线性代数示例

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

解题步骤 1

求特征值。

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解题步骤 1.1

建立公式以求特征方程

$p (λ)$ 。

$p (λ) = 行列式 (A - λ I_{4})$

解题步骤 1.2

大小为

$4$ 的单位矩阵，是主对角线为 1 而其余元素皆为 0 的

$4 \times 4$ 方阵。

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

解题步骤 1.3

将已知值代入

$p (λ) = 行列式 (A - λ I_{4})$ 。

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解题步骤 1.3.1

代入

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

$A$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] - λ I_{4})$

解题步骤 1.3.2

代入

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

$I_{4}$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}])$

解题步骤 1.4

化简。

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解题步骤 1.4.1

化简每一项。

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解题步骤 1.4.1.1

将

$- λ$ 乘以矩阵中的每一个元素。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2

化简矩阵中的每一个元素。

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解题步骤 1.4.1.2.1

将

$- 1$ 乘以

$1$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.2

乘以

$- λ \cdot 0$ 。

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解题步骤 1.4.1.2.2.1

将

$0$ 乘以

$- 1$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.2.2

将

$0$ 乘以

$λ$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.3

乘以

$- λ \cdot 0$ 。

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解题步骤 1.4.1.2.3.1

将

$0$ 乘以

$- 1$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.3.2

将

$0$ 乘以

$λ$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.4

乘以

$- λ \cdot 0$ 。

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解题步骤 1.4.1.2.4.1

将

$0$ 乘以

$- 1$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.4.2

将

$0$ 乘以

$λ$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.5

乘以

$- λ \cdot 0$ 。

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解题步骤 1.4.1.2.5.1

将

$0$ 乘以

$- 1$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 λ & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.5.2

将

$0$ 乘以

$λ$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.6

将

$- 1$ 乘以

$1$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.7

乘以

$- λ \cdot 0$ 。

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解题步骤 1.4.1.2.7.1

将

$0$ 乘以

$- 1$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.7.2

将

$0$ 乘以

$λ$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.8

乘以

$- λ \cdot 0$ 。

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解题步骤 1.4.1.2.8.1

将

$0$ 乘以

$- 1$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.8.2

将

$0$ 乘以

$λ$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.9

乘以

$- λ \cdot 0$ 。

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解题步骤 1.4.1.2.9.1

将

$0$ 乘以

$- 1$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.9.2

将

$0$ 乘以

$λ$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.10

乘以

$- λ \cdot 0$ 。

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解题步骤 1.4.1.2.10.1

将

$0$ 乘以

$- 1$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 λ & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.10.2

将

$0$ 乘以

$λ$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.11

将

$- 1$ 乘以

$1$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.12

乘以

$- λ \cdot 0$ 。

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解题步骤 1.4.1.2.12.1

将

$0$ 乘以

$- 1$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.12.2

将

$0$ 乘以

$λ$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.13

乘以

$- λ \cdot 0$ 。

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解题步骤 1.4.1.2.13.1

将

$0$ 乘以

$- 1$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.13.2

将

$0$ 乘以

$λ$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.14

乘以

$- λ \cdot 0$ 。

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解题步骤 1.4.1.2.14.1

将

$0$ 乘以

$- 1$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.14.2

将

$0$ 乘以

$λ$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.15

乘以

$- λ \cdot 0$ 。

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解题步骤 1.4.1.2.15.1

将

$0$ 乘以

$- 1$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 & 0 λ & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.15.2

将

$0$ 乘以

$λ$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 & 0 & - λ \cdot 1 \end{array}])$

解题步骤 1.4.1.2.16

将

$- 1$ 乘以

$1$ 。

$p (λ) = 行列式 ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 & 0 & - λ \end{array}])$

解题步骤 1.4.2

加上相应元素。

$p (λ) = 行列式 [\begin{array}{c} 0 - λ & 1 + 0 & 0 + 0 & - 1 + 0 \\ 1 + 0 & 0 - λ & - 1 + 0 & 0 + 0 \\ 0 + 0 & - 1 + 0 & 0 - λ & - 1 + 0 \\ - 1 + 0 & 0 + 0 & - 1 + 0 & 0 - λ \end{array}]$

解题步骤 1.4.3

Simplify each element.

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解题步骤 1.4.3.1

从

$0$ 中减去

$λ$ 。

$p (λ) = 行列式 [\begin{array}{c} - λ & 1 + 0 & 0 + 0 & - 1 + 0 \\ 1 + 0 & 0 - λ & - 1 + 0 & 0 + 0 \\ 0 + 0 & - 1 + 0 & 0 - λ & - 1 + 0 \\ - 1 + 0 & 0 + 0 & - 1 + 0 & 0 - λ \end{array}]$

解题步骤 1.4.3.2

将

$1$ 和

$0$ 相加。

$p (λ) = 行列式 [\begin{array}{c} - λ & 1 & 0 + 0 & - 1 + 0 \\ 1 + 0 & 0 - λ & - 1 + 0 & 0 + 0 \\ 0 + 0 & - 1 + 0 & 0 - λ & - 1 + 0 \\ - 1 + 0 & 0 + 0 & - 1 + 0 & 0 - λ \end{array}]$

解题步骤 1.4.3.3

将

$0$ 和

$0$ 相加。

$p (λ) = 行列式 [\begin{array}{c} - λ & 1 & 0 & - 1 + 0 \\ 1 + 0 & 0 - λ & - 1 + 0 & 0 + 0 \\ 0 + 0 & - 1 + 0 & 0 - λ & - 1 + 0 \\ - 1 + 0 & 0 + 0 & - 1 + 0 & 0 - λ \end{array}]$

解题步骤 1.4.3.4

将

$- 1$ 和

$0$ 相加。

$p (λ) = 行列式 [\begin{array}{c} - λ & 1 & 0 & - 1 \\ 1 + 0 & 0 - λ & - 1 + 0 & 0 + 0 \\ 0 + 0 & - 1 + 0 & 0 - λ & - 1 + 0 \\ - 1 + 0 & 0 + 0 & - 1 + 0 & 0 - λ \end{array}]$

解题步骤 1.4.3.5

将

$1$ 和

$0$ 相加。

$p (λ) = 行列式 [\begin{array}{c} - λ & 1 & 0 & - 1 \\ 1 & 0 - λ & - 1 + 0 & 0 + 0 \\ 0 + 0 & - 1 + 0 & 0 - λ & - 1 + 0 \\ - 1 + 0 & 0 + 0 & - 1 + 0 & 0 - λ \end{array}]$

解题步骤 1.4.3.6

从

$0$ 中减去

$λ$ 。

$p (λ) = 行列式 [\begin{array}{c} - λ & 1 & 0 & - 1 \\ 1 & - λ & - 1 + 0 & 0 + 0 \\ 0 + 0 & - 1 + 0 & 0 - λ & - 1 + 0 \\ - 1 + 0 & 0 + 0 & - 1 + 0 & 0 - λ \end{array}]$

解题步骤 1.4.3.7

将

$- 1$ 和

$0$ 相加。

$p (λ) = 行列式 [\begin{array}{c} - λ & 1 & 0 & - 1 \\ 1 & - λ & - 1 & 0 + 0 \\ 0 + 0 & - 1 + 0 & 0 - λ & - 1 + 0 \\ - 1 + 0 & 0 + 0 & - 1 + 0 & 0 - λ \end{array}]$

