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

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

$\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|=0$

 

Solution:

$\Delta=\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|$

By applying $\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}+\mathrm{C}_{2}$, we have:

$\Delta=\left|\begin{array}{lll}1 & b c & a b+b c+c a \\ 1 & c a & a b+b c+c a \\ 1 & a b & a b+b c+c a\end{array}\right|$

Here, two columns C1 and Care proportional.

$\therefore \Delta=0$

 

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