线性代数示例

$2 x - y + 3 z = 8$ , $x - 6 y - z = 0$ , $- 6 x + 3 y - 9 z = 24$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 2 & - 1 & 3 & 8 \\ 1 & - 6 & - 1 & 0 \\ - 6 & 3 & - 9 & 24 \end{array}]$

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 $\frac{1}{2}$ 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 $\frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{2}{2} & \frac{- 1}{2} & \frac{3}{2} & \frac{8}{2} \\ 1 & - 6 & - 1 & 0 \\ - 6 & 3 & - 9 & 24 \end{array}]$

解题步骤 2.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & 4 \\ 1 & - 6 & - 1 & 0 \\ - 6 & 3 & - 9 & 24 \end{array}]$

解题步骤 2.2

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

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解题步骤 2.2.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & 4 \\ 1 - 1 & - 6 + \frac{1}{2} & - 1 - \frac{3}{2} & 0 - 4 \\ - 6 & 3 & - 9 & 24 \end{array}]$

解题步骤 2.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & 4 \\ 0 & - \frac{11}{2} & - \frac{5}{2} & - 4 \\ - 6 & 3 & - 9 & 24 \end{array}]$

解题步骤 2.3

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

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解题步骤 2.3.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & 4 \\ 0 & - \frac{11}{2} & - \frac{5}{2} & - 4 \\ - 6 + 6 \cdot 1 & 3 + 6 (- \frac{1}{2}) & - 9 + 6 (\frac{3}{2}) & 24 + 6 \cdot 4 \end{array}]$

解题步骤 2.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & 4 \\ 0 & - \frac{11}{2} & - \frac{5}{2} & - 4 \\ 0 & 0 & 0 & 48 \end{array}]$

解题步骤 2.4

Multiply each element of $R_{2}$ by $- \frac{2}{11}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 2.4.1

Multiply each element of $R_{2}$ by $- \frac{2}{11}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & 4 \\ - \frac{2}{11} \cdot 0 & - \frac{2}{11} (- \frac{11}{2}) & - \frac{2}{11} (- \frac{5}{2}) & - \frac{2}{11} \cdot - 4 \\ 0 & 0 & 0 & 48 \end{array}]$

解题步骤 2.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & 4 \\ 0 & 1 & \frac{5}{11} & \frac{8}{11} \\ 0 & 0 & 0 & 48 \end{array}]$

解题步骤 2.5

Multiply each element of $R_{3}$ by $\frac{1}{48}$ to make the entry at $3, 4$ a $1$ .

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解题步骤 2.5.1

Multiply each element of $R_{3}$ by $\frac{1}{48}$ to make the entry at $3, 4$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & 4 \\ 0 & 1 & \frac{5}{11} & \frac{8}{11} \\ \frac{0}{48} & \frac{0}{48} & \frac{0}{48} & \frac{48}{48} \end{array}]$

解题步骤 2.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & 4 \\ 0 & 1 & \frac{5}{11} & \frac{8}{11} \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 2.6

Perform the row operation $R_{2} = R_{2} - \frac{8}{11} R_{3}$ to make the entry at $2, 4$ a $0$ .

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解题步骤 2.6.1

Perform the row operation $R_{2} = R_{2} - \frac{8}{11} R_{3}$ to make the entry at $2, 4$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & 4 \\ 0 - \frac{8}{11} \cdot 0 & 1 - \frac{8}{11} \cdot 0 & \frac{5}{11} - \frac{8}{11} \cdot 0 & \frac{8}{11} - \frac{8}{11} \cdot 1 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 2.6.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & 4 \\ 0 & 1 & \frac{5}{11} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 2.7

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

$[\begin{array}{c} 1 - 4 \cdot 0 & - \frac{1}{2} - 4 \cdot 0 & \frac{3}{2} - 4 \cdot 0 & 4 - 4 \cdot 1 \\ 0 & 1 & \frac{5}{11} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 2.7.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & 0 \\ 0 & 1 & \frac{5}{11} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 2.8

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

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解题步骤 2.8.1

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

$[\begin{array}{c} 1 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & \frac{3}{2} + \frac{1}{2} \cdot \frac{5}{11} & 0 + \frac{1}{2} \cdot 0 \\ 0 & 1 & \frac{5}{11} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 2.8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & \frac{19}{11} & 0 \\ 0 & 1 & \frac{5}{11} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 3

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

$x + \frac{19}{11} z = 0$

$y + \frac{5}{11} z = 0$

$0 = 1$

解题步骤 4

The system is inconsistent so there is no solution.

$无解$

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线性代数 示例

线性代数示例