7
$\begingroup$

$$\begin{align}\mathrm D &= \left|\begin{matrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (a+c)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{matrix}\right|\\ &= (a+b+c)\left|\begin{matrix} b+c - a & a^2 & a^2 \\ b - a -c & (a+c)^2 & b^2 \\ 0 & c^2 & (a+b)^2 \end{matrix}\right| \\ &= (a+b+c)^2\left|\begin{matrix} b+c - a & 0 & a^2 \\ b - a -c & a+c - b & b^2 \\ 0 & c - a-b & (a+b)^2 \end{matrix}\right|\\ &= (a+b+c)^2\left|\begin{matrix} b+c - a & 0 & a^2 \\ 0 & a+c - b & b^2 \\ c - a-b & c - a-b & (a+b)^2 \end{matrix}\right|\end{align}$$

Can $\rm D$ be further simplified without expanding ? I feel it should be because this was competition question.

$\endgroup$
  • 1
    $\begingroup$ $$R_3'=R_3-R_2-R_1$$ $\endgroup$ – lab bhattacharjee Aug 14 '17 at 16:20
  • $\begingroup$ @labbhattacharjee Then $R_3 = 2[-b \qquad -a \qquad ab]$. It simplifies things somewhat but still not very helpful. $\endgroup$ – user8277998 Aug 14 '17 at 16:52
  • $\begingroup$ If we'll expand the determinant after this, we obtain $$2[(b+c-a)(a+c-b)ab+b(a+c-b)a^2+ab^2(b+c-a)]=$$ $$2[(b+c-a)ab(a+c)+b(a+c-b)a^2]=$$ $$2ab[(b+c-a)(a+c)+(a+c-b)a]=$$ $$2ab[ab+bc+ac+c^2-a^2-ac+a^2+ac-ab]=$$ $$2ab[bc+ac+c^2]=$$ $$2abc[a+b+c],$$ and the initial determinant equals $2(a+b+c)^3abc$ (I verified this with Mathcad). $\endgroup$ – Alex Ravsky Aug 14 '17 at 20:42
  • $\begingroup$ @AlexRavsky Yes that is the answer but I would like to know if there is a way without these tedious calculations. $\endgroup$ – user8277998 Aug 14 '17 at 23:01
  • $\begingroup$ OK, I’ll think once more about such a way. Nevertheless, a search for it may be a much more lengthy and non-trivial task than these calculations and it may be unsuccessful. So it is not recommended to do it in real competitions. :-) $\endgroup$ – Alex Ravsky Aug 15 '17 at 4:08
0
$\begingroup$

We already reduced the problem to calculate

$$D’=\left|\begin{matrix} b+c - a & 0 & a^2 \\ 0 & a+c - b & b^2 \\ b & a & -ab \end{matrix}\right|$$

If $a=0$ then

$$D’=\left|\begin{matrix} b+c & 0 & 0\\ 0 & c - b & b^2 \\ b & 0 & 0 \end{matrix}\right|=0.$$

If $b=0$ then

$$D’=\left|\begin{matrix} c - a & 0 & a^2 \\ 0 & a+c & 0 \\ 0 & a & 0 \end{matrix}\right|=0.$$

Otherwise put $R’_1=R_1+\frac abR_3$ and $R’_2=R_2+\frac baR_3$. Then

$$D’=\left|\begin{matrix} b+c & \frac {a^2}b & 0\\ \frac {b^2}a & a+c & 0 \\ b & a & -ab \end{matrix}\right|=-ab\left|\begin{matrix} b+c & \frac {a^2}b \\ \frac {b^2}a & a+c \\ \end{matrix}\right|=-ab[(a+c)(b+c)-ab]=-ab[ac+bc+c^2]=-abc(a+b+c).$$

The latter formula holds also when $a=0$ or $b=0$. Finally,

$$D=(a+b+c)(-2)D’=2(a+b+c)^3abc.$$

$\endgroup$
  • $\begingroup$ I don't think we can do better than this. $\endgroup$ – user8277998 Aug 16 '17 at 8:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.