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# Using the property of determinants and without expanding, prove that determinant 1 bc a(b+c) 1 ca b(c+a) 1 ab c(a+b) = 0 Ex 4.2 Q : 2

Q : 4           Using the property of determinants and without expanding, prove that

$\dpi{100} \begin{vmatrix}1 &bc &a(b+c) \\1 &ca &b(c+a) \\1 &ab & c(a+b) \end{vmatrix}=0$

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We have determinant:

$\triangle = \begin{vmatrix} 1 &bc &a(b+c) \\ 1& ca &b(c+a) \\ 1& ab &c(a+b) \end{vmatrix}$

Applying $C_{3} \rightarrow C_{2} + C_{3}$ we have then;

$\triangle = \begin{vmatrix} 1 &bc & ab+bc+ca \\ 1& ca &ab+bc+ca \\ 1& ab &ab+bc+ca \end{vmatrix}$

So, here column 3 and column 1 are proportional.

Therefore, $\triangle = 0$.

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