有限数学示例

$x + 6 y - 3 z = - 5$ , $z = 8$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 1 & 6 & - 3 & - 5 \\ 0 & 0 & 1 & 8 \end{array}]$

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

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解题步骤 2.1.1

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

$[\begin{array}{c} 1 + 3 \cdot 0 & 6 + 3 \cdot 0 & - 3 + 3 \cdot 1 & - 5 + 3 \cdot 8 \\ 0 & 0 & 1 & 8 \end{array}]$

解题步骤 2.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 6 & 0 & 19 \\ 0 & 0 & 1 & 8 \end{array}]$

解题步骤 3

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

$x + 6 y = 19$

$z = 8$

解题步骤 4

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

$(19 - 6 y, y, 8)$

有限数学 示例