有限数学示例

$x - y + 12 z = 11$ , $y - 8 z = - 11$ , $z = 2$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 1 & - 1 & 12 & 11 \\ 0 & 1 & - 8 & - 11 \\ 0 & 0 & 1 & 2 \end{array}]$

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 $R_{2} = R_{2} + 8 R_{3}$ to make the entry at $2, 3$ a $0$ .

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解题步骤 2.1.1

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

$[\begin{array}{c} 1 & - 1 & 12 & 11 \\ 0 + 8 \cdot 0 & 1 + 8 \cdot 0 & - 8 + 8 \cdot 1 & - 11 + 8 \cdot 2 \\ 0 & 0 & 1 & 2 \end{array}]$

解题步骤 2.1.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 1 & 12 & 11 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & 2 \end{array}]$

解题步骤 2.2

Perform the row operation $R_{1} = R_{1} - 12 R_{3}$ to make the entry at $1, 3$ a $0$ .

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解题步骤 2.2.1

Perform the row operation $R_{1} = R_{1} - 12 R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 12 \cdot 0 & - 1 - 12 \cdot 0 & 12 - 12 \cdot 1 & 11 - 12 \cdot 2 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & 2 \end{array}]$

解题步骤 2.2.2

化简 $R_{1}$ 。

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

解题步骤 2.3

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

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解题步骤 2.3.1

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

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

解题步骤 2.3.2

化简 $R_{1}$ 。

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

解题步骤 3

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

$x = - 8$

$y = 5$

$z = 2$

解题步骤 4

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

$(- 8, 5, 2)$

有限数学 示例