Simple question: How do these changes affect the determinant of my matrix?

Question

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

$$\det \left[\begin{array}{ccc} a &2 &d\cr b &3 &e\cr c &4 &f\cr \end{array}\right] = 4?$$

Answer 1

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.

Answer 2

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 $