线性代数示例

$y = 4 x + 3 x - 2$ , $y = 6$

解题步骤 1

Move variables to the left and constant terms to the right.

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解题步骤 1.1

将所有包含变量的项移到等式左边。

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解题步骤 1.1.1

从等式两边同时减去 $4 x$ 。

$y - 4 x = 3 x - 2$

$y = 6$

解题步骤 1.1.2

从等式两边同时减去 $3 x$ 。

$y - 4 x - 3 x = - 2$

$y = 6$

$y - 4 x - 3 x = - 2$

$y = 6$

解题步骤 1.2

从 $- 4 x$ 中减去 $3 x$ 。

$y - 7 x = - 2$

$y = 6$

解题步骤 1.3

将 $y$ 和 $- 7 x$ 重新排序。

$- 7 x + y = - 2$

$y = 6$

$- 7 x + y = - 2$

$y = 6$

解题步骤 2

Write the system as a matrix.

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

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 $- \frac{1}{7}$ 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 $- \frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

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

解题步骤 3.1.2

化简 $R_{1}$ 。

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

解题步骤 3.2

Perform the row operation $R_{1} = R_{1} + \frac{1}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 3.2.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{1}{7} \cdot 0 & - \frac{1}{7} + \frac{1}{7} \cdot 1 & \frac{2}{7} + \frac{1}{7} \cdot 6 \\ 0 & 1 & 6 \end{array}]$

解题步骤 3.2.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & \frac{8}{7} \\ 0 & 1 & 6 \end{array}]$

解题步骤 4

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

$x = \frac{8}{7}$

$y = 6$

解题步骤 5

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

$(\frac{8}{7}, 6)$

线性代数 示例

线性代数示例