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Question Statement:- Show that $$\begin{vmatrix} (b+c)^2 & c^2 & b^2 \\ c^2 & (c+a)^2 & a^2 \\ b^2 & a^2 & (a+b)^2 \\ \end{vmatrix}=2(ab+bc+ca)^3$$

Attempt at a Solution:-

1st attempt(which was in vain):-

LHS:-$$\begin{align}\begin{vmatrix} (b+c)^2 & c^2 & b^2 \\ c^2 & (c+a)^2 & a^2 \\ b^2 & a^2 & (a+b)^2 \\ \end{vmatrix}=\begin{vmatrix} 2bc & c^2 & b^2 \\ (c^2-a^2)-(c+a)^2 & (c+a)^2 & a^2 \\ (b^2-a^2)-(b +a)^2 & a^2 & (a+b)^2 \\ \end{vmatrix}\left[\begin{array}{11} C_1\rightarrow C_1-C_2-C_3\end{array}\right] =\begin{vmatrix} 2bc & c^2 & b^2 \\ -2(a^2+ac) & (c+a)^2 & a^2 \\ -2(a^2+ab) & a^2 & (a+b)^2 \\ \end{vmatrix}= 2\begin{vmatrix} bc & c^2 & b^2 \\ -(a^2+ac) & (c+a)^2 & a^2 \\ -(a^2+ab) & a^2 & (a+b)^2 \\ \end{vmatrix}= \dfrac{2}{bc}\begin{vmatrix} b^2c^2 & c^2 & b^2 \\ -bc(a^2+ac) & (c+a)^2 & a^2 \\ -bc(a^2+ab) & a^2 & (a+b)^2 \\ \end{vmatrix}\left[C_1\rightarrow bc\cdot C_1\right]= \dfrac{2}{bc}\begin{vmatrix} b^2c^2-b^2c^2 & c^2 & b^2 \\ -bc(a^2+ac)-b^2(c+a)^2 & (c+a)^2 & a^2 \\ -bc(a^2+ab)-a^2b^2 & a^2 & (a+b)^2 \\ \end{vmatrix}[C_1\rightarrow C_1-b^2C_2]= \dfrac{2}{bc}\begin{vmatrix}0 & c^2 & b^2 \\ -b(a+c)(ab+bc+ac) & (c+a)^2 & a^2 \\ -ab(ab+bc+ac) & a^2 & (a+b)^2 \\ \end{vmatrix}\\ =-2\left(\dfrac{ab+bc+ac}{c}\right)\begin{vmatrix} 0 & c^2 & b^2 \\ a+c & (c+a)^2 & a^2 \\ a & a^2 & (a+b)^2 \\ \end{vmatrix} \end{align}$$

I was pretty much stuck after this so I tried another approach.

2nd Attempt:- $$\begin{vmatrix} (b+c)^2 & c^2 & b^2 \\ c^2 & (c+a)^2 & a^2 \\ b^2 & a^2 & (a+b)^2 \\ \end{vmatrix}= (abc)^4\begin{vmatrix} \left(\dfrac{1}{b}+\dfrac{1}{c}\right)^2 & \dfrac{1}{b^2} & \dfrac{1}{c^2} \\ \dfrac{1}{a^2} & \left(\dfrac{1}{a}+\dfrac{1}{c}\right)^2 & \dfrac{1}{c^2} \\ \dfrac{1}{a^2} & \dfrac{1}{b^2} & \left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2 \\ \end{vmatrix}\left[\begin{array}{11} R_1\rightarrow \dfrac{R_1}{b^2c^2} \\ R_2\rightarrow \dfrac{R_2}{a^2c^2} \\ R_2\rightarrow \dfrac{R_3}{a^2b^2 }\end{array}\right]$$

And after starting along the route that I have shown in the second attempt I figured it was much more useless than the previous one. So, I thought that the Mathematics Stack Exchange is the only route left. So, your help is very much needed.

user350331
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  • I recommend computing the determinant by cofactor expansion instead of trying to diagonalize it, maybe after some row operations. – Matt Samuel Oct 15 '16 at 13:16

1 Answers1

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Let us continue with your valiant first attempt (note that you incorrectly missed out a factor of $2$, from your third stage to the fourth stage) $$-\color{red}{2}\left(\dfrac{ab+bc+ac}{c}\right)\begin{vmatrix} 0 & c^2 & b^2 \\ a+c & (c+a)^2 & a^2 \\ a & a^2 & (a+b)^2 \\\end{vmatrix}$$ Using a brute force approach to evaluate the determinant of the $3\times 3$ matrix (aided by the fact that the top left entry is $0$), the determinant evaluates to $$0-c^2((a+b)^2(a+c)-a^2)+b^2(a^2(a+c)-(c+a)^2a)$$ which, after a fair amount of algebraic manipulation, simplifies to $$-c(a^2c^2+b^2c^2+a^2b^2+2a^2bc+2abc^2+2ab^2c)=-c(ab+bc+ac)^2$$ We thus have the desired result $$-2\left(\frac{ab+bc+ac}{c}\right)\times-c(ab+bc+ac)^2=2(ab+bc+ac)^3$$

A less brutal approach is to make use of the Matrix determinant lemma see here.

The matrix, denote by $M$, can be expressed as $$M=vv^T-A=-(A+(-vv^T))$$ where $v= \left( \begin{array}{c} b+c \\ a+c \\ a+b \end{array} \right) $ and $A= (ab+bc+ca)\left( \begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array} \right) $

Invoking the Matrix determinant lemma, we have $$\det(M)=-(1-v^TA^{-1}v)\det(A)\tag{Eq.1}$$ Now, it is straightforward to show that $$\det(A)=2(ab+bc+ac)^3$$ and $$A^{-1}=\frac{1}{2(ab+bc+ca)}\left( \begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array} \right)$$ Leading to $v^TA^{-1}v=2$

Substituting all the values into Eq.$1$, we have $$\det(M)=-(1-2)\times2(ab+bc+ac)^3=2(ab+bc+ca)^3$$

Alijah Ahmed
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  • You see I am trying to learn to efficiently use the properties of the determinants to arrive at the $RHS$. Although your solution is very good but I was looking for a solution that does not include brute force. If you can come up with that can you add that too to your solution. – user350331 Oct 15 '16 at 12:49
  • @user350331 Okay, I'll have a look at a more efficient solution. – Alijah Ahmed Oct 15 '16 at 12:57
  • @user350331 I have found a more efficient solution based on the Matrix determinant lemma, and have added it to my answer. – Alijah Ahmed Oct 15 '16 at 14:39
  • I don't intend to devalue your efforts on providing me the solution which didn't include brute force but the second solution just went way over my head. I am a highschooler so I was just looking for a solution which used the basic properties of the determinant. So, if you still have the patience to write another one I will be very thankful. If not I think I would have to go with brute force. – user350331 Oct 16 '16 at 03:43
  • @user350331 No worries. The second solution does go above the high school level, so you might be better off with the brute force approach, but doing it with cofactor expansion (see Matt Samuel's useful comment). – Alijah Ahmed Oct 16 '16 at 10:21