# 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

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

• To summarize the answers: the answer to "Why is it" is that there is no reason why this should be so, and indeed the conclusion fails in many cases. Are you withholding some additional information from us, which might have been used to justify these computations in specific instances? – Marc van Leeuwen Apr 29 '15 at 11:09

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.
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 $\det \left[\begin{array}{ccc} 1 &2 &0\cr 0 &3 &0\cr 0 &4 &1\cr \end{array}\right] =3 \neq 4$