[beginarrayllsin(45^{circ)}&log_{5(25)}e^{4}&8endarray]

Frequently Asked Questions

• Evaluate the determinant of $\left[\begin{array}{ll}\sin \left(45^{\circ }\right)& {\log }_5\left(25\right)\\ e^4& 8\end{array}\right]$

  

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

• Find the matrix inverse of $\left[\begin{array}{ll}\sin \left(45^{\circ }\right)& {\log }_5\left(25\right)\\ e^4& 8\end{array}\right]$

  

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

• Find the matrix transpose of $\left[\begin{array}{ll}\sin \left(45^{\circ }\right)& {\log }_5\left(25\right)\\ e^4& 8\end{array}\right]$

  

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

• Find the eigenvalues of $\left[\begin{array}{ll}\sin \left(45^{\circ }\right)& {\log }_5\left(25\right)\\ e^4& 8\end{array}\right]$

  

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

• Find the row echelon form of $\left[\begin{array}{ll}\sin \left(45^{\circ }\right)& {\log }_5\left(25\right)\\ e^4& 8\end{array}\right]$

  

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

• Find the reduced row echelon form of $\left[\begin{array}{ll}\sin \left(45^{\circ }\right)& {\log }_5\left(25\right)\\ e^4& 8\end{array}\right]$

  

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