[beginarrayl sin45^{circ} & log_5(25)e^4& 8endarray]

常见问题解答

• 求该行列式的值 $\left[\begin{array}{lc}\sin 45^{\circ }& {\log }_5\left(25\right)\\ e^4& 8\end{array}\right]$

  

  $$4\sqrt{2}-2e^4$$

• 求逆矩阵 $\left[\begin{array}{lc}\sin 45^{\circ }& {\log }_5\left(25\right)\\ e^4& 8\end{array}\right]$

  

  $$\left[\begin{array}{ll}-\frac{8\sqrt{2}+4e^4}{e^8-8}& \frac{2\sqrt{2}+e^4}{e^8-8}\\ \frac{2\sqrt{2}\times e^4+e^8}{2e^8-16}& -\frac{4+\sqrt{2}\times e^4}{4e^8-32}\end{array}\right]$$

• 求矩阵转置 $\left[\begin{array}{lc}\sin 45^{\circ }& {\log }_5\left(25\right)\\ e^4& 8\end{array}\right]$

  

  $$\left[\begin{array}{ll}\frac{\sqrt{2}}{2}& e^4\\ 2& 8\end{array}\right]$$

• 求特征值 $\left[\begin{array}{lc}\sin 45^{\circ }& {\log }_5\left(25\right)\\ e^4& 8\end{array}\right]$

  

  $${\lambda }_1=\frac{\sqrt{2}+16-\sqrt{258-32\sqrt{2}+32e^4}}{4},{\lambda }_2=\frac{\sqrt{2}+16+\sqrt{258-32\sqrt{2}+32e^4}}{4}$$

• 求行梯形形式 $\left[\begin{array}{lc}\sin 45^{\circ }& {\log }_5\left(25\right)\\ e^4& 8\end{array}\right]$

  

  $$\left[\begin{array}{ll}\frac{\sqrt{2}}{2}& 2\\ 0& 8-2\sqrt{2}\times e^4\end{array}\right]$$

• 求简化的行梯形形式 $\left[\begin{array}{lc}\sin 45^{\circ }& {\log }_5\left(25\right)\\ e^4& 8\end{array}\right]$

  

  $$\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]$$