Respuesta :
Answer:
We have to prove
[tex]det(A^{-1})=\frac{1}{det(A)}[/tex]
In general
[tex]A.A^{-1}=I\\\\|A.A^{-1}|=|\,I\,|\\\\|\,I\,|=1\\\\\implies|A|.|(A^{-1})|=1\\\\|A^{-1}|=\frac{1}{|A|}[/tex]
Consider a 2 x 2 matrix
[tex]A=\left[\begin{array}{cc}1&2\\4&5\end{array}\right] \\\\|A|=(5)(1)-(2)(4)\\\\|A|=-3---(1)[/tex]
Now we find A⁻¹
[tex]A=\left[\begin{array}{cc}1&2\\4&5\end{array}\right] \\\\|A|=-3\\\\Adj(A)=\left[\begin{array}{cc}5&-2\\-4&1\end{array}\right] \\\\\\A^{-1}=\frac{Adj(A)}{|A|}\\\\A^{-1}=\left[\begin{array}{cc}\frac{5}{-3}&\frac{-2}{-3}\\\frac{-4}{-3}&\frac{1}{-3}\end{array}\right] \\\\\\|A^{-1}|=(\frac{5}{-3})(\frac{1}{-3})-(\frac{-4}{-3})(\frac{-2}{-3})\\\\|A^{-1}|=\frac{5}{9}-\frac{8}{9}\\\\|A^{-1}|=\frac{-1}{3}[/tex]
Hence proved.
