We have that
$-2 \sqrt{2} < x < 2 \sqrt{2}$ : $5$ integers
$-2 \sqrt{2} < y < 2 \sqrt{2}$ : $5$ integers
$-2 < z < 2$ : $3$ integers

To find how many non-trivial points $(x, y, z)$ we multiply $5 \cdot 5 \cdot 3$ and substract $1$, which is the case $(0, 0, 0)$, or not??

Then the result is $74$, but in my notes I have that there are $72$ cases... How do we get this result??

"Now if you choose
the adjugated amounts of other matrix" what do you mean ?

"you pick the next row
and find it's amounts"

@ted I am given $A\left(-8;3\right)$, $B\left(-6;-6\right)$, $C\left(-7;2\right)$.

I want to calculate the equation of the median from C.

If M is the point such that $\left(MC\right)$ is the median from C,

$MC : y=mx+p$

$m=\dfrac{\frac{y_a+y_b}{2}-y_c}{\frac{x_a+x_b}{2}-x_c}$, So

$m=\dfrac{\frac{y_a+y_b}{2}-y_c}{\frac{-8-6}{2}+7}$, So

$m=\dfrac{\frac{y_a+y_b}{2}-y_c}{0}$

And that's impossible ! Although, I now this equation exists... Where does it go wrong is what's just above ?