线性代数示例

$- 3 x - 5 y + z = 3$ , $5 x + y - 6 z = 2$ , $x + 4 y + z = - 6$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} - 3 & - 5 & 1 & 3 \\ 5 & 1 & - 6 & 2 \\ 1 & 4 & 1 & - 6 \end{array}]$

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 $- \frac{1}{3}$ 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 $- \frac{1}{3}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot - 5 & - \frac{1}{3} \cdot 1 & - \frac{1}{3} \cdot 3 \\ 5 & 1 & - 6 & 2 \\ 1 & 4 & 1 & - 6 \end{array}]$

解题步骤 2.1.2

化简 $R_{1}$ 。

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

解题步骤 2.2

Perform the row operation $R_{2} = R_{2} - 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

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解题步骤 2.2.1

Perform the row operation $R_{2} = R_{2} - 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

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

解题步骤 2.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{5}{3} & - \frac{1}{3} & - 1 \\ 0 & - \frac{22}{3} & - \frac{13}{3} & 7 \\ 1 & 4 & 1 & - 6 \end{array}]$

解题步骤 2.3

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 2.3.1

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

$[\begin{array}{c} 1 & \frac{5}{3} & - \frac{1}{3} & - 1 \\ 0 & - \frac{22}{3} & - \frac{13}{3} & 7 \\ 1 - 1 & 4 - \frac{5}{3} & 1 + \frac{1}{3} & - 6 + 1 \end{array}]$

解题步骤 2.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{5}{3} & - \frac{1}{3} & - 1 \\ 0 & - \frac{22}{3} & - \frac{13}{3} & 7 \\ 0 & \frac{7}{3} & \frac{4}{3} & - 5 \end{array}]$

解题步骤 2.4

Multiply each element of $R_{2}$ by $- \frac{3}{22}$ 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 $- \frac{3}{22}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{5}{3} & - \frac{1}{3} & - 1 \\ - \frac{3}{22} \cdot 0 & - \frac{3}{22} (- \frac{22}{3}) & - \frac{3}{22} (- \frac{13}{3}) & - \frac{3}{22} \cdot 7 \\ 0 & \frac{7}{3} & \frac{4}{3} & - 5 \end{array}]$

解题步骤 2.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{5}{3} & - \frac{1}{3} & - 1 \\ 0 & 1 & \frac{13}{22} & - \frac{21}{22} \\ 0 & \frac{7}{3} & \frac{4}{3} & - 5 \end{array}]$

解题步骤 2.5

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

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解题步骤 2.5.1

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

$[\begin{array}{c} 1 & \frac{5}{3} & - \frac{1}{3} & - 1 \\ 0 & 1 & \frac{13}{22} & - \frac{21}{22} \\ 0 - \frac{7}{3} \cdot 0 & \frac{7}{3} - \frac{7}{3} \cdot 1 & \frac{4}{3} - \frac{7}{3} \cdot \frac{13}{22} & - 5 - \frac{7}{3} (- \frac{21}{22}) \end{array}]$

解题步骤 2.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{5}{3} & - \frac{1}{3} & - 1 \\ 0 & 1 & \frac{13}{22} & - \frac{21}{22} \\ 0 & 0 & - \frac{1}{22} & - \frac{61}{22} \end{array}]$

解题步骤 2.6

Multiply each element of $R_{3}$ by $- 22$ to make the entry at $3, 3$ a $1$ .

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解题步骤 2.6.1

Multiply each element of $R_{3}$ by $- 22$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{5}{3} & - \frac{1}{3} & - 1 \\ 0 & 1 & \frac{13}{22} & - \frac{21}{22} \\ - 22 \cdot 0 & - 22 \cdot 0 & - 22 (- \frac{1}{22}) & - 22 (- \frac{61}{22}) \end{array}]$

解题步骤 2.6.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{5}{3} & - \frac{1}{3} & - 1 \\ 0 & 1 & \frac{13}{22} & - \frac{21}{22} \\ 0 & 0 & 1 & 61 \end{array}]$

解题步骤 2.7

Perform the row operation $R_{2} = R_{2} - \frac{13}{22} R_{3}$ to make the entry at $2, 3$ a $0$ .

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解题步骤 2.7.1

Perform the row operation $R_{2} = R_{2} - \frac{13}{22} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{5}{3} & - \frac{1}{3} & - 1 \\ 0 - \frac{13}{22} \cdot 0 & 1 - \frac{13}{22} \cdot 0 & \frac{13}{22} - \frac{13}{22} \cdot 1 & - \frac{21}{22} - \frac{13}{22} \cdot 61 \\ 0 & 0 & 1 & 61 \end{array}]$

解题步骤 2.7.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{5}{3} & - \frac{1}{3} & - 1 \\ 0 & 1 & 0 & - 37 \\ 0 & 0 & 1 & 61 \end{array}]$

解题步骤 2.8

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

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解题步骤 2.8.1

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

$[\begin{array}{c} 1 + \frac{1}{3} \cdot 0 & \frac{5}{3} + \frac{1}{3} \cdot 0 & - \frac{1}{3} + \frac{1}{3} \cdot 1 & - 1 + \frac{1}{3} \cdot 61 \\ 0 & 1 & 0 & - 37 \\ 0 & 0 & 1 & 61 \end{array}]$

解题步骤 2.8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{5}{3} & 0 & \frac{58}{3} \\ 0 & 1 & 0 & - 37 \\ 0 & 0 & 1 & 61 \end{array}]$

解题步骤 2.9

Perform the row operation $R_{1} = R_{1} - \frac{5}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 2.9.1

Perform the row operation $R_{1} = R_{1} - \frac{5}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{5}{3} \cdot 0 & \frac{5}{3} - \frac{5}{3} \cdot 1 & 0 - \frac{5}{3} \cdot 0 & \frac{58}{3} - \frac{5}{3} \cdot - 37 \\ 0 & 1 & 0 & - 37 \\ 0 & 0 & 1 & 61 \end{array}]$

解题步骤 2.9.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & 81 \\ 0 & 1 & 0 & - 37 \\ 0 & 0 & 1 & 61 \end{array}]$

解题步骤 3

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

$x = 81$

$y = - 37$

$z = 61$

解题步骤 4

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

$(81, - 37, 61)$

线性代数 示例

线性代数示例