线性代数示例

$2 x + y - z = 2$ , $x + 3 y + 2 z = 1$ , $x + y + z = 2$

解题步骤 1

Write the system as a matrix.

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

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 $1$ .

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

解题步骤 2.1.2

化简 $R_{1}$ 。

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

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 $0$ .

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

解题步骤 2.2.2

化简 $R_{2}$ 。

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

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 $0$ .

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

解题步骤 2.3.2

化简 $R_{3}$ 。

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

解题步骤 2.4

Multiply each element of $R_{2}$ by $\frac{2}{5}$ 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 $\frac{2}{5}$ to make the entry at $2, 2$ a $1$ .

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

解题步骤 2.4.2

化简 $R_{2}$ 。

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

解题步骤 2.5

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

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解题步骤 2.5.1

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

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

解题步骤 2.5.2

化简 $R_{3}$ 。

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

解题步骤 2.6

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

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解题步骤 2.6.1

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

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

解题步骤 2.6.2

化简 $R_{2}$ 。

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

解题步骤 2.7

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

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解题步骤 2.7.1

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

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

解题步骤 2.7.2

化简 $R_{1}$ 。

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

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 $0$ .

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

解题步骤 2.8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 1 & 1 \end{array}]$

解题步骤 3

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

$x = 2$

$y = - 1$

$z = 1$

解题步骤 4

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

$(2, - 1, 1)$

线性代数 示例

线性代数示例