$$$\left[\begin{array}{cc}- r \sin{\left(\tanh{\left(\eta \right)} \right)} \operatorname{sech}^{2}{\left(\eta \right)} & \cos{\left(\tanh{\left(\eta \right)} \right)}\\r \cos{\left(\tanh{\left(\eta \right)} \right)} \operatorname{sech}^{2}{\left(\eta \right)} & \sin{\left(\tanh{\left(\eta \right)} \right)}\end{array}\right]$$$ 的行列式
相关计算器: 代数余子式矩阵计算器
您的输入
计算 $$$\left|\begin{array}{cc}- r \sin{\left(\tanh{\left(\eta \right)} \right)} \operatorname{sech}^{2}{\left(\eta \right)} & \cos{\left(\tanh{\left(\eta \right)} \right)}\\r \cos{\left(\tanh{\left(\eta \right)} \right)} \operatorname{sech}^{2}{\left(\eta \right)} & \sin{\left(\tanh{\left(\eta \right)} \right)}\end{array}\right|$$$。
解答
2x2 矩阵的行列式为 $$$\left|\begin{array}{cc}a & b\\c & d\end{array}\right| = a d - b c$$$。
$$$\left|\begin{array}{cc}- r \sin{\left(\tanh{\left(\eta \right)} \right)} \operatorname{sech}^{2}{\left(\eta \right)} & \cos{\left(\tanh{\left(\eta \right)} \right)}\\r \cos{\left(\tanh{\left(\eta \right)} \right)} \operatorname{sech}^{2}{\left(\eta \right)} & \sin{\left(\tanh{\left(\eta \right)} \right)}\end{array}\right| = \left(- r \sin{\left(\tanh{\left(\eta \right)} \right)} \operatorname{sech}^{2}{\left(\eta \right)}\right)\cdot \left(\sin{\left(\tanh{\left(\eta \right)} \right)}\right) - \left(\cos{\left(\tanh{\left(\eta \right)} \right)}\right)\cdot \left(r \cos{\left(\tanh{\left(\eta \right)} \right)} \operatorname{sech}^{2}{\left(\eta \right)}\right) = - r \operatorname{sech}^{2}{\left(\eta \right)}$$$
答案
$$$\left|\begin{array}{cc}- r \sin{\left(\tanh{\left(\eta \right)} \right)} \operatorname{sech}^{2}{\left(\eta \right)} & \cos{\left(\tanh{\left(\eta \right)} \right)}\\r \cos{\left(\tanh{\left(\eta \right)} \right)} \operatorname{sech}^{2}{\left(\eta \right)} & \sin{\left(\tanh{\left(\eta \right)} \right)}\end{array}\right| = - r \operatorname{sech}^{2}{\left(\eta \right)}$$$A