线性代数示例

$x + y + 2 z = - 1$ , $x + 3 y - 6 z = 5$ , $2 x + 5 y - 8 z = 7$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 1 & 1 & 2 & - 1 \\ 1 & 3 & - 6 & 5 \\ 2 & 5 & - 8 & 7 \end{array}]$

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

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解题步骤 2.1.1

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

$[\begin{array}{c} 1 & 1 & 2 & - 1 \\ 1 - 1 & 3 - 1 & - 6 - 2 & 5 + 1 \\ 2 & 5 & - 8 & 7 \end{array}]$

解题步骤 2.1.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 1 & 2 & - 1 \\ 0 & 2 & - 8 & 6 \\ 2 & 5 & - 8 & 7 \end{array}]$

解题步骤 2.2

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

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解题步骤 2.2.1

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

$[\begin{array}{c} 1 & 1 & 2 & - 1 \\ 0 & 2 & - 8 & 6 \\ 2 - 2 \cdot 1 & 5 - 2 \cdot 1 & - 8 - 2 \cdot 2 & 7 - 2 \cdot - 1 \end{array}]$

解题步骤 2.2.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & 1 & 2 & - 1 \\ 0 & 2 & - 8 & 6 \\ 0 & 3 & - 12 & 9 \end{array}]$

解题步骤 2.3

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

$[\begin{array}{c} 1 & 1 & 2 & - 1 \\ \frac{0}{2} & \frac{2}{2} & \frac{- 8}{2} & \frac{6}{2} \\ 0 & 3 & - 12 & 9 \end{array}]$

解题步骤 2.3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 1 & 2 & - 1 \\ 0 & 1 & - 4 & 3 \\ 0 & 3 & - 12 & 9 \end{array}]$

解题步骤 2.4

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

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解题步骤 2.4.1

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

$[\begin{array}{c} 1 & 1 & 2 & - 1 \\ 0 & 1 & - 4 & 3 \\ 0 - 3 \cdot 0 & 3 - 3 \cdot 1 & - 12 - 3 \cdot - 4 & 9 - 3 \cdot 3 \end{array}]$

解题步骤 2.4.2

化简 $R_{3}$ 。

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

解题步骤 2.5

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 2.5.1

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

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

解题步骤 2.5.2

化简 $R_{1}$ 。

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

解题步骤 3

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

$x + 6 z = - 4$

$y - 4 z = 3$

$0 = 0$

解题步骤 4

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

$(- 4 - 6 z, 3 + 4 z, z)$

线性代数 示例

线性代数示例