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Using the property of determinants and without expanding, prove that \[ \begin{vmatrix} a-b&b-c&c-a \\b-c&c-a&a-b \\c-a&a-b&b-c \end{vmatrix} = 0\]

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  • (i)If any two rows or column of a determinant are identical (all corresponding elements are same)then the value of the determinant is zero.
  • (ii)If some or all elements of a row or column of a determinant are expressed as sum of two (or more)terms,then the determinant can be expressed as sum of two(or more) determinants.
Prove that $\begin{vmatrix}a-b & b-c & c-a\\b-c & c-a & a-b\\c-a & a-b & b-c\end{vmatrix}=0$
 
By using the property of determinants let us split the determinant as
 
$\begin{vmatrix}a & b & c\\b & c & a\\c & a & b\end{vmatrix}-\begin{vmatrix}b & c & a\\c & a & b\\a & b & c\end{vmatrix}$
 
The second determinant can be rearranged as,
 
$\begin{vmatrix}a & b & c\\b & c & a\\c & a & b\end{vmatrix}$
 
Hence $\begin{vmatrix}a-b & b-c & c-a\\b-c & c-a & a-b\\c-a & a-b & b-c\end{vmatrix}=\begin{vmatrix}a & b & c\\b & c & a\\c & a & b\end{vmatrix}-\begin{vmatrix}a & b & c\\b & c & a\\c & a & b\end{vmatrix}=0$

 

answered Feb 21, 2013 by sreemathi.v
edited Mar 6, 2013 by sreemathi.v
 

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