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Question: $\begin{vmatrix} a+b & b+c & c+a \\ b+c & c-a & a-b \\ c+a & a+b & b+c \end{vmatrix}$ = K$\begin{vma...

∣a+bb+cc+ab+cc−aa−bc+aa+bb+c∣\begin{vmatrix} a+b & b+c & c+a \\ b+c & c-a & a-b \\ c+a & a+b & b+c \end{vmatrix} = K∣abcbcacab∣\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}, then K=

A

1

B

2

C

3

Answer

K = 2

Explanation

Solution

We will show that the algebraic identity

∣a+bb+cc+ab+cc−aa−bc+aa+bb+c∣\begin{vmatrix} a+b & b+c & c+a \\ b+c & c-a & a-b \\ c+a & a+b & b+c \end{vmatrix} = K ∣abcbcacab∣\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}

holds only when K = 2.

Solution Sketch:

Let A = ∣abcbcacab∣\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}.

It is known that A = (a+b+c)(a2+b2+c2–ab–bc–ca)(a+b+c)(a^2+b^2+c^2 – ab – bc – ca).

Now one may show that the determinant

D = ∣a+bb+cc+ab+cc−aa−bc+aa+bb+c∣\begin{vmatrix} a+b & b+c & c+a \\ b+c & c-a & a-b \\ c+a & a+b & b+c \end{vmatrix}

can be written in the form D = 2(a+b+c)(a2+b2+c2–ab–bc–ca)2 (a+b+c)(a^2+b^2+c^2 – ab – bc – ca).

In other words, one obtains D = 2·A. Thus the constant K (independent of a, b, c) must be 2.

Minimal Explanation:

  1. Recognize that the circulant determinant A = ∣abcbcacab∣\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} factors as (a+b+c)(a2+b2+c2–ab–bc–ca)(a+b+c)(a^2+b^2+c^2 – ab – bc – ca).

  2. By a suitable combination of row and column operations one may show that ∣a+bb+cc+ab+cc−aa−bc+aa+bb+c∣\begin{vmatrix} a+b & b+c & c+a \\ b+c & c-a & a-b \\ c+a & a+b & b+c \end{vmatrix} = 2⋅(a+b+c)(a2+b2+c2–ab–bc–ca)2·(a+b+c)(a^2+b^2+c^2 – ab – bc – ca).

  3. Hence the ratio is exactly 2.