5.4 Additional Properties of Determinants

最新推荐文章于 2021-04-17 11:31:58 发布

christangdt

最新推荐文章于 2021-04-17 11:31:58 发布

阅读量532

点赞数 2

CC 4.0 BY-SA版权

分类专栏：线性代数（Hoffman&Kunze）

本文链接：https://blog.youkuaiyun.com/christangdt/article/details/105191625

这篇博客介绍了行列式的性质，包括转置矩阵和特殊分块矩阵的行列式计算，如何利用经典伴随矩阵求逆矩阵，以及行列式在确定矩阵可逆性中的作用。博客强调理解行列式的理论意义而非仅仅关注计算，并给出了Cramer's rule的应用示例。习题部分涉及计算和证明行列式的各种问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

这一节主要说了转置矩阵行列式的计算、特殊的分块矩阵行列式计算，通过classical adjoint计算行列式，以及通过行列式确定矩阵是否可逆及逆矩阵计算方法（Theorem 4）, 最后提了一下Cramer’s rule。最后的最后，作者还苦口婆心的说不要注重行列式的计算，要注重其理论意义和证明，关注how it behaves而不是how to compute。（结果习题上来就是两道compute……）

Exercises

1.Use the classical adjoint formula to compute the inverse of each of the following $3×33\times 3$ real matrices.
$[−23260341−1],[cos⁡θ0−sin⁡θ010sin⁡θ0cos⁡θ]\begin{bmatrix}-2&3&2\\6&0&3\\4&1&-1\end{bmatrix},\qquad \begin{bmatrix}\cos \theta&0&-\sin \theta\\0&1&0\\\sin \theta&0&\cos \theta\end{bmatrix}$
Solution: Let the first matrix be $A$ and the second be $B$ .
We have $det⁡A=72\det A=72$ and
$A−1=172[−35918−618614−18]\text{adj }A=\begin{bmatrix}-3&5&9\\18&-6&18\\6&14&-18\end{bmatrix}\implies A^{-1}=\frac{1}{72}\begin{bmatrix}-3&5&9\\18&-6&18\\6&14&-18\end{bmatrix}$
We have $det⁡B=1\det B=1$ and
$B−1=[cos⁡θ0sin⁡θ010−sin⁡θ0cos⁡θ]\text{adj }B=\begin{bmatrix}\cos\theta&0&\sin\theta\\0&1&0\\-\sin\theta&0&\cos\theta\end{bmatrix}\implies B^{-1}=\begin{bmatrix}\cos\theta&0&\sin\theta\\0&1&0\\-\sin\theta&0&\cos\theta\end{bmatrix}$

2.Use Cramer’s rule to solve each of the following systems of linear equations over the field of rational numbers.
( a ) $x+y+z=112x−6y−z=03x+4y+2z=0\begin{aligned}x&+y+z=11\\2x&-6y-z=0\\3x&+4y+2z=0\end{aligned}$
( b ) $3x−2y=73y−2z=63z−2x=−1\begin{aligned}3x&-2y=7\\3y&-2z=6\\3z&-2x=-1\end{aligned}$
Solution:
( a ) The coefficient matrix $A$ of the system is
$A=[1112−6−1342],det⁡A=11A=\begin{bmatrix}1&1&1\\2&-6&-1\\3&4&2\end{bmatrix},\qquad \det A=11$
Also we have
$B1=[11110−6−1042],B2=[111120−1302],B3=[11112−60340]B_1=\begin{bmatrix}11&1&1\\0&-6&-1\\0&4&2\end{bmatrix},B_2=\begin{bmatrix}1&11&1\\2&0&-1\\3&0&2\end{bmatrix},B_3=\begin{bmatrix}1&1&11\\2&-6&0\\3&4&0\end{bmatrix}$
A simple calculation shows $det⁡B1=−88,det⁡B2=−77,det⁡B3=11×26\det B_1=-88,\det B_2=-77,\det B_3=11\times 26$ , thus the solution is $(x, y, z) = (- 8, - 7, 26)$ .
( b ) The coefficient matrix $A$ of the system is
$A=[3−2003−2−203],det⁡A=19A=\begin{bmatrix}3&-2&0\\0&3&-2\\-2&0&3\end{bmatrix},\qquad \det A=19$
Also we have
$B1=[7−2063−2−103],B2=[37006−2−2−13],B3=[3−27036−20−1]B_1=\begin{bmatrix}7&-2&0\\6&3&-2\\-1&0&3\end{bmatrix},B_2=\begin{bmatrix}3&7&0\\0&6&-2\\-2&-1&3\end{bmatrix},B_3=\begin{bmatrix}3&-2&7\\0&3&6\\-2&0&-1\end{bmatrix}$
A simple calculation shows $det B_1=95,\det B_2=76,\det B_3=57$ , thus the solution is $($