$detA_{6x6} \neq 0$.
$2A+7B=0$
Calculate $6det(2(A^t)^2B^{-1}A^{-1})$
My solution attempt:
$A = -7/2*B$ and $det A^t = det A$ so $6det(2*A*(-7/2B)*B^{-1}A^{-1}) = 6det(-7)= 6*(-7)^6 = 705894$
This seems to be incorrect but I can't find why. I would greatly appreciate if someone could point out what I did wrong.
$6det(2(A^T)^2B^{-1}A^{-1})$
$=6det(2(A^T)^2)det(B^{-1})det(A^{-1})=$
$=6 \cdot 2^6 det(A)^2 det(B)^{-1} det(A)^{-1}=$
$= 6 \cdot 2^6 det(A) det(B)^{-1} = $
$= 6 \cdot 2^6 det(A) det(-\frac{2}{7}A)^{-1} = $
$= 6 \cdot 2^6 det(A) \cdot (-\frac{2}{7})^{-6} \cdot det(A)^{-1} =$
$= 6 \cdot 2^6 \cdot \frac{7^6}{2^6}=$
$= 6 \cdot 7^6=$
$=705894$
So your answer is right, but I see that you want to do the calculations very quickly and you might lose yourself in the process ;-)