# Q : 2      Using the property of determinants and without expanding, prove that               $\dpi{100} \begin{vmatrix}a-b &b-c &c-a \\b-c &c-a &a-b \\c-a &a-b &b-c \end{vmatrix}=0$

Given determinant $\dpi{100} \triangle =\begin{vmatrix}a-b &b-c &c-a \\b-c &c-a &a-b \\c-a &a-b &b-c \end{vmatrix}=0$

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

$\dpi{100} \triangle =\begin{vmatrix}a-c &b-a &c-b \\b-c &c-a &a-b \\-(a-c) &-(b-a) &-(c-b) \end{vmatrix}=0$

$\dpi{100} =-\begin{vmatrix}a-c &b-a &c-b \\b-c &c-a &a-b \\(a-c) &(b-a) &(c-b) \end{vmatrix}=0$

So, we have two rows $\dpi{100} R_{1}$ and $\dpi{100} R_{2}$ identical hence we can say that the value of determinant = 0

Therefore $\dpi{100} \triangle = 0$.

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