#### Explain Solution R.D. Sharma Class 12 Chapter  deteminants Exercise 5.2 Question 2 sub question 10 maths Textbook Solution.

Answer:-$\Delta=\left|\begin{array}{ccc} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array}\right|+\Delta_{1}=\left|\begin{array}{ccc} 1 & 1 & 1 \\ y z & z x & x y \\ x & y & z \end{array}\right|=0$

Hint: First we solve $\Delta$ and after we solve for $\Delta _{1}$

We will try to make some elements of the determinant  zero

Given:$\Delta=\left|\begin{array}{lll} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array}\right| \quad, \quad \Delta_{1}=\left|\begin{array}{ccc} 1 & 1 & 1 \\ y z & z x & x y \\ x & y & z \end{array}\right|$

Solution:Let us solve for $\Delta=\left|\begin{array}{lll} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array}\right|$

$\Delta=\left|\begin{array}{lll} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array}\right|$

\begin{aligned} &\text { On applying } R_{1} \rightarrow R_{1}-R_{2} \text { and } R_{2} \rightarrow R_{2}-R_{3}\\ &\Delta=\left|\begin{array}{ccc} 0 & x-y & x^{2}-y^{2} \\ 0 & y-z & y^{2}-z^{2} \\ 1 & z & z^{2} \end{array}\right|\\ &\Delta=\left|\begin{array}{ccc} 0 & x-y & (x+y)(x-y) \\ 0 & y-z & (y+z)(y-z) \\ 1 & z & z^{2} \end{array}\right| \quad \mathrm{a}^{2}-b^{2}=(a+b)(a-b) \end{aligned}

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

$\Delta=(x-y)(y-z)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1 & 1 & z \\ (x+y) & (y+z) & z^{2} \end{array}\right|$

If we convert all columns into rows

then the value of determinant does not change

$\Delta=(x-y)(y-z)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1 & 1 & z \\ (x+y) & (y+z) & z^{2} \end{array}\right|$

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

Now Lets us Solve for $\Delta_{1}=\left|\begin{array}{ccc} 1 & 1 & 1 \\ y z & z x & x y \\ x & y & z \end{array}\right|$

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

\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}-C_{2} \text { and } \mathrm{C}_{2} \rightarrow C_{2}-C_{3}\\ &\Delta_{1}=\left|\begin{array}{ccc} 0 & 0 & 1 \\ y z-z x & z x-x y & x y \\ x-y & y-z & z \end{array}\right|\\ &\Delta_{1}=\left|\begin{array}{ccc} 0 & 0 & 1 \\ -z(x-y) & -x(y-z) & x y \\ x-y & y-z & z \end{array}\right| \end{aligned}

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

On expanding w.r.t $R_{1}$

\begin{aligned} &\Delta_{1}=(x-y)(y-z)\left[0\left|\begin{array}{cc} -x & x y \\ 1 & z \end{array}\right|-0\left|\begin{array}{cc} -z & x y \\ 1 & z \end{array}\right|+1 \mid \begin{array}{cc} -z & -x \\ 1 & 1 \end{array}\right] \\ &\Delta_{1}=(x-y)(y-z)[0-0+1\{(-z)(1)-(-x)(1)\}] \\ &\Delta_{1}=(x-y)(y-z)(-z+x) \\ &\Delta_{1}=-(x-y)(y-z)(z-x) \end{aligned}

Now L.H.S

$\Delta+\Delta_{1}=(x-y)(y-z)(z-x)+\{-(x-y)(y-z)(z-x)\}$

From   equation (1) and (2)

\begin{aligned} &\Delta+\Delta_{1}=0 \\ &=R \cdot H \cdot S \end{aligned}

Hence it is proved that $\Delta+\Delta_{1}=0$