有限数学示例

$5 x - 5 y + 2 z = - 13$ , $10 x - 10 y + 4 z = - 23$ , $15 x - 15 y + 6 z = - 37$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 5 & - 5 & 2 & - 13 \\ 10 & - 10 & 4 & - 23 \\ 15 & - 15 & 6 & - 37 \end{array}]$

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 $\frac{1}{5}$ 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 $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{5}{5} & \frac{- 5}{5} & \frac{2}{5} & \frac{- 13}{5} \\ 10 & - 10 & 4 & - 23 \\ 15 & - 15 & 6 & - 37 \end{array}]$

解题步骤 2.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - 1 & \frac{2}{5} & - \frac{13}{5} \\ 10 & - 10 & 4 & - 23 \\ 15 & - 15 & 6 & - 37 \end{array}]$

解题步骤 2.2

Perform the row operation $R_{2} = R_{2} - 10 R_{1}$ to make the entry at $2, 1$ a $0$ .

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解题步骤 2.2.1

Perform the row operation $R_{2} = R_{2} - 10 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - 1 & \frac{2}{5} & - \frac{13}{5} \\ 10 - 10 \cdot 1 & - 10 - 10 \cdot - 1 & 4 - 10 (\frac{2}{5}) & - 23 - 10 (- \frac{13}{5}) \\ 15 & - 15 & 6 & - 37 \end{array}]$

解题步骤 2.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 1 & \frac{2}{5} & - \frac{13}{5} \\ 0 & 0 & 0 & 3 \\ 15 & - 15 & 6 & - 37 \end{array}]$

解题步骤 2.3

Perform the row operation $R_{3} = R_{3} - 15 R_{1}$ to make the entry at $3, 1$ a $0$ .

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解题步骤 2.3.1

Perform the row operation $R_{3} = R_{3} - 15 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - 1 & \frac{2}{5} & - \frac{13}{5} \\ 0 & 0 & 0 & 3 \\ 15 - 15 \cdot 1 & - 15 - 15 \cdot - 1 & 6 - 15 (\frac{2}{5}) & - 37 - 15 (- \frac{13}{5}) \end{array}]$

解题步骤 2.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 1 & \frac{2}{5} & - \frac{13}{5} \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 2 \end{array}]$

解题步骤 2.4

Multiply each element of $R_{2}$ by $\frac{1}{3}$ to make the entry at $2, 4$ a $1$ .

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解题步骤 2.4.1

Multiply each element of $R_{2}$ by $\frac{1}{3}$ to make the entry at $2, 4$ a $1$ .

$[\begin{array}{c} 1 & - 1 & \frac{2}{5} & - \frac{13}{5} \\ \frac{0}{3} & \frac{0}{3} & \frac{0}{3} & \frac{3}{3} \\ 0 & 0 & 0 & 2 \end{array}]$

解题步骤 2.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 1 & \frac{2}{5} & - \frac{13}{5} \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 2 \end{array}]$

解题步骤 2.5

Perform the row operation $R_{3} = R_{3} - 2 R_{2}$ to make the entry at $3, 4$ a $0$ .

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解题步骤 2.5.1

Perform the row operation $R_{3} = R_{3} - 2 R_{2}$ to make the entry at $3, 4$ a $0$ .

$[\begin{array}{c} 1 & - 1 & \frac{2}{5} & - \frac{13}{5} \\ 0 & 0 & 0 & 1 \\ 0 - 2 \cdot 0 & 0 - 2 \cdot 0 & 0 - 2 \cdot 0 & 2 - 2 \cdot 1 \end{array}]$

解题步骤 2.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 1 & \frac{2}{5} & - \frac{13}{5} \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}]$

解题步骤 2.6

Perform the row operation $R_{1} = R_{1} + \frac{13}{5} R_{2}$ to make the entry at $1, 4$ a $0$ .

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解题步骤 2.6.1

Perform the row operation $R_{1} = R_{1} + \frac{13}{5} R_{2}$ to make the entry at $1, 4$ a $0$ .

$[\begin{array}{c} 1 + \frac{13}{5} \cdot 0 & - 1 + \frac{13}{5} \cdot 0 & \frac{2}{5} + \frac{13}{5} \cdot 0 & - \frac{13}{5} + \frac{13}{5} \cdot 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}]$

解题步骤 2.6.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - 1 & \frac{2}{5} & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}]$

解题步骤 3

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

$x - y + \frac{2}{5} z = 0$

$0 = 1$

$0 = 0$

解题步骤 4

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

$(y - \frac{2 z}{5}, y, z)$

有限数学 示例

有限数学示例