To show: $\det\left[\begin{smallmatrix} -bc & b^2+bc & c^2+bc\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{smallmatrix}\right]=(ab+bc+ca)^3$

Question

I've been having quite some trouble with this question. I'm required to show that the below equation holds, by using properties of determinants (i.e. not allowed to directly expand the determinant before greatly simplifying it).

$$\begin{vmatrix} -bc & b^2+bc & c^2+bc\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{vmatrix}= (ab+bc+ca)^3$$

I tried everything:

$R_1\to R_1+R_2+R_3$ and similar transformations to extract that $ab+bc+ca$ term, but to no avail. $C_2\to C_1+C_2$ and $C_3\to C_3+C_1$ seemed to be a good lead, but I couldn't follow up.

How can I solve this question?

Answer 1

If $a=b=c=0$ then the equality holds. WLG we assume $a\neq 0$ then $$\begin{vmatrix} -bc & b^2+bc & c^2+bc\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{vmatrix}\\=\frac{1}{a}\begin{vmatrix} a(-bc)+b(a^2+ac)+c(a^2+ab) & b(ac+bc+ca) & c(ac+bc+ca)\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{vmatrix}\\(applying R_1 \to \frac1a(aR_1+bR_2+cR_3))\\ =\frac{1}{a}\begin{vmatrix} a(ab+bc+ca) & b(ac+bc+ca) & c(ac+bc+ca)\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{vmatrix}\\ =\frac{1}{a}(ab+bc+ca)\begin{vmatrix} a & b & c\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{vmatrix}\\ =\frac{1}{a}(ab+bc+ca)\begin{vmatrix} a & b & c\\ 0 & -(ab+bc+ac) & 0 \\ 0 & 0 & -(ab+bc+ca) \end{vmatrix}\\ (applying R_2 \to R_2-(a+c)R_1\, and\, R_3 \to R_3-(a+b)R_1)\\ =(ab+bc+ca)^3$$

Answer 2

I put my solution as an alternative (a little bit simpler). Write the original determinant as

$\begin{vmatrix} -bc & b(b+c) & c(b+c)\\ a(a+c) & -ac & c(a+c) \\ a(a+b) & b(a+b) & -ab \end{vmatrix}$

Multiply the first column with $b+c$ and the second one with c and bring $\frac{1}{c(b+c)}$ in front of the determinant:

$\frac{1}{c(b+c)}\begin{vmatrix} -bc(b+c) & bc(b+c) & c(b+c)\\ a(a+c)(b+c) & -ac^2 & c(a+c) \\ a(a+b)(b+c) & bc(a+b) & -ab \end{vmatrix}$

Multiply the first row by $\frac{1}{c(b+c)}$ in front of the determinant:

$\begin{vmatrix} -b & b & 1\\ a(a+c)(b+c) & -ac^2 & c(a+c) \\ a(a+b)(b+c) & bc(a+b) & -ab \end{vmatrix}$

Zero the first row and third column. Add the second column to the first column, multiply the third column with $-b$ and add it to the first column. Take the factors $ab+ac+bc$ out of the determinant:

$(ab+ac+bc)^2\begin{vmatrix} 0 & 0 & 1\\ a & -c & c(a+c) \\ a+b & b & -ab \end{vmatrix}= (ab+ac+bc)^2\begin{vmatrix} 0 & 0 & 1\\ a & -c & 0 \\ a+b & b & 0 \end{vmatrix}$

Now you may expand or continue the elimination in the same way.

The cases $c=0$ and $b+c=0$ must be solved separately (but these cases are much simpler).

Answer 3

$A:=\begin{vmatrix} -bc & b^2+bc & c^2+bc\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{vmatrix}\enspace\enspace\enspace$ Shift: $\enspace C_2\to C_1$, $C_3\to C_2$, $C_1\to C_3\enspace$ => $\enspace B$

$B:=\begin{vmatrix} b^2+bc & c^2+bc & -bc\\ -ac & c^2+ac & a^2+ac \\ b^2+ab & -ab & a^2+ab \end{vmatrix}$

$C:=\begin{vmatrix} 0 & 0 & (ab+ac+bc)^2\\ (ab+ac+bc)^2 & 0 & 0 \\ 0& (ab+ac+bc)^2 & 0 \end{vmatrix}$

It's $|A|\cdot|B|=|A\cdot B|=|C|$ with $|A|=|B|$ (Sarrus rule without calculating, only a compare).

=> $|A|=\sqrt{|C|}=(ab+ac+bc)^3$

Addition:

In order to calculate the sign of the square root, it's sufficient to put in a value, e.g. $(a;b;c):=(1;1;0)$ .