初级微积分示例

$- 3 x - 2 y = 2$ , $6 x + 6 y = 5$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} - 3 & - 2 & 2 \\ 6 & 6 & 5 \end{array}]$

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 $- \frac{1}{3}$ 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 $- \frac{1}{3}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot - 2 & - \frac{1}{3} \cdot 2 \\ 6 & 6 & 5 \end{array}]$

解题步骤 2.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{2}{3} & - \frac{2}{3} \\ 6 & 6 & 5 \end{array}]$

解题步骤 2.2

Perform the row operation $R_{2} = R_{2} - 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

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解题步骤 2.2.1

Perform the row operation $R_{2} = R_{2} - 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{2}{3} & - \frac{2}{3} \\ 6 - 6 \cdot 1 & 6 - 6 (\frac{2}{3}) & 5 - 6 (- \frac{2}{3}) \end{array}]$

解题步骤 2.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{2}{3} & - \frac{2}{3} \\ 0 & 2 & 9 \end{array}]$

解题步骤 2.3

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

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解题步骤 2.3.1

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

$[\begin{array}{c} 1 & \frac{2}{3} & - \frac{2}{3} \\ \frac{0}{2} & \frac{2}{2} & \frac{9}{2} \end{array}]$

解题步骤 2.3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{2}{3} & - \frac{2}{3} \\ 0 & 1 & \frac{9}{2} \end{array}]$

解题步骤 2.4

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

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解题步骤 2.4.1

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

$[\begin{array}{c} 1 - \frac{2}{3} \cdot 0 & \frac{2}{3} - \frac{2}{3} \cdot 1 & - \frac{2}{3} - \frac{2}{3} \cdot \frac{9}{2} \\ 0 & 1 & \frac{9}{2} \end{array}]$

解题步骤 2.4.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & - \frac{11}{3} \\ 0 & 1 & \frac{9}{2} \end{array}]$

解题步骤 3

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

$x = - \frac{11}{3}$

$y = \frac{9}{2}$

解题步骤 4

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

$(- \frac{11}{3}, \frac{9}{2})$

初级微积分 示例

初级微积分示例