解题步骤 1.4.3.8

将

$0$ 和

$0$ 相加。

$p (λ) = 行列式 [\begin{array}{c} - λ & 1 & 0 & - 1 \\ 1 & - λ & - 1 & 0 \\ 0 + 0 & - 1 + 0 & 0 - λ & - 1 + 0 \\ - 1 + 0 & 0 + 0 & - 1 + 0 & 0 - λ \end{array}]$

解题步骤 1.4.3.9

将

$0$ 和

$0$ 相加。

$p (λ) = 行列式 [\begin{array}{c} - λ & 1 & 0 & - 1 \\ 1 & - λ & - 1 & 0 \\ 0 & - 1 + 0 & 0 - λ & - 1 + 0 \\ - 1 + 0 & 0 + 0 & - 1 + 0 & 0 - λ \end{array}]$

解题步骤 1.4.3.10

将

$- 1$ 和

$0$ 相加。

$p (λ) = 行列式 [\begin{array}{c} - λ & 1 & 0 & - 1 \\ 1 & - λ & - 1 & 0 \\ 0 & - 1 & 0 - λ & - 1 + 0 \\ - 1 + 0 & 0 + 0 & - 1 + 0 & 0 - λ \end{array}]$

解题步骤 1.4.3.11

从

$0$ 中减去

$λ$ 。

$p (λ) = 行列式 [\begin{array}{c} - λ & 1 & 0 & - 1 \\ 1 & - λ & - 1 & 0 \\ 0 & - 1 & - λ & - 1 + 0 \\ - 1 + 0 & 0 + 0 & - 1 + 0 & 0 - λ \end{array}]$

解题步骤 1.4.3.12

将

$- 1$ 和

$0$ 相加。

$p (λ) = 行列式 [\begin{array}{c} - λ & 1 & 0 & - 1 \\ 1 & - λ & - 1 & 0 \\ 0 & - 1 & - λ & - 1 \\ - 1 + 0 & 0 + 0 & - 1 + 0 & 0 - λ \end{array}]$

解题步骤 1.4.3.13

将

$- 1$ 和

$0$ 相加。

$p (λ) = 行列式 [\begin{array}{c} - λ & 1 & 0 & - 1 \\ 1 & - λ & - 1 & 0 \\ 0 & - 1 & - λ & - 1 \\ - 1 & 0 + 0 & - 1 + 0 & 0 - λ \end{array}]$

解题步骤 1.4.3.14

将

$0$ 和

$0$ 相加。

$p (λ) = 行列式 [\begin{array}{c} - λ & 1 & 0 & - 1 \\ 1 & - λ & - 1 & 0 \\ 0 & - 1 & - λ & - 1 \\ - 1 & 0 & - 1 + 0 & 0 - λ \end{array}]$

解题步骤 1.4.3.15

将

$- 1$ 和

$0$ 相加。

$p (λ) = 行列式 [\begin{array}{c} - λ & 1 & 0 & - 1 \\ 1 & - λ & - 1 & 0 \\ 0 & - 1 & - λ & - 1 \\ - 1 & 0 & - 1 & 0 - λ \end{array}]$

解题步骤 1.4.3.16

从

$0$ 中减去

$λ$ 。

$p (λ) = 行列式 [\begin{array}{c} - λ & 1 & 0 & - 1 \\ 1 & - λ & - 1 & 0 \\ 0 & - 1 & - λ & - 1 \\ - 1 & 0 & - 1 & - λ \end{array}]$

解题步骤 1.5

Find the determinant.

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解题步骤 1.5.1

Choose the row or column with the most

$0$ elements. If there are no

$0$ elements choose any row or column. Multiply every element in row

$1$ by its cofactor and add.

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解题步骤 1.5.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + & - \\ - & + & - & + \\ + & - & + & - \\ - & + & - & + \end{array} |$

解题步骤 1.5.1.2

The cofactor is the minor with the sign changed if the indices match a

$-$ position on the sign chart.

解题步骤 1.5.1.3

The minor for

$a_{11}$ is the determinant with row

$1$ and column

$1$ deleted.

$| \begin{array}{c} - λ & - 1 & 0 \\ - 1 & - λ & - 1 \\ 0 & - 1 & - λ \end{array} |$

解题步骤 1.5.1.4

Multiply element

$a_{11}$ by its cofactor.

$- λ | \begin{array}{c} - λ & - 1 & 0 \\ - 1 & - λ & - 1 \\ 0 & - 1 & - λ \end{array} |$

解题步骤 1.5.1.5

The minor for

$a_{12}$ is the determinant with row

$1$ and column

$2$ deleted.

$| \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} |$

解题步骤 1.5.1.6

Multiply element

$a_{12}$ by its cofactor.

$- 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} |$

解题步骤 1.5.1.7

The minor for

$a_{13}$ is the determinant with row

$1$ and column

$3$ deleted.

$| \begin{array}{c} 1 & - λ & 0 \\ 0 & - 1 & - 1 \\ - 1 & 0 & - λ \end{array} |$

解题步骤 1.5.1.8

Multiply element

$a_{13}$ by its cofactor.

$0 | \begin{array}{c} 1 & - λ & 0 \\ 0 & - 1 & - 1 \\ - 1 & 0 & - λ \end{array} |$

解题步骤 1.5.1.9

The minor for

$a_{14}$ is the determinant with row

$1$ and column

$4$ deleted.

$| \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.1.10

Multiply element

$a_{14}$ by its cofactor.

$1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.1.11

Add the terms together.

$p (λ) = - λ | \begin{array}{c} - λ & - 1 & 0 \\ - 1 & - λ & - 1 \\ 0 & - 1 & - λ \end{array} | - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 | \begin{array}{c} 1 & - λ & 0 \\ 0 & - 1 & - 1 \\ - 1 & 0 & - λ \end{array} | + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.2

将

$0$ 乘以

$| \begin{array}{c} 1 & - λ & 0 \\ 0 & - 1 & - 1 \\ - 1 & 0 & - λ \end{array} |$ 。

$p (λ) = - λ | \begin{array}{c} - λ & - 1 & 0 \\ - 1 & - λ & - 1 \\ 0 & - 1 & - λ \end{array} | - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3

计算

$| \begin{array}{c} - λ & - 1 & 0 \\ - 1 & - λ & - 1 \\ 0 & - 1 & - λ \end{array} |$ 。

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解题步骤 1.5.3.1

Choose the row or column with the most

$0$ elements. If there are no

$0$ elements choose any row or column. Multiply every element in row

$1$ by its cofactor and add.

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解题步骤 1.5.3.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

解题步骤 1.5.3.1.2

The cofactor is the minor with the sign changed if the indices match a

$-$ position on the sign chart.

解题步骤 1.5.3.1.3

The minor for

$a_{11}$ is the determinant with row

$1$ and column

$1$ deleted.

$| \begin{array}{c} - λ & - 1 \\ - 1 & - λ \end{array} |$

解题步骤 1.5.3.1.4

Multiply element

$a_{11}$ by its cofactor.

$- λ | \begin{array}{c} - λ & - 1 \\ - 1 & - λ \end{array} |$

解题步骤 1.5.3.1.5

The minor for

$a_{12}$ is the determinant with row

$1$ and column

$2$ deleted.

