Q : 11         By using properties of determinants, show that:

                   (i) \begin{vmatrix} a-b-c &2a &2a \\ 2b &b-c-a &2b \\ 2c &2c &c-a-b \end{vmatrix}=(a+b+c)^3

Answers (1)

Given determinant:

                               \triangle = \begin{vmatrix} a-b-c &2a &2a \\ 2b &b-c-a &2b \\ 2c &2c &c-a-b \end{vmatrix}

We apply row transformation: R_{1} \rightarrow R_{1}+R_{2}+R_{3} we have;

= \begin{vmatrix} a+b+c &a+b+c &a+b+c \\ 2b &b-c-a &2b \\ 2c &2c &c-a-b \end{vmatrix}

Taking common factor (a+b+c) out.

=(a+b+c) \begin{vmatrix} 1 &1 &1 \\ 2b &b-c-a &2b \\ 2c &2c &c-a-b \end{vmatrix}

Now, applying column tranformation C_{1} \rightarrow C_{1}- C_{2}    and then  C_{2} \rightarrow C_{2}- C_{3}

We have;

=(a+b+c) \begin{vmatrix} 0 &0 &1 \\ b+c+a &-b-c-a &2b \\ 0 &c+a+b &c-a-b \end{vmatrix}

=(a+b+c)(a+b+c)(a+b+c) \begin{vmatrix} 0 &0 &1 \\ 1 &-1 &2b \\ 0 &1 &c-a-b \end{vmatrix}

=(a+b+c)(a+b+c)(a+b+c) = (a+b+c)^3

Hence Proved.

 

 

 

 

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