Question

50 pontos para quem responder! - Considere as matrizes quadradas de ordem 2: $A = \left[\begin{array}{ccc}1&0\\2&1\end{array}\right]$ e $B = \left[\begin{array}{ccc}2&1\\0&2\end{array}\right]$ .

Seja $M = A . B^{t} ,$ onde $B^{t}$ é a matriz transposta de B. O determinante da matriz inversa de M é:

Answer 1

$A= \left[\begin{array}{cc}1&0\\2&1\end{array}\right] \ \ \ \ \ \ B= \left[\begin{array}{cc}2&1\\0&2\end{array}\right]\ \ \ \ \ \ B^t= \left[\begin{array}{cc}2&0\\1&2\end{array}\right]\\\\\\\\\\ M=A.B^t=\left[\begin{array}{cc}1&0\\2&1\end{array}\right]\times\left[\begin{array}{cc}2&0\\1&2\end{array}\right]=\left[\begin{array}{cc}2&0\\4&2\end{array}\right]$



$M.M^t=I\ \ \ \left[\begin{array}{cc}2&0\\4&2\end{array}\right]\times\left[\begin{array}{cc}a&b\\c&d\end{array}\right]=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]\\\\\\\\ Resolvendo\ o\ primeiro\ sistema:\\\\ \left \{ {{2a+0c=1} \atop {4a+2c=0}} \right. \\\\ 2a=1\\\\ a=\frac{1}{2}\\\\ Trocando\ a\ na\ segunda\ express\~ao:\\\\ 4(\frac{1}{2})+2c=0\\\\ 2+2c=0\\\\ 2c=-2\\\\ c=-\frac{2}{2}\\\\ c=-1$

$Resolvendo\ o\ segundo\ sistema:\\\\ \left \{ {{2b+0d=0} \atop {4b+2d=1}} \right. \\\\ 2b=0\\ b=0\\\\ Trocando\ b\ na\ segunda\ express\~ao:\\\\ 4(0)+2d=1\\\\ d=\frac{1}{2}$

Montando a matriz determinante, a partir dos valores encontrados:

$M_D= \left[\begin{array}{cc}\frac{1}{2}&0\\-1&\frac{1}{2}\end{array}\right]$

Espero ter ajudado.
Bons estudos! :)