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# Using the property of determinants and without expanding, prove that $\begin{vmatrix} 1&bc&a(b+c) \\ 1&ca&b(c+a) \\ 1&ab&c(a+b) \end{vmatrix} = 0$

Toolbox:
• (i)If any two rows or column of a determinant are identical,then the value of the corresponding determinant is zero.
• 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 $\mid A \mid=\begin{vmatrix}1 & bc & a(b+c)\\1 & ca & b(c+a)\\1 & ab & c(a+b)\end{vmatrix}$

This can be written as,

$A=\begin{vmatrix}1 & bc & ab+ac\\1 & ca & bc+ab\\1 & ab & ca+cb\end{vmatrix}$

By adding $C_2$ and $C_3$ and rewriting $C_3$ ,

Therefore $\mid A\mid= \begin{vmatrix}1 & bc & ab+ac+bc\\1 & ca & bc+ab+ca\\1 & ab & ab+ca+cb\end{vmatrix}$

Now we can see $C_1$ and $C_3$ are identical,

Hence the value of the determinant is 0.

Therefore $\begin{vmatrix}1 & bc &a(b+c)\\1 & ca & b(c+a)\\1 & ab & c(a+b)\end{vmatrix}=0.$

edited Feb 24, 2013