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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Using the property of determinants and without expanding, prove that \[ \begin{vmatrix} 1&bc&a(b+c) \\ 1&ca&b(c+a) \\ 1&ab&c(a+b) \end{vmatrix} = 0\]

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


answered Feb 21, 2013 by sreemathi.v
edited Feb 24, 2013 by vijayalakshmi_ramakrishnans

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