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

Frequently Asked Questions

• Вычислите значение определителя \(\left[\begin{array}{l} {\sin45^{\circ}} & {\log_5\left(25\right)}\\{e^4}& {8}\end{array}\right]\)

  

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

• Найти обратную матрицу \(\left[\begin{array}{l} {\sin45^{\circ}} & {\log_5\left(25\right)}\\{e^4}& {8}\end{array}\right]\)

  

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

• Найдите матрицу транспонирования \(\left[\begin{array}{l} {\sin45^{\circ}} & {\log_5\left(25\right)}\\{e^4}& {8}\end{array}\right]\)

  

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

• Найдите собственные значения \(\left[\begin{array}{l} {\sin45^{\circ}} & {\log_5\left(25\right)}\\{e^4}& {8}\end{array}\right]\)

  

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

• Найдите форму эшелона строки \(\left[\begin{array}{l} {\sin45^{\circ}} & {\log_5\left(25\right)}\\{e^4}& {8}\end{array}\right]\)

  

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

• Найдите форму редуцированного эшелона строк \(\left[\begin{array}{l} {\sin45^{\circ}} & {\log_5\left(25\right)}\\{e^4}& {8}\end{array}\right]\)

  

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