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If $\left|\begin{array}{lll}y & x & y+z \\ z & y & x+y \\ x & z & z+x\end{array}\right|=0$, then which one of the following is
correct?
Options:
correct?
Solution:
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Verified Answer
The correct answer is:
Either $x+y=-z$ or $x=z$
Given, $\left|\begin{array}{ccc}y & x & y+z \\ z & y & x+y \\ x & z & z+x\end{array}\right|=0$
Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}$
$\Rightarrow\left|\begin{array}{ccc}x+y+ & x+y+z & 2(x+y+z) \\ z & y & x+y \\ x & z & z+x\end{array}\right|=0$
$\Rightarrow(x+y+z)\left|\begin{array}{ccc}1 & 1 & 2 \\ z & y & x+y \\ x & z & z+x\end{array}\right|=0$
Applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2}, \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-2 \mathrm{C}_{1}$
$\Rightarrow(x+y+z)\left|\begin{array}{ccc}1 & 0 & 0 \\ z & z-y & x+y-2 z \\ x & z-x & z-x\end{array}\right|=0$
$\Rightarrow(x+y+z)\left|\begin{array}{cc}z-y & x+y-2 z \\ z-x & z-x\end{array}\right|=0$
$\Rightarrow(x+y+z)(z-x)(z-y-x-y+2 z)=0$
$\Rightarrow x+y=-z$ or $z=x$
Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}$
$\Rightarrow\left|\begin{array}{ccc}x+y+ & x+y+z & 2(x+y+z) \\ z & y & x+y \\ x & z & z+x\end{array}\right|=0$
$\Rightarrow(x+y+z)\left|\begin{array}{ccc}1 & 1 & 2 \\ z & y & x+y \\ x & z & z+x\end{array}\right|=0$
Applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2}, \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-2 \mathrm{C}_{1}$
$\Rightarrow(x+y+z)\left|\begin{array}{ccc}1 & 0 & 0 \\ z & z-y & x+y-2 z \\ x & z-x & z-x\end{array}\right|=0$
$\Rightarrow(x+y+z)\left|\begin{array}{cc}z-y & x+y-2 z \\ z-x & z-x\end{array}\right|=0$
$\Rightarrow(x+y+z)(z-x)(z-y-x-y+2 z)=0$
$\Rightarrow x+y=-z$ or $z=x$
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