$| \begin{array}{c} - 1 & - 1 \\ 0 & - λ \end{array} |$

解题步骤 1.5.3.1.6

Multiply element

$a_{12}$ by its cofactor.

$1 | \begin{array}{c} - 1 & - 1 \\ 0 & - λ \end{array} |$

解题步骤 1.5.3.1.7

The minor for

$a_{13}$ is the determinant with row

$1$ and column

$3$ deleted.

$| \begin{array}{c} - 1 & - λ \\ 0 & - 1 \end{array} |$

解题步骤 1.5.3.1.8

Multiply element

$a_{13}$ by its cofactor.

$0 | \begin{array}{c} - 1 & - λ \\ 0 & - 1 \end{array} |$

解题步骤 1.5.3.1.9

Add the terms together.

$p (λ) = - λ (- λ | \begin{array}{c} - λ & - 1 \\ - 1 & - λ \end{array} | + 1 | \begin{array}{c} - 1 & - 1 \\ 0 & - λ \end{array} | + 0 | \begin{array}{c} - 1 & - λ \\ 0 & - 1 \end{array} |) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.2

将

$0$ 乘以

$| \begin{array}{c} - 1 & - λ \\ 0 & - 1 \end{array} |$ 。

$p (λ) = - λ (- λ | \begin{array}{c} - λ & - 1 \\ - 1 & - λ \end{array} | + 1 | \begin{array}{c} - 1 & - 1 \\ 0 & - λ \end{array} | + 0) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.3

计算

$| \begin{array}{c} - λ & - 1 \\ - 1 & - λ \end{array} |$ 。

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解题步骤 1.5.3.3.1

可以使用公式

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 求

$2 \times 2$ 矩阵的行列式。

$p (λ) = - λ (- λ (- λ (- λ) - - - 1) + 1 | \begin{array}{c} - 1 & - 1 \\ 0 & - λ \end{array} | + 0) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.3.2

化简每一项。

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解题步骤 1.5.3.3.2.1

使用乘法的交换性质重写。

$p (λ) = - λ (- λ (- 1 \cdot - 1 λ \cdot λ - - - 1) + 1 | \begin{array}{c} - 1 & - 1 \\ 0 & - λ \end{array} | + 0) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.3.2.2

通过指数相加将

$λ$ 乘以

$λ$ 。

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解题步骤 1.5.3.3.2.2.1

移动

$λ$ 。

$p (λ) = - λ (- λ (- 1 \cdot - 1 (λ \cdot λ) - - - 1) + 1 | \begin{array}{c} - 1 & - 1 \\ 0 & - λ \end{array} | + 0) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.3.2.2.2

将

$λ$ 乘以

$λ$ 。

$p (λ) = - λ (- λ (- 1 \cdot - 1 λ^{2} - - - 1) + 1 | \begin{array}{c} - 1 & - 1 \\ 0 & - λ \end{array} | + 0) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.3.2.3

将

$- 1$ 乘以

$- 1$ 。

$p (λ) = - λ (- λ (1 λ^{2} - - - 1) + 1 | \begin{array}{c} - 1 & - 1 \\ 0 & - λ \end{array} | + 0) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.3.2.4

将

$λ^{2}$ 乘以

$1$ 。

$p (λ) = - λ (- λ (λ^{2} - - - 1) + 1 | \begin{array}{c} - 1 & - 1 \\ 0 & - λ \end{array} | + 0) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.3.2.5

乘以

$- - - 1$ 。

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解题步骤 1.5.3.3.2.5.1

将

$- 1$ 乘以

$- 1$ 。

$p (λ) = - λ (- λ (λ^{2} - 1 \cdot 1) + 1 | \begin{array}{c} - 1 & - 1 \\ 0 & - λ \end{array} | + 0) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.3.2.5.2

将

$- 1$ 乘以

$1$ 。

$p (λ) = - λ (- λ (λ^{2} - 1) + 1 | \begin{array}{c} - 1 & - 1 \\ 0 & - λ \end{array} | + 0) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.4

计算

$| \begin{array}{c} - 1 & - 1 \\ 0 & - λ \end{array} |$ 。

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解题步骤 1.5.3.4.1

可以使用公式

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 求

$2 \times 2$ 矩阵的行列式。

$p (λ) = - λ (- λ (λ^{2} - 1) + 1 (- - λ + 0 \cdot - 1) + 0) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.4.2

化简行列式。

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解题步骤 1.5.3.4.2.1

化简每一项。

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解题步骤 1.5.3.4.2.1.1

乘以

$- - λ$ 。

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解题步骤 1.5.3.4.2.1.1.1

将

$- 1$ 乘以

$- 1$ 。

$p (λ) = - λ (- λ (λ^{2} - 1) + 1 (1 λ + 0 \cdot - 1) + 0) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.4.2.1.1.2

将

$λ$ 乘以

$1$ 。

$p (λ) = - λ (- λ (λ^{2} - 1) + 1 (λ + 0 \cdot - 1) + 0) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.4.2.1.2

将

$0$ 乘以

$- 1$ 。

$p (λ) = - λ (- λ (λ^{2} - 1) + 1 (λ + 0) + 0) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.4.2.2

将

$λ$ 和

$0$ 相加。

$p (λ) = - λ (- λ (λ^{2} - 1) + 1 λ + 0) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.5

化简行列式。

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解题步骤 1.5.3.5.1

将

$- λ (λ^{2} - 1) + 1 λ$ 和

$0$ 相加。

$p (λ) = - λ (- λ (λ^{2} - 1) + 1 λ) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.5.2

化简每一项。

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解题步骤 1.5.3.5.2.1

运用分配律。

$p (λ) = - λ (- λ \cdot λ^{2} - λ \cdot - 1 + 1 λ) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.5.2.2

通过指数相加将

$λ$ 乘以

$λ^{2}$ 。

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解题步骤 1.5.3.5.2.2.1

移动

$λ^{2}$ 。

$p (λ) = - λ (- (λ^{2} λ) - λ \cdot - 1 + 1 λ) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.5.2.2.2

将

$λ^{2}$ 乘以

$λ$ 。

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解题步骤 1.5.3.5.2.2.2.1

对

$λ$ 进行

$1$ 次方运算。

$p (λ) = - λ (- (λ^{2} λ^{1}) - λ \cdot - 1 + 1 λ) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.5.2.2.2.2

使用幂法则

$a^{m} a^{n} = a^{m + n}$ 合并指数。

$p (λ) = - λ (- λ^{2 + 1} - λ \cdot - 1 + 1 λ) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.5.2.2.3

将

$2$ 和

$1$ 相加。

$p (λ) = - λ (- λ^{3} - λ \cdot - 1 + 1 λ) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.5.2.3

乘以

$- λ \cdot - 1$ 。

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解题步骤 1.5.3.5.2.3.1

将

$- 1$ 乘以

$- 1$ 。

$p (λ) = - λ (- λ^{3} + 1 λ + 1 λ) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.5.2.3.2

将

$λ$ 乘以

$1$ 。

$p (λ) = - λ (- λ^{3} + λ + 1 λ) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.5.2.4

将

$λ$ 乘以

$1$ 。

$p (λ) = - λ (- λ^{3} + λ + λ) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.3.5.3

将

$λ$ 和

$λ$ 相加。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 | \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} | + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.4

计算

$| \begin{array}{c} 1 & - 1 & 0 \\ 0 & - λ & - 1 \\ - 1 & - 1 & - λ \end{array} |$ 。

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解题步骤 1.5.4.1

Choose the row or column with the most

$0$ elements. If there are no

$0$ elements choose any row or column. Multiply every element in row

$2$ by its cofactor and add.

