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Using the property of determinants and without expanding prove that determinant a-b b-c c-a b-c c-a a-b c-a a-b b-c = 0 Ex 4.2 Q : 2

Q : 2      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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Given determinant \triangle =\begin{vmatrix}a-b &b-c &c-a \\b-c &c-a &a-b \\c-a &a-b &b-c \end{vmatrix}=0

Applying the rows addition R_{1} \rightarrow R_{1}+R_{2}   then we have;

\triangle =\begin{vmatrix}a-c &b-a &c-b \\b-c &c-a &a-b \\-(a-c) &-(b-a) &-(c-b) \end{vmatrix}=0

=-\begin{vmatrix}a-c &b-a &c-b \\b-c &c-a &a-b \\(a-c) &(b-a) &(c-b) \end{vmatrix}=0

So, we have two rows R_{1} and R_{2} identical hence we can say that the value of determinant = 0 

Therefore \triangle = 0.

 

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