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

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

Answers (1)

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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