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解题步骤 1.5.4.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

解题步骤 1.5.4.1.2

The cofactor is the minor with the sign changed if the indices match a

$-$ position on the sign chart.

解题步骤 1.5.4.1.3

The minor for

$a_{21}$ is the determinant with row

$2$ and column

$1$ deleted.

$| \begin{array}{c} - 1 & 0 \\ - 1 & - λ \end{array} |$

解题步骤 1.5.4.1.4

Multiply element

$a_{21}$ by its cofactor.

$0 | \begin{array}{c} - 1 & 0 \\ - 1 & - λ \end{array} |$

解题步骤 1.5.4.1.5

The minor for

$a_{22}$ is the determinant with row

$2$ and column

$2$ deleted.

$| \begin{array}{c} 1 & 0 \\ - 1 & - λ \end{array} |$

解题步骤 1.5.4.1.6

Multiply element

$a_{22}$ by its cofactor.

$- λ | \begin{array}{c} 1 & 0 \\ - 1 & - λ \end{array} |$

解题步骤 1.5.4.1.7

The minor for

$a_{23}$ is the determinant with row

$2$ and column

$3$ deleted.

$| \begin{array}{c} 1 & - 1 \\ - 1 & - 1 \end{array} |$

解题步骤 1.5.4.1.8

Multiply element

$a_{23}$ by its cofactor.

$1 | \begin{array}{c} 1 & - 1 \\ - 1 & - 1 \end{array} |$

解题步骤 1.5.4.1.9

Add the terms together.

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (0 | \begin{array}{c} - 1 & 0 \\ - 1 & - λ \end{array} | - λ | \begin{array}{c} 1 & 0 \\ - 1 & - λ \end{array} | + 1 | \begin{array}{c} 1 & - 1 \\ - 1 & - 1 \end{array} |) + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.4.2

将

$0$ 乘以

$| \begin{array}{c} - 1 & 0 \\ - 1 & - λ \end{array} |$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (0 - λ | \begin{array}{c} 1 & 0 \\ - 1 & - λ \end{array} | + 1 | \begin{array}{c} 1 & - 1 \\ - 1 & - 1 \end{array} |) + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.4.3

计算

$| \begin{array}{c} 1 & 0 \\ - 1 & - λ \end{array} |$ 。

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解题步骤 1.5.4.3.1

可以使用公式

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 求

$2 \times 2$ 矩阵的行列式。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (0 - λ (1 (- λ) - - 0) + 1 | \begin{array}{c} 1 & - 1 \\ - 1 & - 1 \end{array} |) + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.4.3.2

化简行列式。

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解题步骤 1.5.4.3.2.1

化简每一项。

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解题步骤 1.5.4.3.2.1.1

将

$- λ$ 乘以

$1$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (0 - λ (- λ - - 0) + 1 | \begin{array}{c} 1 & - 1 \\ - 1 & - 1 \end{array} |) + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.4.3.2.1.2

乘以

$- - 0$ 。

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解题步骤 1.5.4.3.2.1.2.1

将

$- 1$ 乘以

$0$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (0 - λ (- λ - 0) + 1 | \begin{array}{c} 1 & - 1 \\ - 1 & - 1 \end{array} |) + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.4.3.2.1.2.2

将

$- 1$ 乘以

$0$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (0 - λ (- λ + 0) + 1 | \begin{array}{c} 1 & - 1 \\ - 1 & - 1 \end{array} |) + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.4.3.2.2

将

$- λ$ 和

$0$ 相加。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (0 - λ (- λ) + 1 | \begin{array}{c} 1 & - 1 \\ - 1 & - 1 \end{array} |) + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.4.4

计算

$| \begin{array}{c} 1 & - 1 \\ - 1 & - 1 \end{array} |$ 。

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解题步骤 1.5.4.4.1

可以使用公式

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 求

$2 \times 2$ 矩阵的行列式。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (0 - λ (- λ) + 1 (1 \cdot - 1 - - - 1)) + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.4.4.2

化简行列式。

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解题步骤 1.5.4.4.2.1

化简每一项。

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解题步骤 1.5.4.4.2.1.1

将

$- 1$ 乘以

$1$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (0 - λ (- λ) + 1 (- 1 - - - 1)) + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.4.4.2.1.2

乘以

$- - - 1$ 。

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解题步骤 1.5.4.4.2.1.2.1

将

$- 1$ 乘以

$- 1$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (0 - λ (- λ) + 1 (- 1 - 1 \cdot 1)) + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.4.4.2.1.2.2

将

$- 1$ 乘以

$1$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (0 - λ (- λ) + 1 (- 1 - 1)) + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.4.4.2.2

从

$- 1$ 中减去

$1$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (0 - λ (- λ) + 1 \cdot - 2) + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.4.5

化简行列式。

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解题步骤 1.5.4.5.1

从

$0$ 中减去

$λ (- λ)$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (- λ (- λ) + 1 \cdot - 2) + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.4.5.2

化简每一项。

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解题步骤 1.5.4.5.2.1

使用乘法的交换性质重写。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (- 1 \cdot - 1 λ \cdot λ + 1 \cdot - 2) + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.4.5.2.2

通过指数相加将

$λ$ 乘以

$λ$ 。

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解题步骤 1.5.4.5.2.2.1

移动

$λ$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (- 1 \cdot - 1 (λ \cdot λ) + 1 \cdot - 2) + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.4.5.2.2.2

将

$λ$ 乘以

$λ$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (- 1 \cdot - 1 λ^{2} + 1 \cdot - 2) + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.4.5.2.3

将

$- 1$ 乘以

$- 1$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (1 λ^{2} + 1 \cdot - 2) + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.4.5.2.4

将

$λ^{2}$ 乘以

$1$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (λ^{2} + 1 \cdot - 2) + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.4.5.2.5

将

$- 2$ 乘以

$1$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (λ^{2} - 2) + 0 + 1 | \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$

解题步骤 1.5.5

计算

$| \begin{array}{c} 1 & - λ & - 1 \\ 0 & - 1 & - λ \\ - 1 & 0 & - 1 \end{array} |$ 。

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解题步骤 1.5.5.1

Choose the row or column with the most

$0$ elements. If there are no

$0$ elements choose any row or column. Multiply every element in row

$2$ by its cofactor and add.

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解题步骤 1.5.5.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

解题步骤 1.5.5.1.2

The cofactor is the minor with the sign changed if the indices match a

$-$ position on the sign chart.

解题步骤 1.5.5.1.3

The minor for

$a_{21}$ is the determinant with row

$2$ and column

$1$ deleted.

$| \begin{array}{c} - λ & - 1 \\ 0 & - 1 \end{array} |$

解题步骤 1.5.5.1.4

Multiply element

$a_{21}$ by its cofactor.

$0 | \begin{array}{c} - λ & - 1 \\ 0 & - 1 \end{array} |$

解题步骤 1.5.5.1.5

The minor for

$a_{22}$ is the determinant with row

$2$ and column

$2$ deleted.

