# If a, b and c are real numbers, and $\Delta = \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c\\ a+b & b+c & c+a \end{vmatrix}=0,$ show that either $a+b+c=0$ or $a=b=c.$

Toolbox:
• If each element of a row(or a column) of a determinant is multiplied by a constant k,then its value gets multiplied by k.
• By this property we can take out any common factor from any one row or any column of a given determinant.
• Elementary transformations can be done by
• 1. Interchanging any two rows or columns. rows.
• 2. Mutiplication of the elements of any row or column by a non-zero number
• 3. The addition of any row or column , the corresponding elemnets of any other row or column multiplied by any non zero number.
Let $\Delta = \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c\\ a+b & b+c & c+a \end{vmatrix}=0$

Let us apply $R_1\rightarrow R_1+R_2+R_3$

$\Delta = \begin{vmatrix} 2a+2b+2c & 2a+2b+2c & 2a+2b+2c \\ c+a & a+b & b+c\\ a+b & b+c & c+a \end{vmatrix}$

Taking 2(a+b+c) as the common factor from $R_1$

$\Delta = 2(a+b+c)\begin{vmatrix} 1 & 1 & 1 \\ c+a & a+b & b+c\\ a+b & b+c & c+a \end{vmatrix}$

Apply $C_1\rightarrow C_1-C_2$ and $C_2\rightarrow C_2-C_3$ we get

$\Delta = 2(a+b+c)\begin{vmatrix} 0 & 0 & 1\\ c+a & a+b & b+c\\ a+b & b+c & c+a \end{vmatrix}$

Now expanding along $R_1$

$\mid \Delta\mid=2(a+b+c)[0-0+1[(c-b)(b-a)-(a-c)(a-c)]$

$\;\;\;\;\;=2(a+b+c)[bc-b^2+ab-ac-a^2+2ac-c^2]$

$\;\;\;\;\;=2(a+b+c)[bc+ab+ac-a^2-b^2-c^2]$

It is given that $\Delta =0.$

Therefore $2(a+b+c)[bc+ab+ac-a^2-b^2-c^2]=0.$

Hence either a+b+c=0 or $ab+bc+ca-a^2-b^2-c^2=0.$

a+b+c=0 is proved.

Now let us consider $ab+bc+ca-a^2-b^2-c^2=0.$

Multiply throughout by 2

$2ab+2bc+2ac-2a^2-2b^2-2c^2=0.$

Or $2a^2+2b^2+2c^2-2ab-2bc-2ac=0.$

This can be written as $a^2+a^2+b^2+b^2+c^2+c^2-2ab-2bc-2ca=0$

This can now be grouped in:

$(a^2+b^2-2ab)+(b^2+c^2-2bc)+(c^2+a^2-2ac)=0.$

$\Rightarrow (a-b)^2+(b-c)^2+(c-a)^2=0.$

$\Rightarrow (a-b)^2=(b-c)^2=(c-a)^2=0.$

[$(a-b)^2,(b-c)^2,(c-a)^2$ are non-negative]

a=b,b=c and c=a.

or a=b=c.

Hence proved.

edited Feb 28, 2013