Simple question: How do these changes affect the determinant of my matrix? I assume there's some simple rule to follow, but I can't seem to see what it is.
Given $$\det \left[\begin{array}{ccc}
a &1 &d\cr
b &1 &e\cr
c &1 &f\cr
\end{array}\right] = 1$$
Why is it that
$$\det \left[\begin{array}{ccc}
a &1 &d\cr
b &2 &e\cr
c &3 &f\cr
\end{array}\right] = 8?$$
Also, given 
$$\det \left[\begin{array}{ccc}
a &8 &d\cr
b &8 &e\cr
c &8 &f\cr
\end{array}\right] =3$$
Why is it that
$$\det \left[\begin{array}{ccc}
a &2 &d\cr
b &3 &e\cr
c &4 &f\cr
\end{array}\right] = 4?$$
 A: We have $\det \left[\begin{array}{ccc}
1 &1 &0\cr
0 &1 &0\cr
0 &1 &1\cr
\end{array}\right] = 1$ but
$\det \left[\begin{array}{ccc}
1 &1 &0\cr
0 &2 &0\cr
0 &3 &1\cr
\end{array}\right] =2 \neq 8 $ and $\det \left[\begin{array}{ccc}
1 &8 &0\cr
0 &8 &0\cr
0 &8 &1\cr
\end{array}\right] =8\neq 3 $ and $\det \left[\begin{array}{ccc}
1 &2 &0\cr
0 &3 &0\cr
0 &4 &1\cr
\end{array}\right] =3 \neq 4 $
A: So you are saying that since
$$
\det \begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix}
= 1,
$$
then
$$
\det \begin{bmatrix}
1 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 1\\
\end{bmatrix}
= 8?
$$
The general reason why your statements are false is that if you take the three $2 \times 2$ minors (with any convenient choice of signs) of 
$$
\begin{bmatrix}
a & d\\
b & e\\
c & f\\
\end{bmatrix},
$$
they will take arbitrary values $x_{1}, x_{2}, x_{3}$. For instance, if $x_{3} \ne 0$, take
$$
\begin{bmatrix}
x_{3} & 0\\
0 & -1\\
-x_{1} & x_{2} x_{3}^{-1}\\
\end{bmatrix}.
$$
So you are assuming $x_{1} + x_{2}+ x_{3} =1$, that is,
$$
\det \begin{bmatrix}
x_{3} & 1 & 0\\
0 & 1 & -1\\
-x_{1} & 1 & x_{2} x_{3}^{-1}\\
\end{bmatrix} = 1.
$$
 and want to deduce $x_{1} + 2 x_{2}+ 3 x_{3} = 8$, that is,
$$
\det \begin{bmatrix}
x_{3} & 1 & 0\\
0 & 2 & -1\\
-x_{1} & 3 & x_{2} x_{3}^{-1}\\
\end{bmatrix} = 8.
$$
 which is visibly an impossible task.