$| \begin{array}{c} 1 & - 1 \\ - 1 & - 1 \end{array} |$

解题步骤 1.5.5.1.6

Multiply element

$a_{22}$ by its cofactor.

$- 1 | \begin{array}{c} 1 & - 1 \\ - 1 & - 1 \end{array} |$

解题步骤 1.5.5.1.7

The minor for

$a_{23}$ is the determinant with row

$2$ and column

$3$ deleted.

$| \begin{array}{c} 1 & - λ \\ - 1 & 0 \end{array} |$

解题步骤 1.5.5.1.8

Multiply element

$a_{23}$ by its cofactor.

$λ | \begin{array}{c} 1 & - λ \\ - 1 & 0 \end{array} |$

解题步骤 1.5.5.1.9

Add the terms together.

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (λ^{2} - 2) + 0 + 1 (0 | \begin{array}{c} - λ & - 1 \\ 0 & - 1 \end{array} | - 1 | \begin{array}{c} 1 & - 1 \\ - 1 & - 1 \end{array} | + λ | \begin{array}{c} 1 & - λ \\ - 1 & 0 \end{array} |)$

解题步骤 1.5.5.2

将

$0$ 乘以

$| \begin{array}{c} - λ & - 1 \\ 0 & - 1 \end{array} |$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (λ^{2} - 2) + 0 + 1 (0 - 1 | \begin{array}{c} 1 & - 1 \\ - 1 & - 1 \end{array} | + λ | \begin{array}{c} 1 & - λ \\ - 1 & 0 \end{array} |)$

解题步骤 1.5.5.3

计算

$| \begin{array}{c} 1 & - 1 \\ - 1 & - 1 \end{array} |$ 。

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解题步骤 1.5.5.3.1

可以使用公式

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 求

$2 \times 2$ 矩阵的行列式。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (λ^{2} - 2) + 0 + 1 (0 - 1 (1 \cdot - 1 - - - 1) + λ | \begin{array}{c} 1 & - λ \\ - 1 & 0 \end{array} |)$

解题步骤 1.5.5.3.2

化简行列式。

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解题步骤 1.5.5.3.2.1

化简每一项。

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解题步骤 1.5.5.3.2.1.1

将

$- 1$ 乘以

$1$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (λ^{2} - 2) + 0 + 1 (0 - 1 (- 1 - - - 1) + λ | \begin{array}{c} 1 & - λ \\ - 1 & 0 \end{array} |)$

解题步骤 1.5.5.3.2.1.2

乘以

$- - - 1$ 。

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解题步骤 1.5.5.3.2.1.2.1

将

$- 1$ 乘以

$- 1$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (λ^{2} - 2) + 0 + 1 (0 - 1 (- 1 - 1 \cdot 1) + λ | \begin{array}{c} 1 & - λ \\ - 1 & 0 \end{array} |)$

解题步骤 1.5.5.3.2.1.2.2

将

$- 1$ 乘以

$1$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (λ^{2} - 2) + 0 + 1 (0 - 1 (- 1 - 1) + λ | \begin{array}{c} 1 & - λ \\ - 1 & 0 \end{array} |)$

解题步骤 1.5.5.3.2.2

从

$- 1$ 中减去

$1$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (λ^{2} - 2) + 0 + 1 (0 - 1 \cdot - 2 + λ | \begin{array}{c} 1 & - λ \\ - 1 & 0 \end{array} |)$

解题步骤 1.5.5.4

计算

$| \begin{array}{c} 1 & - λ \\ - 1 & 0 \end{array} |$ 。

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解题步骤 1.5.5.4.1

可以使用公式

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 求

$2 \times 2$ 矩阵的行列式。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (λ^{2} - 2) + 0 + 1 (0 - 1 \cdot - 2 + λ (1 \cdot 0 - - - λ))$

解题步骤 1.5.5.4.2

化简行列式。

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解题步骤 1.5.5.4.2.1

化简每一项。

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解题步骤 1.5.5.4.2.1.1

将

$0$ 乘以

$1$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (λ^{2} - 2) + 0 + 1 (0 - 1 \cdot - 2 + λ (0 - - - λ))$

解题步骤 1.5.5.4.2.1.2

乘以

$- - λ$ 。

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解题步骤 1.5.5.4.2.1.2.1

将

$- 1$ 乘以

$- 1$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (λ^{2} - 2) + 0 + 1 (0 - 1 \cdot - 2 + λ (0 - (1 λ)))$

解题步骤 1.5.5.4.2.1.2.2

将

$λ$ 乘以

$1$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (λ^{2} - 2) + 0 + 1 (0 - 1 \cdot - 2 + λ (0 - λ))$

解题步骤 1.5.5.4.2.2

从

$0$ 中减去

$λ$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (λ^{2} - 2) + 0 + 1 (0 - 1 \cdot - 2 + λ (- λ))$

解题步骤 1.5.5.5

化简行列式。

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解题步骤 1.5.5.5.1

从

$0$ 中减去

$1 \cdot - 2$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (λ^{2} - 2) + 0 + 1 (- 1 \cdot - 2 + λ (- λ))$

解题步骤 1.5.5.5.2

化简每一项。

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解题步骤 1.5.5.5.2.1

将

$- 1$ 乘以

$- 2$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (λ^{2} - 2) + 0 + 1 (2 + λ (- λ))$

解题步骤 1.5.5.5.2.2

使用乘法的交换性质重写。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (λ^{2} - 2) + 0 + 1 (2 - λ \cdot λ)$

解题步骤 1.5.5.5.2.3

通过指数相加将

$λ$ 乘以 $λ$ 。

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解题步骤 1.5.5.5.2.3.1

移动 $λ$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (λ^{2} - 2) + 0 + 1 (2 - (λ \cdot λ))$

解题步骤 1.5.5.5.2.3.2

将 $λ$ 乘以 $λ$ 。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (λ^{2} - 2) + 0 + 1 (2 - λ^{2})$

解题步骤 1.5.5.5.3

将 $2$ 和 $- λ^{2}$ 重新排序。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (λ^{2} - 2) + 0 + 1 (- λ^{2} + 2)$

解题步骤 1.5.6

化简行列式。

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解题步骤 1.5.6.1

将 $- λ (- λ^{3} + 2 λ) - 1 (λ^{2} - 2)$ 和 $0$ 相加。

$p (λ) = - λ (- λ^{3} + 2 λ) - 1 (λ^{2} - 2) + 1 (- λ^{2} + 2)$

解题步骤 1.5.6.2

化简每一项。

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解题步骤 1.5.6.2.1

运用分配律。

$p (λ) = - λ (- λ^{3}) - λ (2 λ) - 1 (λ^{2} - 2) + 1 (- λ^{2} + 2)$

解题步骤 1.5.6.2.2

使用乘法的交换性质重写。

$p (λ) = - 1 \cdot - 1 λ \cdot λ^{3} - λ (2 λ) - 1 (λ^{2} - 2) + 1 (- λ^{2} + 2)$

解题步骤 1.5.6.2.3

使用乘法的交换性质重写。

$p (λ) = - 1 \cdot - 1 λ \cdot λ^{3} - 1 \cdot 2 λ \cdot λ - 1 (λ^{2} - 2) + 1 (- λ^{2} + 2)$

