有限数学示例

$- 9 x - 27 y - 12 z = 27$ , $3 x + 9 y + 4 z = - 9$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} - 9 & - 27 & - 12 & 27 \\ 3 & 9 & 4 & - 9 \end{array}]$

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 $- \frac{1}{9}$ 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 $- \frac{1}{9}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{9} \cdot - 9 & - \frac{1}{9} \cdot - 27 & - \frac{1}{9} \cdot - 12 & - \frac{1}{9} \cdot 27 \\ 3 & 9 & 4 & - 9 \end{array}]$

解题步骤 2.1.2

化简 $R_{1}$ 。

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

解题步骤 2.2

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

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解题步骤 2.2.1

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

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

解题步骤 2.2.2

化简 $R_{2}$ 。

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

解题步骤 3

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

$x + 3 y + \frac{4}{3} z = - 3$

$0 = 0$

解题步骤 4

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

$(- 3 - 3 y - \frac{4 z}{3}, y, z)$

有限数学 示例

有限数学示例