有限数学示例

$7 x + 4 y - z = 0$ , $5 z + 3 y - z = 0$

解题步骤 1

从 $5 z$ 中减去 $z$ 。

$7 x + 4 y - z = 0$

$3 y + 4 z = 0$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} 7 & 4 & - 1 & 0 \\ 0 & 3 & 4 & 0 \end{array}]$

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 $1$ .

$[\begin{array}{c} \frac{7}{7} & \frac{4}{7} & \frac{- 1}{7} & \frac{0}{7} \\ 0 & 3 & 4 & 0 \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{4}{7} & - \frac{1}{7} & 0 \\ 0 & 3 & 4 & 0 \end{array}]$

解题步骤 3.2

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

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解题步骤 3.2.1

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

$[\begin{array}{c} 1 & \frac{4}{7} & - \frac{1}{7} & 0 \\ \frac{0}{3} & \frac{3}{3} & \frac{4}{3} & \frac{0}{3} \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{4}{7} & - \frac{1}{7} & 0 \\ 0 & 1 & \frac{4}{3} & 0 \end{array}]$

解题步骤 3.3

Perform the row operation $R_{1} = R_{1} - \frac{4}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 3.3.1

Perform the row operation $R_{1} = R_{1} - \frac{4}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{4}{7} \cdot 0 & \frac{4}{7} - \frac{4}{7} \cdot 1 & - \frac{1}{7} - \frac{4}{7} \cdot \frac{4}{3} & 0 - \frac{4}{7} \cdot 0 \\ 0 & 1 & \frac{4}{3} & 0 \end{array}]$

解题步骤 3.3.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & - \frac{19}{21} & 0 \\ 0 & 1 & \frac{4}{3} & 0 \end{array}]$

解题步骤 4

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

$x - \frac{19}{21} z = 0$

$y + \frac{4}{3} z = 0$

解题步骤 5

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

$(\frac{19 z}{21}, - \frac{4 z}{3}, z)$

有限数学 示例

有限数学示例