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

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

Answers (1)

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)

 

 

 

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