#### Explain Solution R.D. Sharma Class 12 Chapter  deteminants Exercise 5.2 Question  19 maths Textbook Solution.

Answer:$\left|\begin{array}{ccc} z & x & y \\ z^{2} & x^{2} & y^{2} \\ z^{4} & x^{4} & y^{4} \end{array}\right|=\left|\begin{array}{ccc} x & y & z \\ x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \end{array}\right|=\left|\begin{array}{ccc} x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \\ x & y & z \end{array}\right|$

$=x y z(x-y)(y-z)(z-x)(x+y+z)$

Hint:We will prove all the determinant equal by using interchange column or row property

Given:\begin{aligned} &\left|\begin{array}{lll} z & x & y \\ z^{2} & x^{2} & y^{2} \\ z^{4} & x^{4} & y^{4} \end{array}\right|=\left|\begin{array}{ccc} x & y & z \\ x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \end{array}\right|=\left|\begin{array}{ccc} x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \\ x & y & z \end{array}\right| \\ &=x y z(x-y)(y-z)(z-x)(x+y+z) \end{aligned}

Solution:

Let us first solve for

$\left|\begin{array}{ccc} z & x & y \\ z^{2} & x^{2} & y^{2} \\ z^{4} & x^{4} & y^{4} \end{array}\right|$

If any two rows or columns of a determinant are interchanged, then sign of the determinant is changed.

$C_{1} \leftrightarrow C_{2}$

$=(-)\left|\begin{array}{ccc}x & z & y \\ x^{2} & z^{2} & y^{2} \\ x^{4} & z^{4} & y^{4}\end{array}\right|$

$C_{1} \leftrightarrow C_{2}$

$=(-)(-)\left|\begin{array}{ccc}x & y & z \\ x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4}\end{array}\right|$

$=\left|\begin{array}{ccc}x & y & z \\ x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4}\end{array}\right|$

$R_{1} \leftrightarrow R_{2} =(-)\left|\begin{array}{ccc}x^{2} & y^{2} & z^{2} \\ x & y & z \\ x^{4} & y^{4} & z^{4}\end{array}\right| R_{2} \leftrightarrow R_{3} =(-)(-)\left|\begin{array}{ccc}x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \\ x & y & z\end{array}\right| =\left|\begin{array}{ccc}x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \\ x & y & z\end{array}\right|$.....2

$From equation (1) and (2) we get \left|\begin{array}{ccc}z & x & y \\ z^{2} & x^{2} & y^{2} \\ z^{4} & x^{4} & y^{4}\end{array}\right|=\left|\begin{array}{ccc}x & y & z \\ x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4}\end{array}\right|=\left|\begin{array}{ccc}x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \\ x & y & z\end{array}\right| \quad \ldots$(3)

$Now consider- \left|\begin{array}{ccc}x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \\ x & y & z\end{array}\right| Taking common x from C_{1}, y from C_{2} and z from C_{3} =x y z\left|\begin{array}{ccc}x & y & z \\ x^{3} & y^{3} & z^{3} \\ 1 & 1 & 1\end{array}\right| On applying C_{1} \rightarrow C_{1}-C_{2} and C_{2} \rightarrow C_{2}-C_{3} =x y z\left|\begin{array}{ccc}x-y & y-z & z \\ x^{3}-y^{3} & y^{3}-z^{3} & z^{3} \\ 0 & 0 & 1\end{array}\right| =x y z\left|\begin{array}{ccc}x-y & y-z & z \\ (x-y)\left(x^{2}+x y+y^{2}\right) & (y-z)\left(y^{2}+y z+z^{2}\right) & z^{3} \\ 0 & 0 & 1\end{array}\right|$

$\because x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$

\begin{aligned} &\text { On taking common }(x-y) \text { common from } \mathrm{C}_{1} \text { and }(y-z) \text { from } \mathrm{C}_{2}\\ &=x y z(x-y)(y-z)\left|\begin{array}{ccc} 1 & 1 & z \\ \left(x^{2}+x y+y^{2}\right) & \left(y^{2}+y z+z^{2}\right) & z^{3} \\ 0 & 0 & 1 \end{array}\right| \end{aligned}

\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{3}\\ &=x y z(x-y)(y-z)\left[0\left|\begin{array}{cc} 1 & z \\ y^{2}+y z+z^{2} & z^{3} \end{array}\right|-0\left|\begin{array}{cc} 1 & z \\ x^{2}+x y+y^{2} & z^{3} \end{array}\right|+1\left|\begin{array}{cc} 1 & 1 \\ x^{2}+x y+y^{2} & y^{2}+y z+z^{2} \end{array}\right|\right]\\ &=x y z(x-y)(y-z)\left[0-0+1\left\{\left(y^{2}+y z+z^{2}\right)-\left(x^{2}+x y+y^{2}\right)\right\}\right]\\ &=x y z(x-y)(y-z)\left(y^{2}+y z+z^{2}-x^{2}-x y-y^{2}\right)\\ &=x y z(x-y)(y-z)\left(z^{2}-x^{2}+y z-x y\right)\\ &=x y z(x-y)(y-z)\{(z+x)(z-x)+y(z-x)\}\\ &=x y z(x-y)(y-z)(z-x)(z+x+y)\\ &=x y z(x-y)(y-z)(z-x)(x+y+z) \end{aligned}

From equation (1), (2),(3) and (4)

Hence it is proved that

$\left|\begin{array}{ccc} z & x & y \\ z^{2} & x^{2} & y^{2} \\ z^{4} & x^{4} & y^{4} \end{array}\right|=\left|\begin{array}{ccc} x & y & z \\ x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \end{array}\right|=\left|\begin{array}{ccc} x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \\ x & y & z \end{array}\right|$

$=x y z(x-y)(y-z)(z-x)(x+y+z)$