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If $\mathrm{A}=\left[\begin{array}{lll}\mathrm{x} & \mathrm{x}^{2} & 1+\mathrm{x}^{2} \\ \mathrm{y} & \mathrm{y}^{2} & 1+\mathrm{y}^{2} \\ \mathrm{z} & \mathrm{z}^{2} & 1+\mathrm{z}^{2}\end{array}\right]$ where $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are distinct what is $\mid \mathrm{A}|$?
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Verified Answer
The correct answer is:
$(x-y)(y-z)(z-x)$
$A=\left[\begin{array}{lll}x & x^{2} & 1+x^{2} \\ y & y^{2} & 1+y^{2} \\ z & z^{2} & 1+z^{2}\end{array}\right]$
$|A|=\left|\begin{array}{lll}x & x^{2} & 1+x^{2} \\ y & y^{2} & 1+y^{2} \\ z & z^{2} & 1+z^{2}\end{array}\right|$
Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}$ and $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}$
$|\mathrm{~A}|=\left|\begin{array}{ccc}\mathrm{x}-\mathrm{y} & (\mathrm{x}-\mathrm{y})(\mathrm{x}+\mathrm{y}) & (\mathrm{x}-\mathrm{y})(\mathrm{x}+\mathrm{y}) \\ \mathrm{y}-\mathrm{z} & (\mathrm{y}-\mathrm{z})(\mathrm{y}+\mathrm{z}) & (\mathrm{y}-\mathrm{z})(\mathrm{y}+\mathrm{z}) \\ \mathrm{z} & \mathrm{z}^{2} & 1+\mathrm{z}^{2}\end{array}\right|$
$=(x-y)(y-z)\left|\begin{array}{ccc}1 & x+y & x+y \\ 1 & y+z & y+z \\ z & z^{2} & 1+z^{2}\end{array}\right|$
Applying $\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{2}$
$|\mathrm{~A}|=(\mathrm{x}-\mathrm{y})(\mathrm{y}-\mathrm{z})\left|\begin{array}{ccc}1 & \mathrm{x}+\mathrm{y} & 0 \\ 1 & \mathrm{y}+\mathrm{z} & 0 \\ \mathrm{z} & \mathrm{z}^{2} & 1\end{array}\right|$
$=(\mathrm{x}-\mathrm{y})(\mathrm{y}-\mathrm{z})[1\{\mathrm{y}+\mathrm{z}-(\mathrm{x}+\mathrm{y})]$
$=(\mathrm{x}-\mathrm{y})(\mathrm{y}-\mathrm{z})(\mathrm{z}-\mathrm{x})$
$|A|=\left|\begin{array}{lll}x & x^{2} & 1+x^{2} \\ y & y^{2} & 1+y^{2} \\ z & z^{2} & 1+z^{2}\end{array}\right|$
Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}$ and $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}$
$|\mathrm{~A}|=\left|\begin{array}{ccc}\mathrm{x}-\mathrm{y} & (\mathrm{x}-\mathrm{y})(\mathrm{x}+\mathrm{y}) & (\mathrm{x}-\mathrm{y})(\mathrm{x}+\mathrm{y}) \\ \mathrm{y}-\mathrm{z} & (\mathrm{y}-\mathrm{z})(\mathrm{y}+\mathrm{z}) & (\mathrm{y}-\mathrm{z})(\mathrm{y}+\mathrm{z}) \\ \mathrm{z} & \mathrm{z}^{2} & 1+\mathrm{z}^{2}\end{array}\right|$
$=(x-y)(y-z)\left|\begin{array}{ccc}1 & x+y & x+y \\ 1 & y+z & y+z \\ z & z^{2} & 1+z^{2}\end{array}\right|$
Applying $\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{2}$
$|\mathrm{~A}|=(\mathrm{x}-\mathrm{y})(\mathrm{y}-\mathrm{z})\left|\begin{array}{ccc}1 & \mathrm{x}+\mathrm{y} & 0 \\ 1 & \mathrm{y}+\mathrm{z} & 0 \\ \mathrm{z} & \mathrm{z}^{2} & 1\end{array}\right|$
$=(\mathrm{x}-\mathrm{y})(\mathrm{y}-\mathrm{z})[1\{\mathrm{y}+\mathrm{z}-(\mathrm{x}+\mathrm{y})]$
$=(\mathrm{x}-\mathrm{y})(\mathrm{y}-\mathrm{z})(\mathrm{z}-\mathrm{x})$
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