac{- 3}{4} & \frac{27}{4} \\ - 4 & - 4 & - 2 & 32 \\ 4 & 9 & 7 & 37 \end{array}]$

解题步骤 2.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{1}{4} & - \frac{3}{4} & \frac{27}{4} \\ - 4 & - 4 & - 2 & 32 \\ 4 & 9 & 7 & 37 \end{array}]$

解题步骤 2.2

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

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解题步骤 2.2.1

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

$[\begin{array}{c} 1 & - \frac{1}{4} & - \frac{3}{4} & \frac{27}{4} \\ - 4 + 4 \cdot 1 & - 4 + 4 (- \frac{1}{4}) & - 2 + 4 (- \frac{3}{4}) & 32 + 4 (\frac{27}{4}) \\ 4 & 9 & 7 & 37 \end{array}]$

解题步骤 2.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{4} & - \frac{3}{4} & \frac{27}{4} \\ 0 & - 5 & - 5 & 59 \\ 4 & 9 & 7 & 37 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & - \frac{3}{4} & \frac{27}{4} \\ 0 & - 5 & - 5 & 59 \\ 4 - 4 \cdot 1 & 9 - 4 (- \frac{1}{4}) & 7 - 4 (- \frac{3}{4}) & 37 - 4 (\frac{27}{4}) \end{array}]$

解题步骤 2.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{1}{4} & - \frac{3}{4} & \frac{27}{4} \\ 0 & - 5 & - 5 & 59 \\ 0 & 10 & 10 & 10 \end{array}]$

解题步骤 2.4

Multiply each element of $R_{2}$ by $- \frac{1}{5}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 2.4.1

Multiply each element of $R_{2}$ by $- \frac{1}{5}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & - \frac{3}{4} & \frac{27}{4} \\ - \frac{1}{5} \cdot 0 & - \frac{1}{5} \cdot - 5 & - \frac{1}{5} \cdot - 5 & - \frac{1}{5} \cdot 59 \\ 0 & 10 & 10 & 10 \end{array}]$

解题步骤 2.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{4} & - \frac{3}{4} & \frac{27}{4} \\ 0 & 1 & 1 & - \frac{59}{5} \\ 0 & 10 & 10 & 10 \end{array}]$

解题步骤 2.5

Perform the row operation $R_{3} = R_{3} - 10 R_{2}$ to make the entry at $3, 2$ a $0$ .

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解题步骤 2.5.1

Perform the row operation $R_{3} = R_{3} - 10 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & - \frac{3}{4} & \frac{27}{4} \\ 0 & 1 & 1 & - \frac{59}{5} \\ 0 - 10 \cdot 0 & 10 - 10 \cdot 1 & 10 - 10 \cdot 1 & 10 - 10 (- \frac{59}{5}) \end{array}]$

解题步骤 2.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{1}{4} & - \frac{3}{4} & \frac{27}{4} \\ 0 & 1 & 1 & - \frac{59}{5} \\ 0 & 0 & 0 & 128 \end{array}]$

解题步骤 2.6

Multiply each element of $R_{3}$ by $\frac{1}{128}$ to make the entry at $3, 4$ a $1$ .

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解题步骤 2.6.1

Multiply each element of $R_{3}$ by $\frac{1}{128}$ to make the entry at $3, 4$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & - \frac{3}{4} & \frac{27}{4} \\ 0 & 1 & 1 & - \frac{59}{5} \\ \frac{0}{128} & \frac{0}{128} & \frac{0}{128} & \frac{128}{128} \end{array}]$

解题步骤 2.6.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{1}{4} & - \frac{3}{4} & \frac{27}{4} \\ 0 & 1 & 1 & - \frac{59}{5} \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 2.7

Perform the row operation $R_{2} = R_{2} + \frac{59}{5} R_{3}$ to make the entry at $2, 4$ a $0$ .

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解题步骤 2.7.1

Perform the row operation $R_{2} = R_{2} + \frac{59}{5} R_{3}$ to make the entry at $2, 4$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & - \frac{3}{4} & \frac{27}{4} \\ 0 + \frac{59}{5} \cdot 0 & 1 + \frac{59}{5} \cdot 0 & 1 + \frac{59}{5} \cdot 0 & - \frac{59}{5} + \frac{59}{5} \cdot 1 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 2.7.2

化简 $R_{2}$ 。

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

解题步骤 2.8

Perform the row operation $R_{1} = R_{1} - \frac{27}{4} R_{3}$ to make the entry at $1, 4$ a $0$ .

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解题步骤 2.8.1

Perform the row operation $R_{1} = R_{1} - \frac{27}{4} R_{3}$ to make the entry at $1, 4$ a $0$ .

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

解题步骤 2.8.2

化简 $R_{1}$ 。

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

解题步骤 2.9

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 2.9.1

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

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

解题步骤 2.9.2

化简 $R_{1}$ 。

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

解题步骤 3

使用结果矩阵定义方程组的最终解。

$a - \frac{c}{2} = 0$

$b + c = 0$

$0 = 1$

解题步骤 4

在等式两边都加上 $\frac{c}{2}$ 。

$a = \frac{c}{2}$

$b + c = 0$

$0 = 1$

解题步骤 5

从等式两边同时减去 $c$ 。

$b = - c$

$a = \frac{c}{2}$

$0 = 1$

解题步骤 6

因为 $0 \neq 1$ ，所以没有解。

无解

线性代数 示例

线性代数示例