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# Using the property of determinants and without expanding prove that determinant b plus c q plus r y plus z c plua a r plus p z plus x a plus b p plus q x plus y equal to 2 determinant a p x b q y c r z

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

$\dpi{100} \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ a+b &p+q & x+y \end{vmatrix}=2\begin{vmatrix} a &p &x \\ b &q &y \\ c &r & z \end{vmatrix}$

Views

Given determinant :

$\dpi{100} \triangle= \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ a+b &p+q & x+y \end{vmatrix}$

Splitting the third row; we get,

$\dpi{100} = \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ a &p & x \end{vmatrix} + \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ b &q & y \end{vmatrix} = \triangle_{1} + \triangle_{2}\ (assume\ that)$.

Then we have,

$\dpi{100} \triangle_{1} = \begin{vmatrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a &p & x \end{vmatrix}$

On Applying row transformation $\dpi{100} R_{2} \rightarrow R_{2} - R_{3}$   and then  $\dpi{100} R_{1} \rightarrow R_{1} - R_{2}$;

we get, $\dpi{100} \triangle_{1} = \begin{vmatrix} b & q & y \\ c & r & z \\ a &p & x \end{vmatrix}$

Applying Rows exchange transformation $\dpi{100} R_{1} \leftrightarrow R_{2}$   and   $\dpi{100} R_{2} \leftrightarrow R_{3}$, we have:

$\dpi{100} \triangle_{1} =(-1)^2 \begin{vmatrix} b & q & y \\ c & r & z \\ a &p & x \end{vmatrix}= \begin{vmatrix} a & p & x\\ b & q&y \\ c& r & z \end{vmatrix}$

also $\dpi{100} \triangle_{2} = \begin{vmatrix} b+c & q+r & y+z \\ c+a&r+p &z+x \\ b & q & y \end{vmatrix}$

On applying rows transformation, $\dpi{100} R_{1} \rightarrow R_{1} - R_{3}$ and then $\dpi{100} R_{2} \rightarrow R_{2} - R_{1}$

$\dpi{100} \triangle_{2} = \begin{vmatrix} c & r & z \\ c+a&r+p &z+x \\ b & q & y \end{vmatrix}$  and then  $\dpi{100} \triangle_{2} = \begin{vmatrix} c & r & z \\ a&p &x \\ b & q & y \end{vmatrix}$

Then applying rows exchange transformation;

$\dpi{100} R_{1} \leftrightarrow R_{2}$   and then $\dpi{100} R_{2} \leftrightarrow R_{3}$. we have then;

$\dpi{100} \triangle_{2} =(-1)^2 \begin{vmatrix} a & p & x \\ b&q &y \\ c & r & z \end{vmatrix}$

So, we now calculate the sum = $\dpi{100} \triangle_{1} + \triangle _{2}$

$\dpi{100} \triangle_{1} + \triangle _{2} = 2 \begin{vmatrix} a &p &x \\ b& q& y\\ c & r& z \end{vmatrix}$

Hence proved.

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