#### Need Solution for R. D. Sharma Maths Class 12 Chapter determinants  Exercise 5.2 Question 30 Maths Textbook Solution.

Answer: $\left|\begin{array}{ccc} 1 & a & a^{2} \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array}\right|=\left(a^{3}-1\right)^{2}$

Hint We will try to convert some elements of determinant into zero

Given:$\left|\begin{array}{ccc} 1 & a & a^{2} \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array}\right|=\left(a^{3}-1\right)^{2}$

Solution:

$\text { L.H.S }\left|\begin{array}{ccc} 1 & a & a^{2} \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array}\right|$

\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}+R_{2}+R_{3} \\ &=\left|\begin{array}{ccc} 1+a^{2}+a & a+1+a^{2} & a^{2}+a+1 \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array}\right| \end{aligned}

\begin{aligned} &\text { On taking }\left(a^{2}+a+1\right) \text { common from } \mathrm{R}_{1}\\ &=\left(a^{2}+a+1\right)\left|\begin{array}{ccc} 1 & 1 & 1 \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array}\right|\\ &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}-C_{2}, C_{2} \rightarrow C_{2}-C_{3}\\ &=\left(a^{2}+a+1\right)\left|\begin{array}{ccc} 0 & 0 & 1 \\ a^{2}-1 & 1-a & a \\ a-a^{2} & a^{2}-1 & 1 \end{array}\right|\\ &=\left(a^{2}+a+1\right)\left|\begin{array}{ccc} 0 & 0 & 1 \\ (a+1)(a-1) & -(a-1) & a \\ -a(a-1) & (a+1)(a-1) & 1 \end{array}\right| \quad \because a^{2}-b^{2}=(a+b)(a-b) \end{aligned}

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

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

Hence it is proved that

$\left|\begin{array}{ccc} 1 & a & a^{2} \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array}\right|=\left(a^{3}-1\right)^{2}$