Q : 5           Using the property of determinants and without expanding, prove that 

                   \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ a+b &p+q & x+y \end{vmatrix}=2\begin{vmatrix} a &p &x \\ b &q &y \\ c &r & z \end{vmatrix}

Answers (1)

Given determinant : 

\triangle= \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ a+b &p+q & x+y \end{vmatrix} 

Splitting the third row; we get,

= \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ a &p & x \end{vmatrix} + \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ b &q & y \end{vmatrix} = \triangle_{1} + \triangle_{2}\ (assume\ that).

Then we have,

\triangle_{1} = \begin{vmatrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a &p & x \end{vmatrix}

On Applying row transformation R_{2} \rightarrow R_{2} - R_{3}   and then  R_{1} \rightarrow R_{1} - R_{2};

we get, \triangle_{1} = \begin{vmatrix} b & q & y \\ c & r & z \\ a &p & x \end{vmatrix}

Applying Rows exchange transformation R_{1} \leftrightarrow R_{2}   and   R_{2} \leftrightarrow R_{3}, we have:

\triangle_{1} =(-1)^2 \begin{vmatrix} b & q & y \\ c & r & z \\ a &p & x \end{vmatrix}= \begin{vmatrix} a & p & x\\ b & q&y \\ c& r & z \end{vmatrix}

also \triangle_{2} = \begin{vmatrix} b+c & q+r & y+z \\ c+a&r+p &z+x \\ b & q & y \end{vmatrix}

On applying rows transformation, R_{1} \rightarrow R_{1} - R_{3} and then R_{2} \rightarrow R_{2} - R_{1}

\triangle_{2} = \begin{vmatrix} c & r & z \\ c+a&r+p &z+x \\ b & q & y \end{vmatrix}  and then  \triangle_{2} = \begin{vmatrix} c & r & z \\ a&p &x \\ b & q & y \end{vmatrix}

Then applying rows exchange transformation;

R_{1} \leftrightarrow R_{2}   and then R_{2} \leftrightarrow R_{3}. we have then;

\triangle_{2} =(-1)^2 \begin{vmatrix} a & p & x \\ b&q &y \\ c & r & z \end{vmatrix}

So, we now calculate the sum = \triangle_{1} + \triangle _{2}

\triangle_{1} + \triangle _{2} = 2 \begin{vmatrix} a &p &x \\ b& q& y\\ c & r& z \end{vmatrix}

Hence proved.

 

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