解题步骤 1.5.6.2.4

化简每一项。

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解题步骤 1.5.6.2.4.1

通过指数相加将 $λ$ 乘以 $λ^{3}$ 。

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解题步骤 1.5.6.2.4.1.1

移动 $λ^{3}$ 。

$p (λ) = - 1 \cdot - 1 (λ^{3} λ) - 1 \cdot 2 λ \cdot λ - 1 (λ^{2} - 2) + 1 (- λ^{2} + 2)$

解题步骤 1.5.6.2.4.1.2

将 $λ^{3}$ 乘以 $λ$ 。

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解题步骤 1.5.6.2.4.1.2.1

对 $λ$ 进行 $1$ 次方运算。

$p (λ) = - 1 \cdot - 1 (λ^{3} λ^{1}) - 1 \cdot 2 λ \cdot λ - 1 (λ^{2} - 2) + 1 (- λ^{2} + 2)$

解题步骤 1.5.6.2.4.1.2.2

使用幂法则 $a^{m} a^{n} = a^{m + n}$ 合并指数。

$p (λ) = - 1 \cdot - 1 λ^{3 + 1} - 1 \cdot 2 λ \cdot λ - 1 (λ^{2} - 2) + 1 (- λ^{2} + 2)$

解题步骤 1.5.6.2.4.1.3

将 $3$ 和 $1$ 相加。

$p (λ) = - 1 \cdot - 1 λ^{4} - 1 \cdot 2 λ \cdot λ - 1 (λ^{2} - 2) + 1 (- λ^{2} + 2)$

解题步骤 1.5.6.2.4.2

将 $- 1$ 乘以 $- 1$ 。

$p (λ) = 1 λ^{4} - 1 \cdot 2 λ \cdot λ - 1 (λ^{2} - 2) + 1 (- λ^{2} + 2)$

解题步骤 1.5.6.2.4.3

将 $λ^{4}$ 乘以 $1$ 。

$p (λ) = λ^{4} - 1 \cdot 2 λ \cdot λ - 1 (λ^{2} - 2) + 1 (- λ^{2} + 2)$

解题步骤 1.5.6.2.4.4

通过指数相加将 $λ$ 乘以 $λ$ 。

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解题步骤 1.5.6.2.4.4.1

移动 $λ$ 。

$p (λ) = λ^{4} - 1 \cdot 2 (λ \cdot λ) - 1 (λ^{2} - 2) + 1 (- λ^{2} + 2)$

解题步骤 1.5.6.2.4.4.2

将 $λ$ 乘以 $λ$ 。

$p (λ) = λ^{4} - 1 \cdot 2 λ^{2} - 1 (λ^{2} - 2) + 1 (- λ^{2} + 2)$

解题步骤 1.5.6.2.4.5

将 $- 1$ 乘以 $2$ 。

$p (λ) = λ^{4} - 2 λ^{2} - 1 (λ^{2} - 2) + 1 (- λ^{2} + 2)$

解题步骤 1.5.6.2.5

运用分配律。

$p (λ) = λ^{4} - 2 λ^{2} - 1 λ^{2} - 1 \cdot - 2 + 1 (- λ^{2} + 2)$

解题步骤 1.5.6.2.6

将 $- 1 λ^{2}$ 重写为 $- λ^{2}$ 。

$p (λ) = λ^{4} - 2 λ^{2} - λ^{2} - 1 \cdot - 2 + 1 (- λ^{2} + 2)$

解题步骤 1.5.6.2.7

将 $- 1$ 乘以 $- 2$ 。

$p (λ) = λ^{4} - 2 λ^{2} - λ^{2} + 2 + 1 (- λ^{2} + 2)$

解题步骤 1.5.6.2.8

将 $- λ^{2} + 2$ 乘以 $1$ 。

$p (λ) = λ^{4} - 2 λ^{2} - λ^{2} + 2 - λ^{2} + 2$

解题步骤 1.5.6.3

从 $- 2 λ^{2}$ 中减去 $λ^{2}$ 。

$p (λ) = λ^{4} - 3 λ^{2} + 2 - λ^{2} + 2$

解题步骤 1.5.6.4

从 $- 3 λ^{2}$ 中减去 $λ^{2}$ 。

$p (λ) = λ^{4} - 4 λ^{2} + 2 + 2$

解题步骤 1.5.6.5

将 $2$ 和 $2$ 相加。

$p (λ) = λ^{4} - 4 λ^{2} + 4$

解题步骤 1.6

使特征多项式等于 $0$ ，以求特征值 $λ$ 。

$λ^{4} - 4 λ^{2} + 4 = 0$

解题步骤 1.7

求解 $λ$ 。

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解题步骤 1.7.1

将 $u = λ^{2}$ 代入方程。这将使得二次公式变得更容易使用。

$u^{2} - 4 u + 4 = 0$

$u = λ^{2}$

解题步骤 1.7.2

使用完全平方法则进行因式分解。

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解题步骤 1.7.2.1

将 $4$ 重写为 $2^{2}$ 。

$u^{2} - 4 u + 2^{2} = 0$

解题步骤 1.7.2.2

请检查中间项是否为第一项被平方数和第三项被平方数的乘积的两倍。

$4 u = 2 \cdot u \cdot 2$

解题步骤 1.7.2.3

重写多项式。

$u^{2} - 2 \cdot u \cdot 2 + 2^{2} = 0$

解题步骤 1.7.2.4

使用完全平方三项式法则对 $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ 进行因式分解，其中 $a = u$ 和 $b = 2$ 。

${(u - 2)}^{2} = 0$

解题步骤 1.7.3

将 $u - 2$ 设为等于 $0$ 。

$u - 2 = 0$

解题步骤 1.7.4

在等式两边都加上 $2$ 。

$u = 2$

解题步骤 1.7.5

将 $u = λ^{2}$ 的真实值代入回已解的方程中。

$λ^{2} = 2$

解题步骤 1.7.6

求解 $λ$ 的方程。

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解题步骤 1.7.6.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$λ = \pm \sqrt{2}$

解题步骤 1.7.6.2

完全解为同时包括解的正数和负数部分的结果。

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解题步骤 1.7.6.2.1

首先，利用 $\pm$ 的正值求第一个解。

$λ = \sqrt{2}$

解题步骤 1.7.6.2.2

下一步，使用 $\pm$ 的负值来求第二个解。

$λ = - \sqrt{2}$

解题步骤 1.7.6.2.3

完全解为同时包括解的正数和负数部分的结果。

$λ = \sqrt{2}, - \sqrt{2}$

解题步骤 2

The eigenvector is equal to the null space of the matrix minus the eigenvalue times the identity matrix where $N$ is the null space and $I$ is the identity matrix.

$ε_{A} = N (A - λ I_{4})$

解题步骤 3

Find the eigenvector using the eigenvalue $λ = \sqrt{2}$ .

