转置矩阵与原矩阵的乘积的逆矩阵怎么求
  • 板块学术版
  • 楼主Eleveslaine
  • 当前回复1
  • 已保存回复1
  • 发布时间2024/4/9 21:30
  • 上次更新2024/4/10 09:32:35
查看原帖
转置矩阵与原矩阵的乘积的逆矩阵怎么求
450246
Eleveslaine楼主2024/4/9 21:30

省流：求 (ATA)−1(A^TA)^{-1}。

最终的目的是求 (ATA)−1ATB(A^TA)^{-1}A^TB。要求人工求解析解。

B=[ux1uy1ux2uy2ux3uy3ux4uy4]B= \left[ \begin{array}{c} u_{x_1} \\ u_{y_1} \\ u_{x_2} \\ u_{y_2} \\ u_{x_3} \\ u_{y_3} \\ u_{x_4} \\ u_{y_4} \\ \end{array} \right] \\ A=[vx1vy11000−ux1vx1−ux1vy1000vx1vy11−uy1vx1−uy1vy1vx2vy21000−ux2vx2−ux2vy2000vx2vy21−uy2vx2−uy2vy2vx3vy31000−ux3vx3−ux3vy3000vx3vy31−uy3vx3−uy3vy3vx4vy41000−ux4vx4−ux4vy4000vx4vy41−uy4vx4−uy4vy4]A=\left[ \begin{array}{cccccccc} v_{x_1} & v_{y_1} & 1 & 0 & 0 & 0 & -u_{x_1}v_{x_1} & -u_{x_1}v_{y_1} \\ 0 & 0 & 0 & v_{x_1} & v_{y_1} & 1 & -u_{y_1}v_{x_1} & -u_{y_1}v_{y_1} \\ v_{x_2} & v_{y_2} & 1 & 0 & 0 & 0 & -u_{x_2}v_{x_2} & -u_{x_2}v_{y_2} \\ 0 & 0 & 0 & v_{x_2} & v_{y_2} & 1 & -u_{y_2}v_{x_2} & -u_{y_2}v_{y_2} \\ v_{x_3} & v_{y_3} & 1 & 0 & 0 & 0 & -u_{x_3}v_{x_3} & -u_{x_3}v_{y_3} \\ 0 & 0 & 0 & v_{x_3} & v_{y_3} & 1 & -u_{y_3}v_{x_3} & -u_{y_3}v_{y_3} \\ v_{x_4} & v_{y_4} & 1 & 0 & 0 & 0 & -u_{x_4}v_{x_4} & -u_{x_4}v_{y_4} \\ 0 & 0 & 0 & v_{x_4} & v_{y_4} & 1 & -u_{y_4}v_{x_4} & -u_{y_4}v_{y_4} \\ \end{array} \right]
2024/4/9 21:30