有限数学示例

$x - 3 y + z = - 9$ , $x + y - 3 z = - 5$ , $3 x - y + z = 11$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 1 & - 3 & 1 & - 9 \\ 1 & 1 & - 3 & - 5 \\ 3 & - 1 & 1 & 11 \end{array}]$

解题步骤 2

求行简化阶梯形矩阵。

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解题步骤 2.1

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

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解题步骤 2.1.1

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

$[\begin{array}{c} 1 & - 3 & 1 & - 9 \\ 1 - 1 & 1 + 3 & - 3 - 1 & - 5 + 9 \\ 3 & - 1 & 1 & 11 \end{array}]$

解题步骤 2.1.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 3 & 1 & - 9 \\ 0 & 4 & - 4 & 4 \\ 3 & - 1 & 1 & 11 \end{array}]$

解题步骤 2.2

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

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解题步骤 2.2.1

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

$[\begin{array}{c} 1 & - 3 & 1 & - 9 \\ 0 & 4 & - 4 & 4 \\ 3 - 3 \cdot 1 & - 1 - 3 \cdot - 3 & 1 - 3 \cdot 1 & 11 - 3 \cdot - 9 \end{array}]$

解题步骤 2.2.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 3 & 1 & - 9 \\ 0 & 4 & - 4 & 4 \\ 0 & 8 & - 2 & 38 \end{array}]$

解题步骤 2.3

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

$[\begin{array}{c} 1 & - 3 & 1 & - 9 \\ \frac{0}{4} & \frac{4}{4} & \frac{- 4}{4} & \frac{4}{4} \\ 0 & 8 & - 2 & 38 \end{array}]$

解题步骤 2.3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 3 & 1 & - 9 \\ 0 & 1 & - 1 & 1 \\ 0 & 8 & - 2 & 38 \end{array}]$

解题步骤 2.4

Perform the row operation $R_{3} = R_{3} - 8 R_{2}$ to make the entry at $3, 2$ a $0$ .

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解题步骤 2.4.1

Perform the row operation $R_{3} = R_{3} - 8 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - 3 & 1 & - 9 \\ 0 & 1 & - 1 & 1 \\ 0 - 8 \cdot 0 & 8 - 8 \cdot 1 & - 2 - 8 \cdot - 1 & 38 - 8 \cdot 1 \end{array}]$

解题步骤 2.4.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 3 & 1 & - 9 \\ 0 & 1 & - 1 & 1 \\ 0 & 0 & 6 & 30 \end{array}]$

解题步骤 2.5

Multiply each element of $R_{3}$ by $\frac{1}{6}$ 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 $\frac{1}{6}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - 3 & 1 & - 9 \\ 0 & 1 & - 1 & 1 \\ \frac{0}{6} & \frac{0}{6} & \frac{6}{6} & \frac{30}{6} \end{array}]$

解题步骤 2.5.2

化简 $R_{3}$ 。

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

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 $0$ .

$[\begin{array}{c} 1 & - 3 & 1 & - 9 \\ 0 + 0 & 1 + 0 & - 1 + 1 \cdot 1 & 1 + 1 \cdot 5 \\ 0 & 0 & 1 & 5 \end{array}]$

解题步骤 2.6.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 3 & 1 & - 9 \\ 0 & 1 & 0 & 6 \\ 0 & 0 & 1 & 5 \end{array}]$

解题步骤 2.7

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

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解题步骤 2.7.1

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

$[\begin{array}{c} 1 - 0 & - 3 - 0 & 1 - 1 & - 9 - 5 \\ 0 & 1 & 0 & 6 \\ 0 & 0 & 1 & 5 \end{array}]$

解题步骤 2.7.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - 3 & 0 & - 14 \\ 0 & 1 & 0 & 6 \\ 0 & 0 & 1 & 5 \end{array}]$

解题步骤 2.8

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

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解题步骤 2.8.1

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

$[\begin{array}{c} 1 + 3 \cdot 0 & - 3 + 3 \cdot 1 & 0 + 3 \cdot 0 & - 14 + 3 \cdot 6 \\ 0 & 1 & 0 & 6 \\ 0 & 0 & 1 & 5 \end{array}]$

解题步骤 2.8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & 6 \\ 0 & 0 & 1 & 5 \end{array}]$

解题步骤 3

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

$x = 4$

$y = 6$

$z = 5$

解题步骤 4

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

$(4, 6, 5)$

有限数学 示例

有限数学示例