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Question Statement:-

Prove that:- $$\begin{vmatrix} b+c-a-d & bc-ad & bc(a+d)-ad(b+c)\\ c+a-b-d & ca-bd & ca(b+d)-bd(c+a)\\ a+b-c-d & ab-cd & ab(c+d)-cd(a+b)\\ \end{vmatrix}\qquad\qquad=2(a-b)(b-c)(c-a)(a-d)(b-d)(c-d)$$

Attempt at a solution:-

I tried to write the given determinant as the product of two other determinants but was not able to come up with anything worthwhile. My attempt was to think on seeing the first column that the determinant would look somehow like this $$\begin{vmatrix} b+c & -1 & ?\\ c+a & -1 & ?\\ a+b & -1 & ?\\ \end{vmatrix}\begin{vmatrix} 1 & a+d & ?\\ 1 & b+d & ?\\ 1 & c+d & ?\\ \end{vmatrix}$$

The question marks represent those entries of the determinant which I was not able to come up with to obtain the original matrix.

It will be very helpful if you could guide me in the right direction to obtain the the determinants whose product gives the original determinant.

If you have an easier method then it would be very helpful too.

user350331
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1 Answers1

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I don't know how to do it your way, but here is another one. It goes by row-reduction.

In a first step, we perform row reduction in the following way: replace $R_1$ with $R_1-R_2$ and $R_3$ with $R_3-R_1$. And we want to replace $R_2$ with $R_1+R_2$. This can be obtained as $$2\left(R_2+\frac{R_1-R_2}2\right)=R_1+R_2.$$ The net effect is that after doing $R_2+(R_1-R_2)/2$ we need to multiply by $2$. This requires us to also multiply by $1/2$. Then our determinant $D$ is equal to $$ D=\frac12\,\begin{vmatrix} 2b-2a& bc+bd-ad-ac&2bcd-2acd\\ 2c-2d& ac+bc-ad-bd& 2abc-2abd\\ 2b-2c&ab+bd-ac-cd&2abd-2acd \end{vmatrix}. $$ Factoring in each entry, we get $$ D=\frac12\,\begin{vmatrix} 2(b-a)& (b-a)(c+d)&2cd(b-a)\\ 2(c-d)& (a+b)(c-d)& 2ab(c-d)\\ 2(b-c)&(a+d)(b-c)&2ad(b-c) \end{vmatrix}. $$ Extracting the common factor in each row, $$ D=\frac12\,(b-a)(c-d)(b-c)\begin{vmatrix} 2& c+d&2cd\\ 2& a+b& 2ab\\ 2&a+d&2ad \end{vmatrix}. $$ Now replace $R_2$ with $R_2-R_3$ and $R_3$ with $R_3-R_1$: $$ D=\frac12\,(b-a)(c-d)(b-c)\begin{vmatrix} 2& c+d&2cd\\ 0& b-d& 2a(b-d)\\ 0&a-c&2d(a-c) \end{vmatrix}. $$ Now we can calculate directly. But if we first extract the common factor from each of the last two rows: $$ D=\frac12\,(b-a)(c-d)(b-c)(b-d)(a-c)\begin{vmatrix} 2& c+d&2cd\\ 0& 1& 2a\\ 0&1&2d \end{vmatrix}. $$ Now the remaining determinant trivially seen to be $4(d-a)$, so $$ D=2(b-a)(c-d)(b-c)(b-d)(a-c)(d-a). $$

Martin Argerami
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