有限数学示例

$- 3 x + y = 8$ , $7 x + y = - 12$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} - 3 & 1 & 8 \\ 7 & 1 & - 12 \end{array}]$

解题步骤 2

求行简化阶梯形矩阵。

点击获取更多步骤...

解题步骤 2.1

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

点击获取更多步骤...

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 $1$ .

$[\begin{array}{c} - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 1 & - \frac{1}{3} \cdot 8 \\ 7 & 1 & - 12 \end{array}]$

解题步骤 2.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{8}{3} \\ 7 & 1 & - 12 \end{array}]$

解题步骤 2.2

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

点击获取更多步骤...

解题步骤 2.2.1

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

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{8}{3} \\ 7 - 7 \cdot 1 & 1 - 7 (- \frac{1}{3}) & - 12 - 7 (- \frac{8}{3}) \end{array}]$

解题步骤 2.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{8}{3} \\ 0 & \frac{10}{3} & \frac{20}{3} \end{array}]$

解题步骤 2.3

Multiply each element of $R_{2}$ by $\frac{3}{10}$ to make the entry at $2, 2$ a $1$ .

点击获取更多步骤...

解题步骤 2.3.1

Multiply each element of $R_{2}$ by $\frac{3}{10}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{8}{3} \\ \frac{3}{10} \cdot 0 & \frac{3}{10} \cdot \frac{10}{3} & \frac{3}{10} \cdot \frac{20}{3} \end{array}]$

解题步骤 2.3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{8}{3} \\ 0 & 1 & 2 \end{array}]$

解题步骤 2.4

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

点击获取更多步骤...

解题步骤 2.4.1

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

$[\begin{array}{c} 1 + \frac{1}{3} \cdot 0 & - \frac{1}{3} + \frac{1}{3} \cdot 1 & - \frac{8}{3} + \frac{1}{3} \cdot 2 \\ 0 & 1 & 2 \end{array}]$

解题步骤 2.4.2

化简 $R_{1}$ 。

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

解题步骤 3

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

$x = - 2$

$y = 2$

解题步骤 4

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

$(- 2, 2)$

有限数学 示例

有限数学示例