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Q : 6        Prove that    \begin{vmatrix} a^2 &bc &ac+a^2 \\ a^2+ab & b^2 & ac\\ ab &b^2+bc &c^2 \end{vmatrix}=4a^2b^2c^2.

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Given matrix \begin{vmatrix} a^2 &bc &ac+a^2 \\ a^2+ab & b^2 & ac\\ ab &b^2+bc &c^2 \end{vmatrix}

Taking common factors a,b and c from the column C_{1}, C_{2}, and\ C_{3} respectively.

we have; 

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

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

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

Then applying R_{2} \rightarrow R_{2}+R_{1} , we get;

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

Applying R_{3} \rightarrow R_{3}+R_{2}, we have;

\triangle = abc\begin{vmatrix} a &c &a+c \\ a+b & b &a\\ 2b &2b &0 \end{vmatrix} = 2ab^2c\begin{vmatrix} a &c &a+c \\ a+b & b &a\\ 1 &1 &0 \end{vmatrix}

Now, applying column transformation; C_{2} \rightarrow C_{2 }-C_{1}, we have

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

So we can now expand the remaining determinant along R_{3} we have;

\triangle = 2ab^2c\left [ a(c-a)+a(a+c) \right ]

= 2ab^2c\left [ ac-a^2+a^2+ac) \right ] = 2ab^2c\left [ 2ac \right ]

= 4a^2b^2c^2

Hence proved.

Posted by

Divya Prakash Singh

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