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Q : 13        Using properties  of determinants, prove that

                  \begin{vmatrix} 3a &-a+b &-a+c \\ -b+c &3b &-b+c \\ -c+a &-c+b &3c \end{vmatrix}=3(a+b+c)(ab+bc+ca)

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Given determinant \triangle = \begin{vmatrix} 3a &-a+b &-a+c \\ -b+c &3b &-b+c \\ -c+a &-c+b &3c \end{vmatrix}

Applying the column transformation, C_{1} \rightarrow C_{1} +C_{2}+C_{3} we have then;

\triangle = \begin{vmatrix} a+b+c &-a+b &-a+c \\ a+b+c &3b &-b+c \\ a+b+c &-c+b &3c \end{vmatrix}

Taking common factor (a+b+c) out from the column first;

=(a+b+c) \begin{vmatrix} 1 &-a+b &-a+c \\ 1 &3b &-b+c \\1 &-c+b &3c \end{vmatrix}

Applying R_{2} \rightarrow R_{2}-R_{1}  and  R_{3} \rightarrow R_{3}-R_{1}, we have then;

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

Now we can expand the remaining determinant along C_{1} we have;

 \triangle=(a+b+c) [(2b+a)(2c+a)-(a-b)(a-c)]

=(a+b+c) [4bc+2ab+2ac+a^2-a^2+ac+ba-bc]



Hence proved.


Posted by

Divya Prakash Singh

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