有限数学示例

$8 x - 5 y + 5 z = - 5$ , $8 x - 5 y + 3 z = 4$ , $9 y - 9 z = - 9$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 8 & - 5 & 5 & - 5 \\ 8 & - 5 & 3 & 4 \\ 0 & 9 & - 9 & - 9 \end{array}]$

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 $\frac{1}{8}$ 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 $\frac{1}{8}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{8}{8} & \frac{- 5}{8} & \frac{5}{8} & \frac{- 5}{8} \\ 8 & - 5 & 3 & 4 \\ 0 & 9 & - 9 & - 9 \end{array}]$

解题步骤 2.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{5}{8} & \frac{5}{8} & - \frac{5}{8} \\ 8 & - 5 & 3 & 4 \\ 0 & 9 & - 9 & - 9 \end{array}]$

解题步骤 2.2

Perform the row operation $R_{2} = R_{2} - 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

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解题步骤 2.2.1

Perform the row operation $R_{2} = R_{2} - 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{5}{8} & \frac{5}{8} & - \frac{5}{8} \\ 8 - 8 \cdot 1 & - 5 - 8 (- \frac{5}{8}) & 3 - 8 (\frac{5}{8}) & 4 - 8 (- \frac{5}{8}) \\ 0 & 9 & - 9 & - 9 \end{array}]$

解题步骤 2.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{5}{8} & \frac{5}{8} & - \frac{5}{8} \\ 0 & 0 & - 2 & 9 \\ 0 & 9 & - 9 & - 9 \end{array}]$

解题步骤 2.3

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 2$ .

$[\begin{array}{c} 1 & - \frac{5}{8} & \frac{5}{8} & - \frac{5}{8} \\ 0 & 9 & - 9 & - 9 \\ 0 & 0 & - 2 & 9 \end{array}]$

解题步骤 2.4

Multiply each element of $R_{2}$ by $\frac{1}{9}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 2.4.1

Multiply each element of $R_{2}$ by $\frac{1}{9}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{5}{8} & \frac{5}{8} & - \frac{5}{8} \\ \frac{0}{9} & \frac{9}{9} & \frac{- 9}{9} & \frac{- 9}{9} \\ 0 & 0 & - 2 & 9 \end{array}]$

解题步骤 2.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{5}{8} & \frac{5}{8} & - \frac{5}{8} \\ 0 & 1 & - 1 & - 1 \\ 0 & 0 & - 2 & 9 \end{array}]$

解题步骤 2.5

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

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解题步骤 2.5.1

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

$[\begin{array}{c} 1 & - \frac{5}{8} & \frac{5}{8} & - \frac{5}{8} \\ 0 & 1 & - 1 & - 1 \\ - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot 9 \end{array}]$

解题步骤 2.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{5}{8} & \frac{5}{8} & - \frac{5}{8} \\ 0 & 1 & - 1 & - 1 \\ 0 & 0 & 1 & - \frac{9}{2} \end{array}]$

解题步骤 2.6

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

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解题步骤 2.6.1

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

$[\begin{array}{c} 1 & - \frac{5}{8} & \frac{5}{8} & - \frac{5}{8} \\ 0 + 0 & 1 + 0 & - 1 + 1 \cdot 1 & - 1 - \frac{9}{2} \\ 0 & 0 & 1 & - \frac{9}{2} \end{array}]$

解题步骤 2.6.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{5}{8} & \frac{5}{8} & - \frac{5}{8} \\ 0 & 1 & 0 & - \frac{11}{2} \\ 0 & 0 & 1 & - \frac{9}{2} \end{array}]$

解题步骤 2.7

Perform the row operation $R_{1} = R_{1} - \frac{5}{8} R_{3}$ to make the entry at $1, 3$ a $0$ .

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解题步骤 2.7.1

Perform the row operation $R_{1} = R_{1} - \frac{5}{8} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{5}{8} \cdot 0 & - \frac{5}{8} - \frac{5}{8} \cdot 0 & \frac{5}{8} - \frac{5}{8} \cdot 1 & - \frac{5}{8} - \frac{5}{8} (- \frac{9}{2}) \\ 0 & 1 & 0 & - \frac{11}{2} \\ 0 & 0 & 1 & - \frac{9}{2} \end{array}]$

解题步骤 2.7.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{5}{8} & 0 & \frac{35}{16} \\ 0 & 1 & 0 & - \frac{11}{2} \\ 0 & 0 & 1 & - \frac{9}{2} \end{array}]$

解题步骤 2.8

Perform the row operation $R_{1} = R_{1} + \frac{5}{8} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 2.8.1

Perform the row operation $R_{1} = R_{1} + \frac{5}{8} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{5}{8} \cdot 0 & - \frac{5}{8} + \frac{5}{8} \cdot 1 & 0 + \frac{5}{8} \cdot 0 & \frac{35}{16} + \frac{5}{8} (- \frac{11}{2}) \\ 0 & 1 & 0 & - \frac{11}{2} \\ 0 & 0 & 1 & - \frac{9}{2} \end{array}]$

解题步骤 2.8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & - \frac{5}{4} \\ 0 & 1 & 0 & - \frac{11}{2} \\ 0 & 0 & 1 & - \frac{9}{2} \end{array}]$

解题步骤 3

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

$x = - \frac{5}{4}$

$y = - \frac{11}{2}$

$z = - \frac{9}{2}$

解题步骤 4

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

$(- \frac{5}{4}, - \frac{11}{2}, - \frac{9}{2})$

有限数学 示例

有限数学示例