线性代数基础，求解如图逆矩阵问题
2个回答
瑞地测控
2024-08-12 广告
在苏州瑞地测控技术有限公司，我们深知频率同步与相位同步的重要性。频率同步确保两个或多个设备的时钟频率变化一致，但相位（即时间点）可保持相对固定差值。而相位同步，即时间同步，要求不仅频率一致，相位也必须完全一致，即时间差恒定为零。相位同步的精...
点击进入详情页
本回答由瑞地测控提供
展开全部
A =
[B O]
[O C]
其中 B =
[1 2]
[2 -1]
C =
[1 0]
[3 2]
B^(-1) = (-1/5)*
[-1 -2]
[-2 1]
=
[1/5 2/5]
[2/5 -1/5]
B^2 = 5E, B^4 = 25E
C^(-1) = (1/2)*
[ 2 0]
[-3 1]
=
[ 1 0]
[-3/2 1/2]
C^2 =
[1 0]
[9 4]
C^4 =
[1 0]
[45 16]
A^(-1) =
[B^(-1) O]
[O C^(-1)]
=
[1/5 2/5 0 0]
[2/5 -1/5 0 0]
[ 0 0 1 0]
[ 0 0 -3/2 1/2]
|A^T| = |A| = |B||C| = (-5)2 = -10
A^4 =
[B^4 O]
[O C^4]
[B O]
[O C]
其中 B =
[1 2]
[2 -1]
C =
[1 0]
[3 2]
B^(-1) = (-1/5)*
[-1 -2]
[-2 1]
=
[1/5 2/5]
[2/5 -1/5]
B^2 = 5E, B^4 = 25E
C^(-1) = (1/2)*
[ 2 0]
[-3 1]
=
[ 1 0]
[-3/2 1/2]
C^2 =
[1 0]
[9 4]
C^4 =
[1 0]
[45 16]
A^(-1) =
[B^(-1) O]
[O C^(-1)]
=
[1/5 2/5 0 0]
[2/5 -1/5 0 0]
[ 0 0 1 0]
[ 0 0 -3/2 1/2]
|A^T| = |A| = |B||C| = (-5)2 = -10
A^4 =
[B^4 O]
[O C^4]
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
