线性代数示例

$[\begin{array}{c} 1 & - 1 & 3 \\ 5 & - 4 & - 4 \\ 7 & - 6 & 2 \end{array}]$

解题步骤 1

Write as an augmented matrix for $A x = 0$ .

$[\begin{array}{c} 1 & - 1 & 3 & 0 \\ 5 & - 4 & - 4 & 0 \\ 7 & - 6 & 2 & 0 \end{array}]$

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

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解题步骤 2.1.1

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

$[\begin{array}{c} 1 & - 1 & 3 & 0 \\ 5 - 5 \cdot 1 & - 4 - 5 \cdot - 1 & - 4 - 5 \cdot 3 & 0 - 5 \cdot 0 \\ 7 & - 6 & 2 & 0 \end{array}]$

解题步骤 2.1.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 1 & 3 & 0 \\ 0 & 1 & - 19 & 0 \\ 7 & - 6 & 2 & 0 \end{array}]$

解题步骤 2.2

Perform the row operation $R_{3} = R_{3} - 7 R_{1}$ to make the entry at $3, 1$ a $0$ .

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解题步骤 2.2.1

Perform the row operation $R_{3} = R_{3} - 7 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - 1 & 3 & 0 \\ 0 & 1 & - 19 & 0 \\ 7 - 7 \cdot 1 & - 6 - 7 \cdot - 1 & 2 - 7 \cdot 3 & 0 - 7 \cdot 0 \end{array}]$

解题步骤 2.2.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 1 & 3 & 0 \\ 0 & 1 & - 19 & 0 \\ 0 & 1 & - 19 & 0 \end{array}]$

解题步骤 2.3

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

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解题步骤 2.3.1

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

$[\begin{array}{c} 1 & - 1 & 3 & 0 \\ 0 & 1 & - 19 & 0 \\ 0 - 0 & 1 - 1 & - 19 + 19 & 0 - 0 \end{array}]$

解题步骤 2.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 1 & 3 & 0 \\ 0 & 1 & - 19 & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

解题步骤 2.4

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 2.4.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 3 - 19 & 0 + 0 \\ 0 & 1 & - 19 & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

解题步骤 2.4.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & - 16 & 0 \\ 0 & 1 & - 19 & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

解题步骤 3

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

$x - 16 z = 0$

$y - 19 z = 0$

$0 = 0$

解题步骤 4

Write a solution vector by solving in terms of the free variables in each row.

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 16 z \\ 19 z \\ z \end{array}]$

解题步骤 5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \\ z \end{array}] = z [\begin{array}{c} 16 \\ 19 \\ 1 \end{array}]$

解题步骤 6

Write as a solution set.

${z [\begin{array}{c} 16 \\ 19 \\ 1 \end{array}] | z \in R}$

线性代数 示例

线性代数示例