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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Using properties of determinants,prove the following $\begin{vmatrix}3x& -x+y & -x+z\\x-y & 3y & z-y\\x-z & y-z & 3z\end{vmatrix}=3(x+y+z)(xy+yz+zx)$

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$\bigtriangleup=\begin{vmatrix}3x& -x+y & -x+z\\x-y & 3y & z-y\\x-z & y-z & 3z\end{vmatrix}$
Apply $c_1 \to \;c_1+c_2+c_3$
$\bigtriangleup=\begin{vmatrix}x+y+z & -x+y & -x+z \\ x+y+z & 3y & z-y \\ x+y+z & y-z & 3z\end{vmatrix}$
$\bigtriangleup=(x+y+z)\begin{vmatrix} 1 & -x+y & -x+z \\ 1 & 3y & z-y \\ 1 & y-z & 3z\end{vmatrix}$
Apply $R_2 \to \;R_2-R_3\;and\;R_3 \to \;R_3-R_1 $
$\bigtriangleup=(x+y+z)\begin{vmatrix} 1 & -x+y & -x+z \\ 0 & 2y+z & -y-2z \\ 0 & x-z & 2z+x\end{vmatrix}$
$\bigtriangleup=(x+y+z)\bigg[1\{(2y+z)(2z+x)+(y+2z)(x-z)\}\bigg]$
$=(x+y+z) \bigg[2yz+2yx+2z^2+xz+yx-yz+2xz-2z^2\bigg]$
$=(x+y+z) \bigg[3xy+3yx+3xz\bigg]$
$3(x+y+z) \bigg[xy+yz+xz\bigg]$
Hence proved
answered Mar 24, 2013 by meena.p
 

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