有限数学示例

$x - y + 6 z = 4$ , $y - 9 z = - 4$ , $z = 3$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 1 & - 1 & 6 & 4 \\ 0 & 1 & - 9 & - 4 \\ 0 & 0 & 1 & 3 \end{array}]$

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 $R_{2} = R_{2} + 9 R_{3}$ to make the entry at $2, 3$ a $0$ .

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解题步骤 2.1.1

Perform the row operation $R_{2} = R_{2} + 9 R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - 1 & 6 & 4 \\ 0 + 9 \cdot 0 & 1 + 9 \cdot 0 & - 9 + 9 \cdot 1 & - 4 + 9 \cdot 3 \\ 0 & 0 & 1 & 3 \end{array}]$

解题步骤 2.1.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 1 & 6 & 4 \\ 0 & 1 & 0 & 23 \\ 0 & 0 & 1 & 3 \end{array}]$

解题步骤 2.2

Perform the row operation $R_{1} = R_{1} - 6 R_{3}$ to make the entry at $1, 3$ a $0$ .

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解题步骤 2.2.1

Perform the row operation $R_{1} = R_{1} - 6 R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 6 \cdot 0 & - 1 - 6 \cdot 0 & 6 - 6 \cdot 1 & 4 - 6 \cdot 3 \\ 0 & 1 & 0 & 23 \\ 0 & 0 & 1 & 3 \end{array}]$

解题步骤 2.2.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - 1 & 0 & - 14 \\ 0 & 1 & 0 & 23 \\ 0 & 0 & 1 & 3 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 0 + 0 & - 14 + 1 \cdot 23 \\ 0 & 1 & 0 & 23 \\ 0 & 0 & 1 & 3 \end{array}]$

解题步骤 2.3.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & 9 \\ 0 & 1 & 0 & 23 \\ 0 & 0 & 1 & 3 \end{array}]$

解题步骤 3

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

$x = 9$

$y = 23$

$z = 3$

解题步骤 4

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

$(9, 23, 3)$

有限数学 示例