线性代数示例

$x - y + 4 z = 6$ , $2 x + y + z = 1$ , $x + 5 y + z = - 9$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 1 & - 1 & 4 & 6 \\ 2 & 1 & 1 & 1 \\ 1 & 5 & 1 & - 9 \end{array}]$

解题步骤 2

求行简化阶梯形矩阵。

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解题步骤 2.1

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

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解题步骤 2.1.1

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

$[\begin{array}{c} 1 & - 1 & 4 & 6 \\ 2 - 2 \cdot 1 & 1 - 2 \cdot - 1 & 1 - 2 \cdot 4 & 1 - 2 \cdot 6 \\ 1 & 5 & 1 & - 9 \end{array}]$

解题步骤 2.1.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 1 & 4 & 6 \\ 0 & 3 & - 7 & - 11 \\ 1 & 5 & 1 & - 9 \end{array}]$

解题步骤 2.2

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

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解题步骤 2.2.1

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

$[\begin{array}{c} 1 & - 1 & 4 & 6 \\ 0 & 3 & - 7 & - 11 \\ 1 - 1 & 5 + 1 & 1 - 4 & - 9 - 6 \end{array}]$

解题步骤 2.2.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 1 & 4 & 6 \\ 0 & 3 & - 7 & - 11 \\ 0 & 6 & - 3 & - 15 \end{array}]$

解题步骤 2.3

Multiply each element of $R_{2}$ by $\frac{1}{3}$ 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 $\frac{1}{3}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 1 & 4 & 6 \\ \frac{0}{3} & \frac{3}{3} & \frac{- 7}{3} & \frac{- 11}{3} \\ 0 & 6 & - 3 & - 15 \end{array}]$

解题步骤 2.3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 1 & 4 & 6 \\ 0 & 1 & - \frac{7}{3} & - \frac{11}{3} \\ 0 & 6 & - 3 & - 15 \end{array}]$

解题步骤 2.4

Perform the row operation $R_{3} = R_{3} - 6 R_{2}$ to make the entry at $3, 2$ a $0$ .

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解题步骤 2.4.1

Perform the row operation $R_{3} = R_{3} - 6 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - 1 & 4 & 6 \\ 0 & 1 & - \frac{7}{3} & - \frac{11}{3} \\ 0 - 6 \cdot 0 & 6 - 6 \cdot 1 & - 3 - 6 (- \frac{7}{3}) & - 15 - 6 (- \frac{11}{3}) \end{array}]$

解题步骤 2.4.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 1 & 4 & 6 \\ 0 & 1 & - \frac{7}{3} & - \frac{11}{3} \\ 0 & 0 & 11 & 7 \end{array}]$

解题步骤 2.5

Multiply each element of $R_{3}$ by $\frac{1}{11}$ 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 $\frac{1}{11}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - 1 & 4 & 6 \\ 0 & 1 & - \frac{7}{3} & - \frac{11}{3} \\ \frac{0}{11} & \frac{0}{11} & \frac{11}{11} & \frac{7}{11} \end{array}]$

解题步骤 2.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 1 & 4 & 6 \\ 0 & 1 & - \frac{7}{3} & - \frac{11}{3} \\ 0 & 0 & 1 & \frac{7}{11} \end{array}]$

解题步骤 2.6

Perform the row operation $R_{2} = R_{2} + \frac{7}{3} R_{3}$ to make the entry at $2, 3$ a $0$ .

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解题步骤 2.6.1

Perform the row operation $R_{2} = R_{2} + \frac{7}{3} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - 1 & 4 & 6 \\ 0 + \frac{7}{3} \cdot 0 & 1 + \frac{7}{3} \cdot 0 & - \frac{7}{3} + \frac{7}{3} \cdot 1 & - \frac{11}{3} + \frac{7}{3} \cdot \frac{7}{11} \\ 0 & 0 & 1 & \frac{7}{11} \end{array}]$

解题步骤 2.6.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 1 & 4 & 6 \\ 0 & 1 & 0 & - \frac{24}{11} \\ 0 & 0 & 1 & \frac{7}{11} \end{array}]$

解题步骤 2.7

Perform the row operation $R_{1} = R_{1} - 4 R_{3}$ to make the entry at $1, 3$ a $0$ .

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解题步骤 2.7.1

Perform the row operation $R_{1} = R_{1} - 4 R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 4 \cdot 0 & - 1 - 4 \cdot 0 & 4 - 4 \cdot 1 & 6 - 4 (\frac{7}{11}) \\ 0 & 1 & 0 & - \frac{24}{11} \\ 0 & 0 & 1 & \frac{7}{11} \end{array}]$

解题步骤 2.7.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - 1 & 0 & \frac{38}{11} \\ 0 & 1 & 0 & - \frac{24}{11} \\ 0 & 0 & 1 & \frac{7}{11} \end{array}]$

解题步骤 2.8

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 2.8.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 0 + 0 & \frac{38}{11} - \frac{24}{11} \\ 0 & 1 & 0 & - \frac{24}{11} \\ 0 & 0 & 1 & \frac{7}{11} \end{array}]$

解题步骤 2.8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & \frac{14}{11} \\ 0 & 1 & 0 & - \frac{24}{11} \\ 0 & 0 & 1 & \frac{7}{11} \end{array}]$

解题步骤 3

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

$x = \frac{14}{11}$

$y = - \frac{24}{11}$

$z = \frac{7}{11}$

解题步骤 4

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

$(\frac{14}{11}, - \frac{24}{11}, \frac{7}{11})$

线性代数 示例

线性代数示例