91影视

$$\begin{gathered} A=\left(\begin{array}{ccc}\cos 胃& -\sin 胃& 0\\ \sin 胃& \cos 胃& 0\\ 0& 0& 1\end{array}\right) \\ B=\left(\begin{array}{ccc}\cos 胃& -\sin 胃& 0\\ \sin 胃& \cos 胃& 0\\ 0& 0& -1\end{array}\right) \end{gathered}$$

A Diagonal Matrix is a square matrix in which all elements are zero except the principal diagonal element.

The trace by the sum of the elements on the diagonals is $TrA=2\cos 胃+1$and $Tr\text{B}=2\cos 胃-1$. This can be verified by noting that the trace of a matrix equals the sum of its eigenvalues. The eigenvalues of A is obtained as shown below.

$TrA=2\cos 胃+1Tr\text{B}=2\cos 胃-1$

$$\begin{gathered} =\left|\begin{array}{ccc}\cos 胃-位& -\sin 胃& 0\\ \sin 胃& \cos 胃-位& 0\\ 0& 0& 1-位\end{array}\right| \\ ={(\cos 胃-位)}^2(1-位)+{\sin }^2胃(1-位) \\ =(1-位)(1+{位}^2-2.\cos 胃.x)=0 \end{gathered}$$

One eigenvalue is 1 , and the other two can be obtained from the quadratic equation $\cos 胃卤i\sin 胃$. The sum of the eigenvalues is equal to $2\cos 胃+1$, as expected. For matrix $B$, the only thing that is different is the element $b_{33}$, and the characteristic equation can be changed as $(-1-位)\left(1+{位}^2-2路\cos 胃路位\right)=0$. The first eigenvalue changes to $-1$, while the other two remains the same. Therefore, the sum of the eigenvalues will be $2\cos 胃-1$, as expected.

$(-1-位)\left(1+{位}^2-2路\cos 胃路位\right)=0$

-1$2\cos 胃-1$