# Computing determinant without expansion

\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.

• $$R_3'=R_3-R_2-R_1$$ – lab bhattacharjee Aug 14 '17 at 16:20
• @labbhattacharjee Then $R_3 = 2[-b \qquad -a \qquad ab]$. It simplifies things somewhat but still not very helpful. – user8277998 Aug 14 '17 at 16:52
• 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). – Alex Ravsky Aug 14 '17 at 20:42
• @AlexRavsky Yes that is the answer but I would like to know if there is a way without these tedious calculations. – user8277998 Aug 14 '17 at 23:01
• 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. :-) – Alex Ravsky Aug 15 '17 at 4:08

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,

$$D=(a+b+c)(-2)D’=2(a+b+c)^3abc.$$

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