Using property of determinants and without expanding, prove that: left|begin{array}{lll}a-b & b-

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3 and 4 .Determinants and Matrices

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Using the property of determinants and without expanding, prove that:

$\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|=0$

Option A

Option B

Option C

Option D

Solution

$\Delta=\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|$

Applying $R_{1} \rightarrow R_{1}+R_{2},$ we have:

$\Delta=\left|\begin{array}{ccc}a-c & b-a & c-b \\ b-c & c-a & a-b \\ -(a-c) & -(b-a) & -(c-b)\end{array}\right|$

$=-\left|\begin{array}{ccc}a-c & b-a & c-b \\ b-c & c-a & a-b \\ a-c & b-a & c-b\end{array}\right|$

Here, the two rows $R_{1}$ and $R_{3}$ are identical.

$\therefore \Delta=0$

Standard 12

Mathematics

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