From 55ec4d9bd9c134300e464764c6dec06f2b03dde2 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= Date: Fri, 23 Jan 2026 13:46:14 +0800 Subject: [PATCH] vault backup: 2026-01-23 13:46:14 --- 素材/整合素材/线代素材/一个计算题的解答.md | 4 +++- 1 file changed, 3 insertions(+), 1 deletion(-) diff --git a/素材/整合素材/线代素材/一个计算题的解答.md b/素材/整合素材/线代素材/一个计算题的解答.md index 62b7aad..85359d2 100644 --- a/素材/整合素材/线代素材/一个计算题的解答.md +++ b/素材/整合素材/线代素材/一个计算题的解答.md @@ -1,7 +1,9 @@ ## 解答 其特征多项式为 -$$\det(A - \lambda E) = \begin{vmatrix} 2-\lambda & 0 & 0 \\ 0 & 3-\lambda & 2 \\ 0 & 2 & 3-\lambda \end{vmatrix} = (2-\lambda)\left[(3-\lambda)^2 - 4\right] = (2-\lambda)(\lambda^2 - 6\lambda + 5) = (2-\lambda)(\lambda-1)(\lambda-5).$$ +$$\begin{aligned}det(A - \lambda E) &= \begin{vmatrix} 2-\lambda & 0 & 0 \\ 0 & 3-\lambda & 2 \\ 0 & 2 & 3-\lambda \end{vmatrix} \\ +&= (2-\lambda)\left[(3-\lambda)^2 - 4\right] \\ +&= (2-\lambda)(\lambda^2 - 6\lambda + 5)\\&=(2-\lambda)(\lambda-1)(\lambda-5).\end{aligned}$$ 特征值为 $\lambda_1 = 1,\quad \lambda_2 = 2,\quad \lambda_3 = 5.$ - 对于 $\lambda_1 = 1$：解 $(A - E)\mathbf{X} = \mathbf{0}$， $A - E = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 2 \\ 0 & 2 & 2 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix},$ 得基础解系 $\mathbf{x}_1 = (0, -1, 1)^\mathrm{T}$。