线性代数示例

$\frac{x}{5} = \frac{y}{4}$ , $\frac{y}{3} = \frac{x}{3} - 1$

解题步骤 1

从等式两边同时减去 $\frac{y}{4}$ 。

$\frac{x}{5} - \frac{y}{4} = 0, \frac{y}{3} = \frac{x}{3} - 1$

解题步骤 2

从等式两边同时减去 $\frac{x}{3}$ 。

$\frac{x}{5} - \frac{y}{4} = 0, \frac{y}{3} - \frac{x}{3} = - 1$

解题步骤 3

以矩阵形式书写方程组。

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

解题步骤 4

求行简化阶梯形矩阵。

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解题步骤 4.1

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

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解题步骤 4.1.1

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

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

解题步骤 4.1.2

化简 $R_{1}$ 。

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

解题步骤 4.2

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

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解题步骤 4.2.1

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

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

解题步骤 4.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{5}{4} & 0 \\ 0 & - \frac{1}{12} & - 1 \end{array}]$

解题步骤 4.3

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

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解题步骤 4.3.1

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

$[\begin{array}{c} 1 & - \frac{5}{4} & 0 \\ - 12 \cdot 0 & - 12 (- \frac{1}{12}) & - 12 \cdot - 1 \end{array}]$

解题步骤 4.3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{5}{4} & 0 \\ 0 & 1 & 12 \end{array}]$

解题步骤 4.4

Perform the row operation $R_{1} = R_{1} + \frac{5}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 4.4.1

Perform the row operation $R_{1} = R_{1} + \frac{5}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{5}{4} \cdot 0 & - \frac{5}{4} + \frac{5}{4} \cdot 1 & 0 + \frac{5}{4} \cdot 12 \\ 0 & 1 & 12 \end{array}]$

解题步骤 4.4.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 15 \\ 0 & 1 & 12 \end{array}]$

解题步骤 5

使用结果矩阵定义方程组的最终解。

$x = 15$

$y = 12$

解题步骤 6

解为使方程组成立的有序对集合。

$(15, 12)$

解题步骤 7

通过重新安排增广矩阵的行简化式中的每一个方程对解向量进行分解，而简化式是通过求解每一行中的因变量得出。

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 15 \\ 12 \end{array}]$

线性代数 示例

线性代数示例