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解题步骤 3.1

将已知值代入公式中。

$N ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] - \sqrt{2} [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}])$

解题步骤 3.2

化简。

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解题步骤 3.2.1

化简每一项。

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解题步骤 3.2.1.1

将 $- \sqrt{2}$ 乘以矩阵中的每一个元素。

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

解题步骤 3.2.1.2

化简矩阵中的每一个元素。

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解题步骤 3.2.1.2.1

将 $- 1$ 乘以 $1$ 。

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

解题步骤 3.2.1.2.2

乘以 $- \sqrt{2} \cdot 0$ 。

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解题步骤 3.2.1.2.2.1

将 $0$ 乘以 $- 1$ 。

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

解题步骤 3.2.1.2.2.2

将 $0$ 乘以 $\sqrt{2}$ 。

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

解题步骤 3.2.1.2.3

乘以 $- \sqrt{2} \cdot 0$ 。

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解题步骤 3.2.1.2.3.1

将 $0$ 乘以 $- 1$ 。

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

解题步骤 3.2.1.2.3.2

将 $0$ 乘以 $\sqrt{2}$ 。

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

解题步骤 3.2.1.2.4

乘以 $- \sqrt{2} \cdot 0$ 。

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解题步骤 3.2.1.2.4.1

将 $0$ 乘以 $- 1$ 。

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

解题步骤 3.2.1.2.4.2

将 $0$ 乘以 $\sqrt{2}$ 。

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

解题步骤 3.2.1.2.5

乘以 $- \sqrt{2} \cdot 0$ 。

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解题步骤 3.2.1.2.5.1

将 $0$ 乘以 $- 1$ 。

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

解题步骤 3.2.1.2.5.2

将 $0$ 乘以 $\sqrt{2}$ 。

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

解题步骤 3.2.1.2.6

将 $- 1$ 乘以 $1$ 。

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

解题步骤 3.2.1.2.7

乘以 $- \sqrt{2} \cdot 0$ 。

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解题步骤 3.2.1.2.7.1

将 $0$ 乘以 $- 1$ 。

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

解题步骤 3.2.1.2.7.2

将 $0$ 乘以 $\sqrt{2}$ 。

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

解题步骤 3.2.1.2.8

乘以 $- \sqrt{2} \cdot 0$ 。

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解题步骤 3.2.1.2.8.1

将 $0$ 乘以 $- 1$ 。

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

解题步骤 3.2.1.2.8.2

将 $0$ 乘以 $\sqrt{2}$ 。

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

解题步骤 3.2.1.2.9

乘以 $- \sqrt{2} \cdot 0$ 。

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解题步骤 3.2.1.2.9.1

将 $0$ 乘以 $- 1$ 。

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

解题步骤 3.2.1.2.9.2

将 $0$ 乘以 $\sqrt{2}$ 。

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

解题步骤 3.2.1.2.10

乘以 $- \sqrt{2} \cdot 0$ 。

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解题步骤 3.2.1.2.10.1

将 $0$ 乘以 $- 1$ 。

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

解题步骤 3.2.1.2.10.2

将 $0$ 乘以 $\sqrt{2}$ 。

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

解题步骤 3.2.1.2.11

将 $- 1$ 乘以 $1$ 。

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

解题步骤 3.2.1.2.12

乘以 $- \sqrt{2} \cdot 0$ 。

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解题步骤 3.2.1.2.12.1

将 $0$ 乘以 $- 1$ 。

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

解题步骤 3.2.1.2.12.2

将 $0$ 乘以 $\sqrt{2}$ 。

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

解题步骤 3.2.1.2.13

乘以 $- \sqrt{2} \cdot 0$ 。

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解题步骤 3.2.1.2.13.1

将 $0$ 乘以 $- 1$ 。

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

解题步骤 3.2.1.2.13.2

将 $0$ 乘以 $\sqrt{2}$ 。

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

解题步骤 3.2.1.2.14

乘以 $- \sqrt{2} \cdot 0$ 。

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解题步骤 3.2.1.2.14.1

将 $0$ 乘以 $- 1$ 。

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

解题步骤 3.2.1.2.14.2

将 $0$ 乘以 $\sqrt{2}$ 。

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

解题步骤 3.2.1.2.15

乘以 $- \sqrt{2} \cdot 0$ 。

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解题步骤 3.2.1.2.15.1

将 $0$ 乘以 $- 1$ 。

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

解题步骤 3.2.1.2.15.2

将 $0$ 乘以 $\sqrt{2}$ 。

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

解题步骤 3.2.1.2.16

将 $- 1$ 乘以 $1$ 。

解题步骤 3.2.2

加上相应元素。

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

解题步骤 3.2.3

Simplify each element.

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解题步骤 3.2.3.1

从 $0$ 中减去 $\sqrt{2}$ 。

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

解题步骤 3.2.3.2

将 $1$ 和 $0$ 相加。

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

解题步骤 3.2.3.3

将 $0$ 和 $0$ 相加。

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

解题步骤 3.2.3.4

将 $- 1$ 和 $0$ 相加。

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

解题步骤 3.2.3.5

将 $1$ 和 $0$ 相加。

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

解题步骤 3.2.3.6

从 $0$ 中减去 $\sqrt{2}$ 。

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

解题步骤 3.2.3.7

将 $- 1$ 和 $0$ 相加。

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

解题步骤 3.2.3.8

将 $0$ 和 $0$ 相加。

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

解题步骤 3.2.3.9

将 $0$ 和 $0$ 相加。

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

解题步骤 3.2.3.10

将 $- 1$ 和 $0$ 相加。

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

解题步骤 3.2.3.11

从 $0$ 中减去 $\sqrt{2}$ 。

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

解题步骤 3.2.3.12

将 $- 1$ 和 $0$ 相加。

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

解题步骤 3.2.3.13

将 $- 1$ 和 $0$ 相加。

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

解题步骤 3.2.3.14

将 $0$ 和 $0$ 相加。

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

解题步骤 3.2.3.15

将 $- 1$ 和 $0$ 相加。

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

解题步骤 3.2.3.16

从 $0$ 中减去 $\sqrt{2}$ 。

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

解题步骤 3.3

Find the null space when $λ = \sqrt{2}$ .

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解题步骤 3.3.1

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

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

解题步骤 3.3.2

求行简化阶梯形矩阵。

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解题步骤 3.3.2.1

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

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解题步骤 3.3.2.1.1

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

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

解题步骤 3.3.2.1.2

化简 $R_{1}$ 。

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

解题步骤 3.3.2.2

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 3.3.2.2.1

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

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

解题步骤 3.3.2.2.2

化简 $R_{2}$ 。

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

解题步骤 3.3.2.3

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

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解题步骤 3.3.2.3.1

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

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

解题步骤 3.3.2.3.2

化简 $R_{4}$ 。

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

解题步骤 3.3.2.4

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

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解题步骤 3.3.2.4.1

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

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

解题步骤 3.3.2.4.2

化简 $R_{2}$ 。

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

解题步骤 3.3.2.5

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 3.3.2.5.1

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

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

解题步骤 3.3.2.5.2

化简 $R_{3}$ 。

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

解题步骤 3.3.2.6

Perform the row operation $R_{4} = R_{4} + \frac{\sqrt{2}}{2} R_{2}$ to make the entry at $4, 2$ a $0$ .

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解题步骤 3.3.2.6.1

Perform the row operation $R_{4} = R_{4} + \frac{\sqrt{2}}{2} R_{2}$ to make the entry at $4, 2$ a $0$ .

