Prove the following
Question:

$\left|\begin{array}{ccc}y^{2} z^{2} & y z & y+z \\ z^{2} x^{2} & z x & z+x \\ x^{2} y^{2} & x y & x+y\end{array}\right|=0$

Solution:

From the given,

[Multiplying $R_{1}, R_{2}, R_{3}$ by $x, y, z$ respectively]

$=\frac{1}{x y z}\left|\begin{array}{ccc}x y^{2} z^{2} & x y z & x y+x z \\ x^{2} y z^{2} & x y z & y z+x y \\ x^{2} y^{2} z & x y z & x z+y z\end{array}\right|$

Next

[Taking $(x y z)$ common from $C_{1}$ and $C_{2}$ ]

$=\frac{1}{x y z}(x y z)^{2}\left|\begin{array}{lll}y z & 1 & x y+x z \\ x z & 1 & y z+x y \\ x y & 1 & x z+y z\end{array}\right|$

Then,

[Applying $C_{3} \rightarrow C_{3}+C_{1}$ ]

$=x y z\left|\begin{array}{lll}y z & 1 & x y+y z+z x \\ x z & 1 & x y+y z+z x \\ x y & 1 & x y+y z+z x\end{array}\right|$

Lastly,

[Taking $(x y+y z+z x)$ common from $C_{3}$ ]

$=x y z(x y+y z+z x)\left|\begin{array}{lll}y z & 1 & 1 \\ x z & 1 & 1 \\ x y & 1 & 1\end{array}\right|$

$=0$

$\left[\because C_{2}\right.$ and $C_{3}$ are identical $]$

Administrator

Leave a comment

Please enter comment.
Please enter your name.