Question

Se A= e M= então o determinante da matriz M é igual a :

Answer 1

Primeiro, encontramos os valores de A transposta e inversa:

$A = \left[\begin{array}{ccc}2&-3\\-5&7\end{array}\right] =\ \textgreater \ A^{t} = \left[\begin{array}{ccc}2&-5\\-3&7\end{array}\right]$

$A*A^{-1}=I_{2} =\ \textgreater \ \left[\begin{array}{ccc}2&-3\\-5&7\end{array}\right] * \left[\begin{array}{ccc}a&b\\c&d\end{array}\right]= \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

$=\ \textgreater \ \left[\begin{array}{ccc}2a-3c&2b-3d\\7c-5a&7d-5b\end{array}\right]=\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

$=\ \textgreater \ \left \{ {{2a-3c=1} \atop {7c-5a=0}} \right. e \left \{ {{2b-3d=0} \atop {7d-5b=1}} \right.$

Resolvendo o sistema, obtemos a = -7, b = -3, c = -5 e d = -2, achando assim a matriz inversa de A:

$A^{-1}= \left[\begin{array}{ccc}-7&-3\\-5&-2\end{array}\right]$

Agora, encontramos a matriz M:

$M = A^{t}+ A^{-1} =\ \textgreater \ \left[\begin{array}{ccc}2&-5\\-3&7\end{array}\right]+\left[\begin{array}{ccc}-7&-3\\-5&-2\end{array}\right] M = \left[\begin{array}{ccc}-5&-8\\-8&5\end{array}\right]$

Em seguida, seu determinante:

$det(M)=\left[\begin{array}{ccc}-5&-8\\-8&5\end{array}\right]=(-5)*5-[(-8)*(-8)]=\ \textgreater \ -25-64=-89$