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

解题步骤 3.3.2.6.2

化简 $R_{4}$ 。

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

解题步骤 3.3.2.7

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

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解题步骤 3.3.2.7.1

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

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

解题步骤 3.3.2.7.2

化简 $R_{1}$ 。

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

解题步骤 3.3.3

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

$x_{1} + x_{3} + \sqrt{2} x_{4} = 0$

$x_{2} + \sqrt{2} x_{3} + x_{4} = 0$

$0 = 0$

解题步骤 3.3.4

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

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = [\begin{array}{c} - x_{3} - \sqrt{2} x_{4} \\ - \sqrt{2} x_{3} - x_{4} \\ x_{3} \\ x_{4} \end{array}]$

解题步骤 3.3.5

Write the solution as a linear combination of vectors.

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

解题步骤 3.3.6

Write as a solution set.

${x_{3} [\begin{array}{c} - 1 \\ - \sqrt{2} \\ 1 \\ 0 \end{array}] + x_{4} [\begin{array}{c} - \sqrt{2} \\ - 1 \\ 0 \\ 1 \end{array}] | x_{3}, x_{4} \in R}$

解题步骤 3.3.7

The solution is the set of vectors created from the free variables of the system.

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

解题步骤 4

Find the eigenvector using the eigenvalue $λ = - \sqrt{2}$ .

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解题步骤 4.1

将已知值代入公式中。

$N ([\begin{array}{c} 0 & 1 & 0 & - 1 \\ 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & - 1 \\ - 1 & 0 & - 1 & 0 \end{array}] + \sqrt{2} [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}])$

解题步骤 4.2

化简。

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解题步骤 4.2.1

化简每一项。

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解题步骤 4.2.1.1

将 $\sqrt{2}$ 乘以矩阵中的每一个元素。

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

解题步骤 4.2.1.2

化简矩阵中的每一个元素。

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解题步骤 4.2.1.2.1

将 $\sqrt{2}$ 乘以 $1$ 。

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

解题步骤 4.2.1.2.2

将 $\sqrt{2}$ 乘以 $0$ 。

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

解题步骤 4.2.1.2.3

将 $\sqrt{2}$ 乘以 $0$ 。

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

解题步骤 4.2.1.2.4

将 $\sqrt{2}$ 乘以 $0$ 。

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

解题步骤 4.2.1.2.5

将 $\sqrt{2}$ 乘以 $0$ 。

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

解题步骤 4.2.1.2.6

将 $\sqrt{2}$ 乘以 $1$ 。

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

解题步骤 4.2.1.2.7

将 $\sqrt{2}$ 乘以 $0$ 。

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

解题步骤 4.2.1.2.8

将 $\sqrt{2}$ 乘以 $0$ 。

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

解题步骤 4.2.1.2.9

将 $\sqrt{2}$ 乘以 $0$ 。

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

解题步骤 4.2.1.2.10

将 $\sqrt{2}$ 乘以 $0$ 。

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

解题步骤 4.2.1.2.11

将 $\sqrt{2}$ 乘以 $1$ 。

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

解题步骤 4.2.1.2.12

将 $\sqrt{2}$ 乘以 $0$ 。

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

解题步骤 4.2.1.2.13

将 $\sqrt{2}$ 乘以 $0$ 。

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

解题步骤 4.2.1.2.14

将 $\sqrt{2}$ 乘以 $0$ 。

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

解题步骤 4.2.1.2.15

将 $\sqrt{2}$ 乘以 $0$ 。

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

解题步骤 4.2.1.2.16

将 $\sqrt{2}$ 乘以 $1$ 。

解题步骤 4.2.2

加上相应元素。

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

解题步骤 4.2.3

Simplify each element.

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解题步骤 4.2.3.1

将 $0$ 和 $\sqrt{2}$ 相加。

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

解题步骤 4.2.3.2

将 $1$ 和 $0$ 相加。

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

解题步骤 4.2.3.3

将 $0$ 和 $0$ 相加。

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

解题步骤 4.2.3.4

将 $- 1$ 和 $0$ 相加。

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

解题步骤 4.2.3.5

将 $1$ 和 $0$ 相加。

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

解题步骤 4.2.3.6

将 $0$ 和 $\sqrt{2}$ 相加。

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

解题步骤 4.2.3.7

将 $- 1$ 和 $0$ 相加。

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

解题步骤 4.2.3.8

将 $0$ 和 $0$ 相加。

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

解题步骤 4.2.3.9

将 $0$ 和 $0$ 相加。

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

解题步骤 4.2.3.10

将 $- 1$ 和 $0$ 相加。

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

解题步骤 4.2.3.11

将 $0$ 和 $\sqrt{2}$ 相加。

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

解题步骤 4.2.3.12

将 $- 1$ 和 $0$ 相加。

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

解题步骤 4.2.3.13

将 $- 1$ 和 $0$ 相加。

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

解题步骤 4.2.3.14

将 $0$ 和 $0$ 相加。

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

解题步骤 4.2.3.15

将 $- 1$ 和 $0$ 相加。

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

解题步骤 4.2.3.16

将 $0$ 和 $\sqrt{2}$ 相加。

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

解题步骤 4.3

Find the null space when $λ = - \sqrt{2}$ .

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解题步骤 4.3.1

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

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

解题步骤 4.3.2

求行简化阶梯形矩阵。

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解题步骤 4.3.2.1

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

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解题步骤 4.3.2.1.1

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

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

解题步骤 4.3.2.1.2

化简 $R_{1}$ 。

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

解题步骤 4.3.2.2

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 4.3.2.2.1

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

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

解题步骤 4.3.2.2.2

化简 $R_{2}$ 。

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

解题步骤 4.3.2.3

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

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解题步骤 4.3.2.3.1

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

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

解题步骤 4.3.2.3.2

化简 $R_{4}$ 。

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

解题步骤 4.3.2.4

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

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解题步骤 4.3.2.4.1

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

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

解题步骤 4.3.2.4.2

化简 $R_{2}$ 。

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

解题步骤 4.3.2.5

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 4.3.2.5.1

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

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

解题步骤 4.3.2.5.2

化简 $R_{3}$ 。

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

解题步骤 4.3.2.6

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

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解题步骤 4.3.2.6.1

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

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

解题步骤 4.3.2.6.2

化简 $R_{4}$ 。

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

解题步骤 4.3.2.7

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

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解题步骤 4.3.2.7.1

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

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

解题步骤 4.3.2.7.2

化简 $R_{1}$ 。

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

解题步骤 4.3.3

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

$x_{1} + x_{3} - \sqrt{2} x_{4} = 0$

$x_{2} - \sqrt{2} x_{3} + x_{4} = 0$

$0 = 0$

解题步骤 4.3.4

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

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = [\begin{array}{c} - x_{3} + \sqrt{2} x_{4} \\ \sqrt{2} x_{3} - x_{4} \\ x_{3} \\ x_{4} \end{array}]$

解题步骤 4.3.5

Write the solution as a linear combination of vectors.

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

解题步骤 4.3.6

Write as a solution set.

${x_{3} [\begin{array}{c} - 1 \\ \sqrt{2} \\ 1 \\ 0 \end{array}] + x_{4} [\begin{array}{c} \sqrt{2} \\ - 1 \\ 0 \\ 1 \end{array}] | x_{3}, x_{4} \in R}$

解题步骤 4.3.7

The solution is the set of vectors created from the free variables of the system.

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

解题步骤 5

The eigenspace of $A$ is the list of the vector space for each eigenvalue.

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

线性代数 示例

线性代数示例